Replace NAs in split lists

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Replace NAs in split lists

Ek Esawi
Hi all--

I stumbled on this problem online. I did not like the solution given
there which was a long UDF. I thought why cannot split and l/s apply
work here. My aim is to split the data frame, use l/sapply, make
changes on the split lists and combine the split lists to new data
frame with the desired changes/output.

The data frame shown below has a column named ID which has 2 variables
a and b; i want to replace the NAs on the Value column by 2, which is
the only numeric entry, for ID=a and by 5 for ID=b.

I worked out the solution but could not replace the results in the split lists.

Original dataframe , df1
  ID ID_2 Firist Value
1  a   aa   TRUE     2
2  a   ab  FALSE    NA
3  a   ac  FALSE    NA
4  b   aa   TRUE     5
5  b   ab  FALSE    NA
Sdf1
$a
ID ID_2 Firist Value
1  a   aa   TRUE     2
2  a   ab  FALSE    NA
3  a   ac  FALSE    NA
$b
  ID ID_2 Firist Value
4  b   aa   TRUE     5
5  b   ab  FALSE    NA
Desired results
ID ID_2 Firist Value
1  a   aa   TRUE    2
2  a   ab  FALSE    2
3  a   ac  FALSE    2

$b
  ID ID_2 Firist Value
4  b   aa   TRUE     5
5  b   ab  FALSE     5

My code

sdf <- split(df1,df$ID)
lapply(sdf, function(z) ifelse(is.na(z$Value),z$Value[!is.na(z$Value)],z$Value))
result:
$ a: num [1:3] 2 2 2
$ b: num [1:2] 5 5

How could I put these two lists back in the split data frame, sdf1?
Then I could use do.call to reassemble a data frame from the split
lists,

Thanks,
EK

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Re: Replace NAs in split lists

Ek Esawi
I just came up with a solution right after i posted the question, but
i figured there must be a better and shorter one.than my solution
sdf1[[1]][1,4]<-lapplyresults[[1]]
sdf1[[2]][1,4]<-lapplyresults[[2]]

EK

On Sun, Jan 7, 2018 at 10:13 PM, Ek Esawi <[hidden email]> wrote:

> Hi all--
>
> I stumbled on this problem online. I did not like the solution given
> there which was a long UDF. I thought why cannot split and l/s apply
> work here. My aim is to split the data frame, use l/sapply, make
> changes on the split lists and combine the split lists to new data
> frame with the desired changes/output.
>
> The data frame shown below has a column named ID which has 2 variables
> a and b; i want to replace the NAs on the Value column by 2, which is
> the only numeric entry, for ID=a and by 5 for ID=b.
>
> I worked out the solution but could not replace the results in the split lists.
>
> Original dataframe , df1
>   ID ID_2 Firist Value
> 1  a   aa   TRUE     2
> 2  a   ab  FALSE    NA
> 3  a   ac  FALSE    NA
> 4  b   aa   TRUE     5
> 5  b   ab  FALSE    NA
> Sdf1
> $a
> ID ID_2 Firist Value
> 1  a   aa   TRUE     2
> 2  a   ab  FALSE    NA
> 3  a   ac  FALSE    NA
> $b
>   ID ID_2 Firist Value
> 4  b   aa   TRUE     5
> 5  b   ab  FALSE    NA
> Desired results
> ID ID_2 Firist Value
> 1  a   aa   TRUE    2
> 2  a   ab  FALSE    2
> 3  a   ac  FALSE    2
>
> $b
>   ID ID_2 Firist Value
> 4  b   aa   TRUE     5
> 5  b   ab  FALSE     5
>
> My code
>
> sdf <- split(df1,df$ID)
> lapply(sdf, function(z) ifelse(is.na(z$Value),z$Value[!is.na(z$Value)],z$Value))
> result:
> $ a: num [1:3] 2 2 2
> $ b: num [1:2] 5 5
>
> How could I put these two lists back in the split data frame, sdf1?
> Then I could use do.call to reassemble a data frame from the split
> lists,
>
> Thanks,
> EK

______________________________________________
[hidden email] mailing list -- To UNSUBSCRIBE and more, see
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Re: Replace NAs in split lists

PIKAL Petr
Hi

library(zoo)

has function na.locf, which probably can do what you want.

so something like (untested)

sdf1.fill<-lapply(sdf1, na.locf)
do.call(rbind, sdf1.fill)

Cheers.
Petr

> -----Original Message-----
> From: R-help [mailto:[hidden email]] On Behalf Of Ek Esawi
> Sent: Monday, January 8, 2018 4:36 AM
> To: [hidden email]
> Subject: Re: [R] Replace NAs in split lists
>
> I just came up with a solution right after i posted the question, but i figured
> there must be a better and shorter one.than my solution sdf1[[1]][1,4]<-
> lapplyresults[[1]]
> sdf1[[2]][1,4]<-lapplyresults[[2]]
>
> EK
>
> On Sun, Jan 7, 2018 at 10:13 PM, Ek Esawi <[hidden email]> wrote:
> > Hi all--
> >
> > I stumbled on this problem online. I did not like the solution given
> > there which was a long UDF. I thought why cannot split and l/s apply
> > work here. My aim is to split the data frame, use l/sapply, make
> > changes on the split lists and combine the split lists to new data
> > frame with the desired changes/output.
> >
> > The data frame shown below has a column named ID which has 2 variables
> > a and b; i want to replace the NAs on the Value column by 2, which is
> > the only numeric entry, for ID=a and by 5 for ID=b.
> >
> > I worked out the solution but could not replace the results in the split lists.
> >
> > Original dataframe , df1
> >   ID ID_2 Firist Value
> > 1  a   aa   TRUE     2
> > 2  a   ab  FALSE    NA
> > 3  a   ac  FALSE    NA
> > 4  b   aa   TRUE     5
> > 5  b   ab  FALSE    NA
> > Sdf1
> > $a
> > ID ID_2 Firist Value
> > 1  a   aa   TRUE     2
> > 2  a   ab  FALSE    NA
> > 3  a   ac  FALSE    NA
> > $b
> >   ID ID_2 Firist Value
> > 4  b   aa   TRUE     5
> > 5  b   ab  FALSE    NA
> > Desired results
> > ID ID_2 Firist Value
> > 1  a   aa   TRUE    2
> > 2  a   ab  FALSE    2
> > 3  a   ac  FALSE    2
> >
> > $b
> >   ID ID_2 Firist Value
> > 4  b   aa   TRUE     5
> > 5  b   ab  FALSE     5
> >
> > My code
> >
> > sdf <- split(df1,df$ID)
> > lapply(sdf, function(z)
> > ifelse(is.na(z$Value),z$Value[!is.na(z$Value)],z$Value))
> > result:
> > $ a: num [1:3] 2 2 2
> > $ b: num [1:2] 5 5
> >
> > How could I put these two lists back in the split data frame, sdf1?
> > Then I could use do.call to reassemble a data frame from the split
> > lists,
> >
> > Thanks,
> > EK
>
> ______________________________________________
> [hidden email] mailing list -- To UNSUBSCRIBE and more, see
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.

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Re: Replace NAs in split lists

Jeff Newmiller
In reply to this post by Ek Esawi
Why do you want to modify df1?

Why not just reassemble the parts as a new data frame and use that going forward in your calculations? That is generally the preferred approach in R so you can re-do your calculations easily if you find a mistake later.
--
Sent from my phone. Please excuse my brevity.

On January 7, 2018 7:35:59 PM PST, Ek Esawi <[hidden email]> wrote:

>I just came up with a solution right after i posted the question, but
>i figured there must be a better and shorter one.than my solution
>sdf1[[1]][1,4]<-lapplyresults[[1]]
>sdf1[[2]][1,4]<-lapplyresults[[2]]
>
>EK
>
>On Sun, Jan 7, 2018 at 10:13 PM, Ek Esawi <[hidden email]> wrote:
>> Hi all--
>>
>> I stumbled on this problem online. I did not like the solution given
>> there which was a long UDF. I thought why cannot split and l/s apply
>> work here. My aim is to split the data frame, use l/sapply, make
>> changes on the split lists and combine the split lists to new data
>> frame with the desired changes/output.
>>
>> The data frame shown below has a column named ID which has 2
>variables
>> a and b; i want to replace the NAs on the Value column by 2, which is
>> the only numeric entry, for ID=a and by 5 for ID=b.
>>
>> I worked out the solution but could not replace the results in the
>split lists.
>>
>> Original dataframe , df1
>>   ID ID_2 Firist Value
>> 1  a   aa   TRUE     2
>> 2  a   ab  FALSE    NA
>> 3  a   ac  FALSE    NA
>> 4  b   aa   TRUE     5
>> 5  b   ab  FALSE    NA
>> Sdf1
>> $a
>> ID ID_2 Firist Value
>> 1  a   aa   TRUE     2
>> 2  a   ab  FALSE    NA
>> 3  a   ac  FALSE    NA
>> $b
>>   ID ID_2 Firist Value
>> 4  b   aa   TRUE     5
>> 5  b   ab  FALSE    NA
>> Desired results
>> ID ID_2 Firist Value
>> 1  a   aa   TRUE    2
>> 2  a   ab  FALSE    2
>> 3  a   ac  FALSE    2
>>
>> $b
>>   ID ID_2 Firist Value
>> 4  b   aa   TRUE     5
>> 5  b   ab  FALSE     5
>>
>> My code
>>
>> sdf <- split(df1,df$ID)
>> lapply(sdf, function(z)
>ifelse(is.na(z$Value),z$Value[!is.na(z$Value)],z$Value))
>> result:
>> $ a: num [1:3] 2 2 2
>> $ b: num [1:2] 5 5
>>
>> How could I put these two lists back in the split data frame, sdf1?
>> Then I could use do.call to reassemble a data frame from the split
>> lists,
>>
>> Thanks,
>> EK
>
>______________________________________________
>[hidden email] mailing list -- To UNSUBSCRIBE and more, see
>https://stat.ethz.ch/mailman/listinfo/r-help
>PLEASE do read the posting guide
>http://www.R-project.org/posting-guide.html
>and provide commented, minimal, self-contained, reproducible code.

______________________________________________
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Re: Replace NAs in split lists

Jeff Newmiller
Upon closer examination I see that you are not using the split version of
df1 as I usually would, so here is a reproducible example:

#----
df1 <- read.table( text=
"ID ID_2 Firist Value
1  a   aa   TRUE     2
2  a   ab  FALSE    NA
3  a   ac  FALSE    NA
4  b   aa   TRUE     5
5  b   ab  FALSE    NA
", header=TRUE, as.is=TRUE )

sdf <- split( df1, df1$ID )
# note the extra [ 1 ] in case you have more than one non-NA value
# per ID
sdf2 <- lapply( sdf
               , function( z ) {
                  z$Value <- ifelse( is.na( z$Value )
                                   , z$Value[ !is.na( z$Value ) ][ 1 ]
                                   , z$Value
                                   )
                  z
                 }
               )
df2 <- do.call( rbind, sdf2 )
df2
#>     ID ID_2 Firist Value
#> a.1  a   aa   TRUE     2
#> a.2  a   ab  FALSE     2
#> a.3  a   ac  FALSE     2
#> b.4  b   aa   TRUE     5
#> b.5  b   ab  FALSE     5

# or using tidyverse methods

library(dplyr)
#>
#> Attaching package: 'dplyr'
#> The following objects are masked from 'package:stats':
#>
#>     filter, lag
#> The following objects are masked from 'package:base':
#>
#>     intersect, setdiff, setequal, union
df3 <- (   df1
        %>% group_by( ID )
        %>% do({
               mutate( .
                     , Value = ifelse( is.na( Value )
                                     , Value[ !is.na( Value ) ][ 1 ]
                                     , Value
                                     )
                     )
            })
        %>% ungroup
        )
df3
#> # A tibble: 5 x 4
#>   ID    ID_2  Firist Value
#>   <chr> <chr> <lgl>  <int>
#> 1 a     aa    T          2
#> 2 a     ab    F          2
#> 3 a     ac    F          2
#> 4 b     aa    T          5
#> 5 b     ab    F          5
#----

On Sun, 7 Jan 2018, Jeff Newmiller wrote:

> Why do you want to modify df1?
>
> Why not just reassemble the parts as a new data frame and use that going
> forward in your calculations? That is generally the preferred approach
> in R so you can re-do your calculations easily if you find a mistake
> later.
> --
> Sent from my phone. Please excuse my brevity.
>
> On January 7, 2018 7:35:59 PM PST, Ek Esawi <[hidden email]> wrote:
>> I just came up with a solution right after i posted the question, but
>> i figured there must be a better and shorter one.than my solution
>> sdf1[[1]][1,4]<-lapplyresults[[1]]
>> sdf1[[2]][1,4]<-lapplyresults[[2]]
>>
>> EK
>>
>> On Sun, Jan 7, 2018 at 10:13 PM, Ek Esawi <[hidden email]> wrote:
>>> Hi all--
>>>
>>> I stumbled on this problem online. I did not like the solution given
>>> there which was a long UDF. I thought why cannot split and l/s apply
>>> work here. My aim is to split the data frame, use l/sapply, make
>>> changes on the split lists and combine the split lists to new data
>>> frame with the desired changes/output.
>>>
>>> The data frame shown below has a column named ID which has 2
>> variables
>>> a and b; i want to replace the NAs on the Value column by 2, which is
>>> the only numeric entry, for ID=a and by 5 for ID=b.
>>>
>>> I worked out the solution but could not replace the results in the
>> split lists.
>>>
>>> Original dataframe , df1
>>>   ID ID_2 Firist Value
>>> 1  a   aa   TRUE     2
>>> 2  a   ab  FALSE    NA
>>> 3  a   ac  FALSE    NA
>>> 4  b   aa   TRUE     5
>>> 5  b   ab  FALSE    NA
>>> Sdf1
>>> $a
>>> ID ID_2 Firist Value
>>> 1  a   aa   TRUE     2
>>> 2  a   ab  FALSE    NA
>>> 3  a   ac  FALSE    NA
>>> $b
>>>   ID ID_2 Firist Value
>>> 4  b   aa   TRUE     5
>>> 5  b   ab  FALSE    NA
>>> Desired results
>>> ID ID_2 Firist Value
>>> 1  a   aa   TRUE    2
>>> 2  a   ab  FALSE    2
>>> 3  a   ac  FALSE    2
>>>
>>> $b
>>>   ID ID_2 Firist Value
>>> 4  b   aa   TRUE     5
>>> 5  b   ab  FALSE     5
>>>
>>> My code
>>>
>>> sdf <- split(df1,df$ID)
>>> lapply(sdf, function(z)
>> ifelse(is.na(z$Value),z$Value[!is.na(z$Value)],z$Value))
>>> result:
>>> $ a: num [1:3] 2 2 2
>>> $ b: num [1:2] 5 5
>>>
>>> How could I put these two lists back in the split data frame, sdf1?
>>> Then I could use do.call to reassemble a data frame from the split
>>> lists,
>>>
>>> Thanks,
>>> EK
>>
>> ______________________________________________
>> [hidden email] mailing list -- To UNSUBSCRIBE and more, see
>> https://stat.ethz.ch/mailman/listinfo/r-help
>> PLEASE do read the posting guide
>> http://www.R-project.org/posting-guide.html
>> and provide commented, minimal, self-contained, reproducible code.
>
> ______________________________________________
> [hidden email] mailing list -- To UNSUBSCRIBE and more, see
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>

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______________________________________________
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Re: Replace NAs in split lists

PIKAL Petr
Hi

With the example, na.locf seems to be the easiest way.
> library(zoo)

> na.locf(df1)
  ID ID_2 Firist Value
1  a   aa   TRUE     2
2  a   ab  FALSE     2
3  a   ac  FALSE     2
4  b   aa   TRUE     5
5  b   ab  FALSE     5

Cheers
Petr

> -----Original Message-----
> From: R-help [mailto:[hidden email]] On Behalf Of Jeff
> Newmiller
> Sent: Monday, January 8, 2018 9:13 AM
> To: [hidden email]; Ek Esawi <[hidden email]>
> Subject: Re: [R] Replace NAs in split lists
>
> Upon closer examination I see that you are not using the split version of
> df1 as I usually would, so here is a reproducible example:
>
> #----
> df1 <- read.table( text=
> "ID ID_2 Firist Value
> 1  a   aa   TRUE     2
> 2  a   ab  FALSE    NA
> 3  a   ac  FALSE    NA
> 4  b   aa   TRUE     5
> 5  b   ab  FALSE    NA
> ", header=TRUE, as.is=TRUE )
>
> sdf <- split( df1, df1$ID )
> # note the extra [ 1 ] in case you have more than one non-NA value # per ID
> sdf2 <- lapply( sdf
>                , function( z ) {
>                   z$Value <- ifelse( is.na( z$Value )
>                                    , z$Value[ !is.na( z$Value ) ][ 1 ]
>                                    , z$Value
>                                    )
>                   z
>                  }
>                )
> df2 <- do.call( rbind, sdf2 )
> df2
> #>     ID ID_2 Firist Value
> #> a.1  a   aa   TRUE     2
> #> a.2  a   ab  FALSE     2
> #> a.3  a   ac  FALSE     2
> #> b.4  b   aa   TRUE     5
> #> b.5  b   ab  FALSE     5
>
> # or using tidyverse methods
>
> library(dplyr)
> #>
> #> Attaching package: 'dplyr'
> #> The following objects are masked from 'package:stats':
> #>
> #>     filter, lag
> #> The following objects are masked from 'package:base':
> #>
> #>     intersect, setdiff, setequal, union
> df3 <- (   df1
>         %>% group_by( ID )
>         %>% do({
>                mutate( .
>                      , Value = ifelse( is.na( Value )
>                                      , Value[ !is.na( Value ) ][ 1 ]
>                                      , Value
>                                      )
>                      )
>             })
>         %>% ungroup
>         )
> df3
> #> # A tibble: 5 x 4
> #>   ID    ID_2  Firist Value
> #>   <chr> <chr> <lgl>  <int>
> #> 1 a     aa    T          2
> #> 2 a     ab    F          2
> #> 3 a     ac    F          2
> #> 4 b     aa    T          5
> #> 5 b     ab    F          5
> #----
>
> On Sun, 7 Jan 2018, Jeff Newmiller wrote:
>
> > Why do you want to modify df1?
> >
> > Why not just reassemble the parts as a new data frame and use that
> > going forward in your calculations? That is generally the preferred
> > approach in R so you can re-do your calculations easily if you find a
> > mistake later.
> > --
> > Sent from my phone. Please excuse my brevity.
> >
> > On January 7, 2018 7:35:59 PM PST, Ek Esawi <[hidden email]> wrote:
> >> I just came up with a solution right after i posted the question, but
> >> i figured there must be a better and shorter one.than my solution
> >> sdf1[[1]][1,4]<-lapplyresults[[1]]
> >> sdf1[[2]][1,4]<-lapplyresults[[2]]
> >>
> >> EK
> >>
> >> On Sun, Jan 7, 2018 at 10:13 PM, Ek Esawi <[hidden email]> wrote:
> >>> Hi all--
> >>>
> >>> I stumbled on this problem online. I did not like the solution given
> >>> there which was a long UDF. I thought why cannot split and l/s apply
> >>> work here. My aim is to split the data frame, use l/sapply, make
> >>> changes on the split lists and combine the split lists to new data
> >>> frame with the desired changes/output.
> >>>
> >>> The data frame shown below has a column named ID which has 2
> >> variables
> >>> a and b; i want to replace the NAs on the Value column by 2, which
> >>> is the only numeric entry, for ID=a and by 5 for ID=b.
> >>>
> >>> I worked out the solution but could not replace the results in the
> >> split lists.
> >>>
> >>> Original dataframe , df1
> >>>   ID ID_2 Firist Value
> >>> 1  a   aa   TRUE     2
> >>> 2  a   ab  FALSE    NA
> >>> 3  a   ac  FALSE    NA
> >>> 4  b   aa   TRUE     5
> >>> 5  b   ab  FALSE    NA
> >>> Sdf1
> >>> $a
> >>> ID ID_2 Firist Value
> >>> 1  a   aa   TRUE     2
> >>> 2  a   ab  FALSE    NA
> >>> 3  a   ac  FALSE    NA
> >>> $b
> >>>   ID ID_2 Firist Value
> >>> 4  b   aa   TRUE     5
> >>> 5  b   ab  FALSE    NA
> >>> Desired results
> >>> ID ID_2 Firist Value
> >>> 1  a   aa   TRUE    2
> >>> 2  a   ab  FALSE    2
> >>> 3  a   ac  FALSE    2
> >>>
> >>> $b
> >>>   ID ID_2 Firist Value
> >>> 4  b   aa   TRUE     5
> >>> 5  b   ab  FALSE     5
> >>>
> >>> My code
> >>>
> >>> sdf <- split(df1,df$ID)
> >>> lapply(sdf, function(z)
> >> ifelse(is.na(z$Value),z$Value[!is.na(z$Value)],z$Value))
> >>> result:
> >>> $ a: num [1:3] 2 2 2
> >>> $ b: num [1:2] 5 5
> >>>
> >>> How could I put these two lists back in the split data frame, sdf1?
> >>> Then I could use do.call to reassemble a data frame from the split
> >>> lists,
> >>>
> >>> Thanks,
> >>> EK
> >>
> >> ______________________________________________
> >> [hidden email] mailing list -- To UNSUBSCRIBE and more, see
> >> https://stat.ethz.ch/mailman/listinfo/r-help
> >> PLEASE do read the posting guide
> >> http://www.R-project.org/posting-guide.html
> >> and provide commented, minimal, self-contained, reproducible code.
> >
> > ______________________________________________
> > [hidden email] mailing list -- To UNSUBSCRIBE and more, see
> > https://stat.ethz.ch/mailman/listinfo/r-help
> > PLEASE do read the posting guide
> > http://www.R-project.org/posting-guide.html
> > and provide commented, minimal, self-contained, reproducible code.
> >
>
> ---------------------------------------------------------------------------
> Jeff Newmiller                        The     .....       .....  Go Live...
> DCN:<[hidden email]>        Basics: ##.#.       ##.#.  Live Go...
>                                        Live:   OO#.. Dead: OO#..  Playing
> Research Engineer (Solar/Batteries            O.O#.       #.O#.  with
> /Software/Embedded Controllers)               .OO#.       .OO#.  rocks...1k
>
> ______________________________________________
> [hidden email] mailing list -- To UNSUBSCRIBE and more, see
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.

________________________________
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- a obsahuje-li nabídku, je adresát oprávněn nabídku bezodkladně přijmout; Odesílatel tohoto e-mailu (nabídky) vylučuje přijetí nabídky ze strany příjemce s dodatkem či odchylkou.
- trvá odesílatel na tom, že příslušná smlouva je uzavřena teprve výslovným dosažením shody na všech jejích náležitostech.
- odesílatel tohoto emailu informuje, že není oprávněn uzavírat za společnost žádné smlouvy s výjimkou případů, kdy k tomu byl písemně zmocněn nebo písemně pověřen a takové pověření nebo plná moc byly adresátovi tohoto emailu případně osobě, kterou adresát zastupuje, předloženy nebo jejich existence je adresátovi či osobě jím zastoupené známá.

This e-mail and any documents attached to it may be confidential and are intended only for its intended recipients.
If you received this e-mail by mistake, please immediately inform its sender. Delete the contents of this e-mail with all attachments and its copies from your system.
If you are not the intended recipient of this e-mail, you are not authorized to use, disseminate, copy or disclose this e-mail in any manner.
The sender of this e-mail shall not be liable for any possible damage caused by modifications of the e-mail or by delay with transfer of the email.

In case that this e-mail forms part of business dealings:
- the sender reserves the right to end negotiations about entering into a contract in any time, for any reason, and without stating any reasoning.
- if the e-mail contains an offer, the recipient is entitled to immediately accept such offer; The sender of this e-mail (offer) excludes any acceptance of the offer on the part of the recipient containing any amendment or variation.
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- the sender of this e-mail informs that he/she is not authorized to enter into any contracts on behalf of the company except for cases in which he/she is expressly authorized to do so in writing, and such authorization or power of attorney is submitted to the recipient or the person represented by the recipient, or the existence of such authorization is known to the recipient of the person represented by the recipient.
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Re: Replace NAs in split lists

Jeff Newmiller
Yes, you are right if the IDs are always sequentially-adjacent and the first non-NA value appears in the first record for each ID.
--
Sent from my phone. Please excuse my brevity.

On January 8, 2018 2:29:40 AM PST, PIKAL Petr <[hidden email]> wrote:

>Hi
>
>With the example, na.locf seems to be the easiest way.
>> library(zoo)
>
>> na.locf(df1)
>  ID ID_2 Firist Value
>1  a   aa   TRUE     2
>2  a   ab  FALSE     2
>3  a   ac  FALSE     2
>4  b   aa   TRUE     5
>5  b   ab  FALSE     5
>
>Cheers
>Petr
>
>> -----Original Message-----
>> From: R-help [mailto:[hidden email]] On Behalf Of Jeff
>> Newmiller
>> Sent: Monday, January 8, 2018 9:13 AM
>> To: [hidden email]; Ek Esawi <[hidden email]>
>> Subject: Re: [R] Replace NAs in split lists
>>
>> Upon closer examination I see that you are not using the split
>version of
>> df1 as I usually would, so here is a reproducible example:
>>
>> #----
>> df1 <- read.table( text=
>> "ID ID_2 Firist Value
>> 1  a   aa   TRUE     2
>> 2  a   ab  FALSE    NA
>> 3  a   ac  FALSE    NA
>> 4  b   aa   TRUE     5
>> 5  b   ab  FALSE    NA
>> ", header=TRUE, as.is=TRUE )
>>
>> sdf <- split( df1, df1$ID )
>> # note the extra [ 1 ] in case you have more than one non-NA value #
>per ID
>> sdf2 <- lapply( sdf
>>                , function( z ) {
>>                   z$Value <- ifelse( is.na( z$Value )
>>                                    , z$Value[ !is.na( z$Value ) ][ 1
>]
>>                                    , z$Value
>>                                    )
>>                   z
>>                  }
>>                )
>> df2 <- do.call( rbind, sdf2 )
>> df2
>> #>     ID ID_2 Firist Value
>> #> a.1  a   aa   TRUE     2
>> #> a.2  a   ab  FALSE     2
>> #> a.3  a   ac  FALSE     2
>> #> b.4  b   aa   TRUE     5
>> #> b.5  b   ab  FALSE     5
>>
>> # or using tidyverse methods
>>
>> library(dplyr)
>> #>
>> #> Attaching package: 'dplyr'
>> #> The following objects are masked from 'package:stats':
>> #>
>> #>     filter, lag
>> #> The following objects are masked from 'package:base':
>> #>
>> #>     intersect, setdiff, setequal, union
>> df3 <- (   df1
>>         %>% group_by( ID )
>>         %>% do({
>>                mutate( .
>>                      , Value = ifelse( is.na( Value )
>>                                      , Value[ !is.na( Value ) ][ 1 ]
>>                                      , Value
>>                                      )
>>                      )
>>             })
>>         %>% ungroup
>>         )
>> df3
>> #> # A tibble: 5 x 4
>> #>   ID    ID_2  Firist Value
>> #>   <chr> <chr> <lgl>  <int>
>> #> 1 a     aa    T          2
>> #> 2 a     ab    F          2
>> #> 3 a     ac    F          2
>> #> 4 b     aa    T          5
>> #> 5 b     ab    F          5
>> #----
>>
>> On Sun, 7 Jan 2018, Jeff Newmiller wrote:
>>
>> > Why do you want to modify df1?
>> >
>> > Why not just reassemble the parts as a new data frame and use that
>> > going forward in your calculations? That is generally the preferred
>> > approach in R so you can re-do your calculations easily if you find
>a
>> > mistake later.
>> > --
>> > Sent from my phone. Please excuse my brevity.
>> >
>> > On January 7, 2018 7:35:59 PM PST, Ek Esawi <[hidden email]>
>wrote:
>> >> I just came up with a solution right after i posted the question,
>but
>> >> i figured there must be a better and shorter one.than my solution
>> >> sdf1[[1]][1,4]<-lapplyresults[[1]]
>> >> sdf1[[2]][1,4]<-lapplyresults[[2]]
>> >>
>> >> EK
>> >>
>> >> On Sun, Jan 7, 2018 at 10:13 PM, Ek Esawi <[hidden email]>
>wrote:
>> >>> Hi all--
>> >>>
>> >>> I stumbled on this problem online. I did not like the solution
>given
>> >>> there which was a long UDF. I thought why cannot split and l/s
>apply
>> >>> work here. My aim is to split the data frame, use l/sapply, make
>> >>> changes on the split lists and combine the split lists to new
>data
>> >>> frame with the desired changes/output.
>> >>>
>> >>> The data frame shown below has a column named ID which has 2
>> >> variables
>> >>> a and b; i want to replace the NAs on the Value column by 2,
>which
>> >>> is the only numeric entry, for ID=a and by 5 for ID=b.
>> >>>
>> >>> I worked out the solution but could not replace the results in
>the
>> >> split lists.
>> >>>
>> >>> Original dataframe , df1
>> >>>   ID ID_2 Firist Value
>> >>> 1  a   aa   TRUE     2
>> >>> 2  a   ab  FALSE    NA
>> >>> 3  a   ac  FALSE    NA
>> >>> 4  b   aa   TRUE     5
>> >>> 5  b   ab  FALSE    NA
>> >>> Sdf1
>> >>> $a
>> >>> ID ID_2 Firist Value
>> >>> 1  a   aa   TRUE     2
>> >>> 2  a   ab  FALSE    NA
>> >>> 3  a   ac  FALSE    NA
>> >>> $b
>> >>>   ID ID_2 Firist Value
>> >>> 4  b   aa   TRUE     5
>> >>> 5  b   ab  FALSE    NA
>> >>> Desired results
>> >>> ID ID_2 Firist Value
>> >>> 1  a   aa   TRUE    2
>> >>> 2  a   ab  FALSE    2
>> >>> 3  a   ac  FALSE    2
>> >>>
>> >>> $b
>> >>>   ID ID_2 Firist Value
>> >>> 4  b   aa   TRUE     5
>> >>> 5  b   ab  FALSE     5
>> >>>
>> >>> My code
>> >>>
>> >>> sdf <- split(df1,df$ID)
>> >>> lapply(sdf, function(z)
>> >> ifelse(is.na(z$Value),z$Value[!is.na(z$Value)],z$Value))
>> >>> result:
>> >>> $ a: num [1:3] 2 2 2
>> >>> $ b: num [1:2] 5 5
>> >>>
>> >>> How could I put these two lists back in the split data frame,
>sdf1?
>> >>> Then I could use do.call to reassemble a data frame from the
>split
>> >>> lists,
>> >>>
>> >>> Thanks,
>> >>> EK
>> >>
>> >> ______________________________________________
>> >> [hidden email] mailing list -- To UNSUBSCRIBE and more, see
>> >> https://stat.ethz.ch/mailman/listinfo/r-help
>> >> PLEASE do read the posting guide
>> >> http://www.R-project.org/posting-guide.html
>> >> and provide commented, minimal, self-contained, reproducible code.
>> >
>> > ______________________________________________
>> > [hidden email] mailing list -- To UNSUBSCRIBE and more, see
>> > https://stat.ethz.ch/mailman/listinfo/r-help
>> > PLEASE do read the posting guide
>> > http://www.R-project.org/posting-guide.html
>> > and provide commented, minimal, self-contained, reproducible code.
>> >
>>
>>
>---------------------------------------------------------------------------
>> Jeff Newmiller                        The     .....       .....  Go
>Live...
>> DCN:<[hidden email]>        Basics: ##.#.       ##.#.  Live
>Go...
>>                                        Live:   OO#.. Dead: OO#..
>Playing
>> Research Engineer (Solar/Batteries            O.O#.       #.O#.  with
>> /Software/Embedded Controllers)               .OO#.       .OO#.
>rocks...1k
>>
>> ______________________________________________
>> [hidden email] mailing list -- To UNSUBSCRIBE and more, see
>> https://stat.ethz.ch/mailman/listinfo/r-help
>> PLEASE do read the posting guide
>http://www.R-project.org/posting-guide.html
>> and provide commented, minimal, self-contained, reproducible code.
>
>________________________________
>Tento e-mail a jakékoliv k němu připojené dokumenty jsou důvěrné a jsou
>určeny pouze jeho adresátům.
>Jestliže jste obdržel(a) tento e-mail omylem, informujte laskavě
>neprodleně jeho odesílatele. Obsah tohoto emailu i s přílohami a jeho
>kopie vymažte ze svého systému.
>Nejste-li zamýšleným adresátem tohoto emailu, nejste oprávněni tento
>email jakkoliv užívat, rozšiřovat, kopírovat či zveřejňovat.
>Odesílatel e-mailu neodpovídá za eventuální škodu způsobenou
>modifikacemi či zpožděním přenosu e-mailu.
>
>V případě, že je tento e-mail součástí obchodního jednání:
>- vyhrazuje si odesílatel právo ukončit kdykoliv jednání o uzavření
>smlouvy, a to z jakéhokoliv důvodu i bez uvedení důvodu.
>- a obsahuje-li nabídku, je adresát oprávněn nabídku bezodkladně
>přijmout; Odesílatel tohoto e-mailu (nabídky) vylučuje přijetí nabídky
>ze strany příjemce s dodatkem či odchylkou.
>- trvá odesílatel na tom, že příslušná smlouva je uzavřena teprve
>výslovným dosažením shody na všech jejích náležitostech.
>- odesílatel tohoto emailu informuje, že není oprávněn uzavírat za
>společnost žádné smlouvy s výjimkou případů, kdy k tomu byl písemně
>zmocněn nebo písemně pověřen a takové pověření nebo plná moc byly
>adresátovi tohoto emailu případně osobě, kterou adresát zastupuje,
>předloženy nebo jejich existence je adresátovi či osobě jím zastoupené
>známá.
>
>This e-mail and any documents attached to it may be confidential and
>are intended only for its intended recipients.
>If you received this e-mail by mistake, please immediately inform its
>sender. Delete the contents of this e-mail with all attachments and its
>copies from your system.
>If you are not the intended recipient of this e-mail, you are not
>authorized to use, disseminate, copy or disclose this e-mail in any
>manner.
>The sender of this e-mail shall not be liable for any possible damage
>caused by modifications of the e-mail or by delay with transfer of the
>email.
>
>In case that this e-mail forms part of business dealings:
>- the sender reserves the right to end negotiations about entering into
>a contract in any time, for any reason, and without stating any
>reasoning.
>- if the e-mail contains an offer, the recipient is entitled to
>immediately accept such offer; The sender of this e-mail (offer)
>excludes any acceptance of the offer on the part of the recipient
>containing any amendment or variation.
>- the sender insists on that the respective contract is concluded only
>upon an express mutual agreement on all its aspects.
>- the sender of this e-mail informs that he/she is not authorized to
>enter into any contracts on behalf of the company except for cases in
>which he/she is expressly authorized to do so in writing, and such
>authorization or power of attorney is submitted to the recipient or the
>person represented by the recipient, or the existence of such
>authorization is known to the recipient of the person represented by
>the recipient.

______________________________________________
[hidden email] mailing list -- To UNSUBSCRIBE and more, see
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
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Re: Replace NAs in split lists

Eric Berger
You can enforce these assumptions by sorting on multiple columns, which
leads to

na.locf(df1[ order(df1$ID,df1$Value), ])



On Mon, Jan 8, 2018 at 4:19 PM, Jeff Newmiller <[hidden email]>
wrote:

> Yes, you are right if the IDs are always sequentially-adjacent and the
> first non-NA value appears in the first record for each ID.
> --
> Sent from my phone. Please excuse my brevity.
>
> On January 8, 2018 2:29:40 AM PST, PIKAL Petr <[hidden email]>
> wrote:
> >Hi
> >
> >With the example, na.locf seems to be the easiest way.
> >> library(zoo)
> >
> >> na.locf(df1)
> >  ID ID_2 Firist Value
> >1  a   aa   TRUE     2
> >2  a   ab  FALSE     2
> >3  a   ac  FALSE     2
> >4  b   aa   TRUE     5
> >5  b   ab  FALSE     5
> >
> >Cheers
> >Petr
> >
> >> -----Original Message-----
> >> From: R-help [mailto:[hidden email]] On Behalf Of Jeff
> >> Newmiller
> >> Sent: Monday, January 8, 2018 9:13 AM
> >> To: [hidden email]; Ek Esawi <[hidden email]>
> >> Subject: Re: [R] Replace NAs in split lists
> >>
> >> Upon closer examination I see that you are not using the split
> >version of
> >> df1 as I usually would, so here is a reproducible example:
> >>
> >> #----
> >> df1 <- read.table( text=
> >> "ID ID_2 Firist Value
> >> 1  a   aa   TRUE     2
> >> 2  a   ab  FALSE    NA
> >> 3  a   ac  FALSE    NA
> >> 4  b   aa   TRUE     5
> >> 5  b   ab  FALSE    NA
> >> ", header=TRUE, as.is=TRUE )
> >>
> >> sdf <- split( df1, df1$ID )
> >> # note the extra [ 1 ] in case you have more than one non-NA value #
> >per ID
> >> sdf2 <- lapply( sdf
> >>                , function( z ) {
> >>                   z$Value <- ifelse( is.na( z$Value )
> >>                                    , z$Value[ !is.na( z$Value ) ][ 1
> >]
> >>                                    , z$Value
> >>                                    )
> >>                   z
> >>                  }
> >>                )
> >> df2 <- do.call( rbind, sdf2 )
> >> df2
> >> #>     ID ID_2 Firist Value
> >> #> a.1  a   aa   TRUE     2
> >> #> a.2  a   ab  FALSE     2
> >> #> a.3  a   ac  FALSE     2
> >> #> b.4  b   aa   TRUE     5
> >> #> b.5  b   ab  FALSE     5
> >>
> >> # or using tidyverse methods
> >>
> >> library(dplyr)
> >> #>
> >> #> Attaching package: 'dplyr'
> >> #> The following objects are masked from 'package:stats':
> >> #>
> >> #>     filter, lag
> >> #> The following objects are masked from 'package:base':
> >> #>
> >> #>     intersect, setdiff, setequal, union
> >> df3 <- (   df1
> >>         %>% group_by( ID )
> >>         %>% do({
> >>                mutate( .
> >>                      , Value = ifelse( is.na( Value )
> >>                                      , Value[ !is.na( Value ) ][ 1 ]
> >>                                      , Value
> >>                                      )
> >>                      )
> >>             })
> >>         %>% ungroup
> >>         )
> >> df3
> >> #> # A tibble: 5 x 4
> >> #>   ID    ID_2  Firist Value
> >> #>   <chr> <chr> <lgl>  <int>
> >> #> 1 a     aa    T          2
> >> #> 2 a     ab    F          2
> >> #> 3 a     ac    F          2
> >> #> 4 b     aa    T          5
> >> #> 5 b     ab    F          5
> >> #----
> >>
> >> On Sun, 7 Jan 2018, Jeff Newmiller wrote:
> >>
> >> > Why do you want to modify df1?
> >> >
> >> > Why not just reassemble the parts as a new data frame and use that
> >> > going forward in your calculations? That is generally the preferred
> >> > approach in R so you can re-do your calculations easily if you find
> >a
> >> > mistake later.
> >> > --
> >> > Sent from my phone. Please excuse my brevity.
> >> >
> >> > On January 7, 2018 7:35:59 PM PST, Ek Esawi <[hidden email]>
> >wrote:
> >> >> I just came up with a solution right after i posted the question,
> >but
> >> >> i figured there must be a better and shorter one.than my solution
> >> >> sdf1[[1]][1,4]<-lapplyresults[[1]]
> >> >> sdf1[[2]][1,4]<-lapplyresults[[2]]
> >> >>
> >> >> EK
> >> >>
> >> >> On Sun, Jan 7, 2018 at 10:13 PM, Ek Esawi <[hidden email]>
> >wrote:
> >> >>> Hi all--
> >> >>>
> >> >>> I stumbled on this problem online. I did not like the solution
> >given
> >> >>> there which was a long UDF. I thought why cannot split and l/s
> >apply
> >> >>> work here. My aim is to split the data frame, use l/sapply, make
> >> >>> changes on the split lists and combine the split lists to new
> >data
> >> >>> frame with the desired changes/output.
> >> >>>
> >> >>> The data frame shown below has a column named ID which has 2
> >> >> variables
> >> >>> a and b; i want to replace the NAs on the Value column by 2,
> >which
> >> >>> is the only numeric entry, for ID=a and by 5 for ID=b.
> >> >>>
> >> >>> I worked out the solution but could not replace the results in
> >the
> >> >> split lists.
> >> >>>
> >> >>> Original dataframe , df1
> >> >>>   ID ID_2 Firist Value
> >> >>> 1  a   aa   TRUE     2
> >> >>> 2  a   ab  FALSE    NA
> >> >>> 3  a   ac  FALSE    NA
> >> >>> 4  b   aa   TRUE     5
> >> >>> 5  b   ab  FALSE    NA
> >> >>> Sdf1
> >> >>> $a
> >> >>> ID ID_2 Firist Value
> >> >>> 1  a   aa   TRUE     2
> >> >>> 2  a   ab  FALSE    NA
> >> >>> 3  a   ac  FALSE    NA
> >> >>> $b
> >> >>>   ID ID_2 Firist Value
> >> >>> 4  b   aa   TRUE     5
> >> >>> 5  b   ab  FALSE    NA
> >> >>> Desired results
> >> >>> ID ID_2 Firist Value
> >> >>> 1  a   aa   TRUE    2
> >> >>> 2  a   ab  FALSE    2
> >> >>> 3  a   ac  FALSE    2
> >> >>>
> >> >>> $b
> >> >>>   ID ID_2 Firist Value
> >> >>> 4  b   aa   TRUE     5
> >> >>> 5  b   ab  FALSE     5
> >> >>>
> >> >>> My code
> >> >>>
> >> >>> sdf <- split(df1,df$ID)
> >> >>> lapply(sdf, function(z)
> >> >> ifelse(is.na(z$Value),z$Value[!is.na(z$Value)],z$Value))
> >> >>> result:
> >> >>> $ a: num [1:3] 2 2 2
> >> >>> $ b: num [1:2] 5 5
> >> >>>
> >> >>> How could I put these two lists back in the split data frame,
> >sdf1?
> >> >>> Then I could use do.call to reassemble a data frame from the
> >split
> >> >>> lists,
> >> >>>
> >> >>> Thanks,
> >> >>> EK
> >> >>
> >> >> ______________________________________________
> >> >> [hidden email] mailing list -- To UNSUBSCRIBE and more, see
> >> >> https://stat.ethz.ch/mailman/listinfo/r-help
> >> >> PLEASE do read the posting guide
> >> >> http://www.R-project.org/posting-guide.html
> >> >> and provide commented, minimal, self-contained, reproducible code.
> >> >
> >> > ______________________________________________
> >> > [hidden email] mailing list -- To UNSUBSCRIBE and more, see
> >> > https://stat.ethz.ch/mailman/listinfo/r-help
> >> > PLEASE do read the posting guide
> >> > http://www.R-project.org/posting-guide.html
> >> > and provide commented, minimal, self-contained, reproducible code.
> >> >
> >>
> >>
> >-----------------------------------------------------------
> ----------------
> >> Jeff Newmiller                        The     .....       .....  Go
> >Live...
> >> DCN:<[hidden email]>        Basics: ##.#.       ##.#.  Live
> >Go...
> >>                                        Live:   OO#.. Dead: OO#..
> >Playing
> >> Research Engineer (Solar/Batteries            O.O#.       #.O#.  with
> >> /Software/Embedded Controllers)               .OO#.       .OO#.
> >rocks...1k
> >>
> >> ______________________________________________
> >> [hidden email] mailing list -- To UNSUBSCRIBE and more, see
> >> https://stat.ethz.ch/mailman/listinfo/r-help
> >> PLEASE do read the posting guide
> >http://www.R-project.org/posting-guide.html
> >> and provide commented, minimal, self-contained, reproducible code.
> >
> >________________________________
> >Tento e-mail a jakékoliv k němu připojené dokumenty jsou důvěrné a jsou
> >určeny pouze jeho adresátům.
> >Jestliže jste obdržel(a) tento e-mail omylem, informujte laskavě
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> >V případě, že je tento e-mail součástí obchodního jednání:
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> >smlouvy, a to z jakéhokoliv důvodu i bez uvedení důvodu.
> >- a obsahuje-li nabídku, je adresát oprávněn nabídku bezodkladně
> >přijmout; Odesílatel tohoto e-mailu (nabídky) vylučuje přijetí nabídky
> >ze strany příjemce s dodatkem či odchylkou.
> >- trvá odesílatel na tom, že příslušná smlouva je uzavřena teprve
> >výslovným dosažením shody na všech jejích náležitostech.
> >- odesílatel tohoto emailu informuje, že není oprávněn uzavírat za
> >společnost žádné smlouvy s výjimkou případů, kdy k tomu byl písemně
> >zmocněn nebo písemně pověřen a takové pověření nebo plná moc byly
> >adresátovi tohoto emailu případně osobě, kterou adresát zastupuje,
> >předloženy nebo jejich existence je adresátovi či osobě jím zastoupené
> >známá.
> >
> >This e-mail and any documents attached to it may be confidential and
> >are intended only for its intended recipients.
> >If you received this e-mail by mistake, please immediately inform its
> >sender. Delete the contents of this e-mail with all attachments and its
> >copies from your system.
> >If you are not the intended recipient of this e-mail, you are not
> >authorized to use, disseminate, copy or disclose this e-mail in any
> >manner.
> >The sender of this e-mail shall not be liable for any possible damage
> >caused by modifications of the e-mail or by delay with transfer of the
> >email.
> >
> >In case that this e-mail forms part of business dealings:
> >- the sender reserves the right to end negotiations about entering into
> >a contract in any time, for any reason, and without stating any
> >reasoning.
> >- if the e-mail contains an offer, the recipient is entitled to
> >immediately accept such offer; The sender of this e-mail (offer)
> >excludes any acceptance of the offer on the part of the recipient
> >containing any amendment or variation.
> >- the sender insists on that the respective contract is concluded only
> >upon an express mutual agreement on all its aspects.
> >- the sender of this e-mail informs that he/she is not authorized to
> >enter into any contracts on behalf of the company except for cases in
> >which he/she is expressly authorized to do so in writing, and such
> >authorization or power of attorney is submitted to the recipient or the
> >person represented by the recipient, or the existence of such
> >authorization is known to the recipient of the person represented by
> >the recipient.
>
> ______________________________________________
> [hidden email] mailing list -- To UNSUBSCRIBE and more, see
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/
> posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>

        [[alternative HTML version deleted]]

______________________________________________
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Re: Replace NAs in split lists

Jeff Newmiller
"Enforce" is overstating it... results will differ if there are no non-NA values for a given ID, and there is a potential further discrepancy if there are multiple non-NA values. But these issues were not identified by the OP, so may not be relevant in their case.
--
Sent from my phone. Please excuse my brevity.

On January 8, 2018 6:41:33 AM PST, Eric Berger <[hidden email]> wrote:

>You can enforce these assumptions by sorting on multiple columns, which
>leads to
>
>na.locf(df1[ order(df1$ID,df1$Value), ])
>
>
>
>On Mon, Jan 8, 2018 at 4:19 PM, Jeff Newmiller
><[hidden email]>
>wrote:
>
>> Yes, you are right if the IDs are always sequentially-adjacent and
>the
>> first non-NA value appears in the first record for each ID.
>> --
>> Sent from my phone. Please excuse my brevity.
>>
>> On January 8, 2018 2:29:40 AM PST, PIKAL Petr
><[hidden email]>
>> wrote:
>> >Hi
>> >
>> >With the example, na.locf seems to be the easiest way.
>> >> library(zoo)
>> >
>> >> na.locf(df1)
>> >  ID ID_2 Firist Value
>> >1  a   aa   TRUE     2
>> >2  a   ab  FALSE     2
>> >3  a   ac  FALSE     2
>> >4  b   aa   TRUE     5
>> >5  b   ab  FALSE     5
>> >
>> >Cheers
>> >Petr
>> >
>> >> -----Original Message-----
>> >> From: R-help [mailto:[hidden email]] On Behalf Of
>Jeff
>> >> Newmiller
>> >> Sent: Monday, January 8, 2018 9:13 AM
>> >> To: [hidden email]; Ek Esawi <[hidden email]>
>> >> Subject: Re: [R] Replace NAs in split lists
>> >>
>> >> Upon closer examination I see that you are not using the split
>> >version of
>> >> df1 as I usually would, so here is a reproducible example:
>> >>
>> >> #----
>> >> df1 <- read.table( text=
>> >> "ID ID_2 Firist Value
>> >> 1  a   aa   TRUE     2
>> >> 2  a   ab  FALSE    NA
>> >> 3  a   ac  FALSE    NA
>> >> 4  b   aa   TRUE     5
>> >> 5  b   ab  FALSE    NA
>> >> ", header=TRUE, as.is=TRUE )
>> >>
>> >> sdf <- split( df1, df1$ID )
>> >> # note the extra [ 1 ] in case you have more than one non-NA value
>#
>> >per ID
>> >> sdf2 <- lapply( sdf
>> >>                , function( z ) {
>> >>                   z$Value <- ifelse( is.na( z$Value )
>> >>                                    , z$Value[ !is.na( z$Value ) ][
>1
>> >]
>> >>                                    , z$Value
>> >>                                    )
>> >>                   z
>> >>                  }
>> >>                )
>> >> df2 <- do.call( rbind, sdf2 )
>> >> df2
>> >> #>     ID ID_2 Firist Value
>> >> #> a.1  a   aa   TRUE     2
>> >> #> a.2  a   ab  FALSE     2
>> >> #> a.3  a   ac  FALSE     2
>> >> #> b.4  b   aa   TRUE     5
>> >> #> b.5  b   ab  FALSE     5
>> >>
>> >> # or using tidyverse methods
>> >>
>> >> library(dplyr)
>> >> #>
>> >> #> Attaching package: 'dplyr'
>> >> #> The following objects are masked from 'package:stats':
>> >> #>
>> >> #>     filter, lag
>> >> #> The following objects are masked from 'package:base':
>> >> #>
>> >> #>     intersect, setdiff, setequal, union
>> >> df3 <- (   df1
>> >>         %>% group_by( ID )
>> >>         %>% do({
>> >>                mutate( .
>> >>                      , Value = ifelse( is.na( Value )
>> >>                                      , Value[ !is.na( Value ) ][ 1
>]
>> >>                                      , Value
>> >>                                      )
>> >>                      )
>> >>             })
>> >>         %>% ungroup
>> >>         )
>> >> df3
>> >> #> # A tibble: 5 x 4
>> >> #>   ID    ID_2  Firist Value
>> >> #>   <chr> <chr> <lgl>  <int>
>> >> #> 1 a     aa    T          2
>> >> #> 2 a     ab    F          2
>> >> #> 3 a     ac    F          2
>> >> #> 4 b     aa    T          5
>> >> #> 5 b     ab    F          5
>> >> #----
>> >>
>> >> On Sun, 7 Jan 2018, Jeff Newmiller wrote:
>> >>
>> >> > Why do you want to modify df1?
>> >> >
>> >> > Why not just reassemble the parts as a new data frame and use
>that
>> >> > going forward in your calculations? That is generally the
>preferred
>> >> > approach in R so you can re-do your calculations easily if you
>find
>> >a
>> >> > mistake later.
>> >> > --
>> >> > Sent from my phone. Please excuse my brevity.
>> >> >
>> >> > On January 7, 2018 7:35:59 PM PST, Ek Esawi <[hidden email]>
>> >wrote:
>> >> >> I just came up with a solution right after i posted the
>question,
>> >but
>> >> >> i figured there must be a better and shorter one.than my
>solution
>> >> >> sdf1[[1]][1,4]<-lapplyresults[[1]]
>> >> >> sdf1[[2]][1,4]<-lapplyresults[[2]]
>> >> >>
>> >> >> EK
>> >> >>
>> >> >> On Sun, Jan 7, 2018 at 10:13 PM, Ek Esawi <[hidden email]>
>> >wrote:
>> >> >>> Hi all--
>> >> >>>
>> >> >>> I stumbled on this problem online. I did not like the solution
>> >given
>> >> >>> there which was a long UDF. I thought why cannot split and l/s
>> >apply
>> >> >>> work here. My aim is to split the data frame, use l/sapply,
>make
>> >> >>> changes on the split lists and combine the split lists to new
>> >data
>> >> >>> frame with the desired changes/output.
>> >> >>>
>> >> >>> The data frame shown below has a column named ID which has 2
>> >> >> variables
>> >> >>> a and b; i want to replace the NAs on the Value column by 2,
>> >which
>> >> >>> is the only numeric entry, for ID=a and by 5 for ID=b.
>> >> >>>
>> >> >>> I worked out the solution but could not replace the results in
>> >the
>> >> >> split lists.
>> >> >>>
>> >> >>> Original dataframe , df1
>> >> >>>   ID ID_2 Firist Value
>> >> >>> 1  a   aa   TRUE     2
>> >> >>> 2  a   ab  FALSE    NA
>> >> >>> 3  a   ac  FALSE    NA
>> >> >>> 4  b   aa   TRUE     5
>> >> >>> 5  b   ab  FALSE    NA
>> >> >>> Sdf1
>> >> >>> $a
>> >> >>> ID ID_2 Firist Value
>> >> >>> 1  a   aa   TRUE     2
>> >> >>> 2  a   ab  FALSE    NA
>> >> >>> 3  a   ac  FALSE    NA
>> >> >>> $b
>> >> >>>   ID ID_2 Firist Value
>> >> >>> 4  b   aa   TRUE     5
>> >> >>> 5  b   ab  FALSE    NA
>> >> >>> Desired results
>> >> >>> ID ID_2 Firist Value
>> >> >>> 1  a   aa   TRUE    2
>> >> >>> 2  a   ab  FALSE    2
>> >> >>> 3  a   ac  FALSE    2
>> >> >>>
>> >> >>> $b
>> >> >>>   ID ID_2 Firist Value
>> >> >>> 4  b   aa   TRUE     5
>> >> >>> 5  b   ab  FALSE     5
>> >> >>>
>> >> >>> My code
>> >> >>>
>> >> >>> sdf <- split(df1,df$ID)
>> >> >>> lapply(sdf, function(z)
>> >> >> ifelse(is.na(z$Value),z$Value[!is.na(z$Value)],z$Value))
>> >> >>> result:
>> >> >>> $ a: num [1:3] 2 2 2
>> >> >>> $ b: num [1:2] 5 5
>> >> >>>
>> >> >>> How could I put these two lists back in the split data frame,
>> >sdf1?
>> >> >>> Then I could use do.call to reassemble a data frame from the
>> >split
>> >> >>> lists,
>> >> >>>
>> >> >>> Thanks,
>> >> >>> EK
>> >> >>
>> >> >> ______________________________________________
>> >> >> [hidden email] mailing list -- To UNSUBSCRIBE and more,
>see
>> >> >> https://stat.ethz.ch/mailman/listinfo/r-help
>> >> >> PLEASE do read the posting guide
>> >> >> http://www.R-project.org/posting-guide.html
>> >> >> and provide commented, minimal, self-contained, reproducible
>code.
>> >> >
>> >> > ______________________________________________
>> >> > [hidden email] mailing list -- To UNSUBSCRIBE and more,
>see
>> >> > https://stat.ethz.ch/mailman/listinfo/r-help
>> >> > PLEASE do read the posting guide
>> >> > http://www.R-project.org/posting-guide.html
>> >> > and provide commented, minimal, self-contained, reproducible
>code.
>> >> >
>> >>
>> >>
>> >-----------------------------------------------------------
>> ----------------
>> >> Jeff Newmiller                        The     .....       .....
>Go
>> >Live...
>> >> DCN:<[hidden email]>        Basics: ##.#.       ##.#.
>Live
>> >Go...
>> >>                                        Live:   OO#.. Dead: OO#..
>> >Playing
>> >> Research Engineer (Solar/Batteries            O.O#.       #.O#.
>with
>> >> /Software/Embedded Controllers)               .OO#.       .OO#.
>> >rocks...1k
>> >>
>> >> ______________________________________________
>> >> [hidden email] mailing list -- To UNSUBSCRIBE and more, see
>> >> https://stat.ethz.ch/mailman/listinfo/r-help
>> >> PLEASE do read the posting guide
>> >http://www.R-project.org/posting-guide.html
>> >> and provide commented, minimal, self-contained, reproducible code.
>> >
>> >________________________________
>> >Tento e-mail a jakékoliv k němu připojené dokumenty jsou důvěrné a
>jsou
>> >určeny pouze jeho adresátům.
>> >Jestliže jste obdržel(a) tento e-mail omylem, informujte laskavě
>> >neprodleně jeho odesílatele. Obsah tohoto emailu i s přílohami a
>jeho
>> >kopie vymažte ze svého systému.
>> >Nejste-li zamýšleným adresátem tohoto emailu, nejste oprávněni tento
>> >email jakkoliv užívat, rozšiřovat, kopírovat či zveřejňovat.
>> >Odesílatel e-mailu neodpovídá za eventuální škodu způsobenou
>> >modifikacemi či zpožděním přenosu e-mailu.
>> >
>> >V případě, že je tento e-mail součástí obchodního jednání:
>> >- vyhrazuje si odesílatel právo ukončit kdykoliv jednání o uzavření
>> >smlouvy, a to z jakéhokoliv důvodu i bez uvedení důvodu.
>> >- a obsahuje-li nabídku, je adresát oprávněn nabídku bezodkladně
>> >přijmout; Odesílatel tohoto e-mailu (nabídky) vylučuje přijetí
>nabídky
>> >ze strany příjemce s dodatkem či odchylkou.
>> >- trvá odesílatel na tom, že příslušná smlouva je uzavřena teprve
>> >výslovným dosažením shody na všech jejích náležitostech.
>> >- odesílatel tohoto emailu informuje, že není oprávněn uzavírat za
>> >společnost žádné smlouvy s výjimkou případů, kdy k tomu byl písemně
>> >zmocněn nebo písemně pověřen a takové pověření nebo plná moc byly
>> >adresátovi tohoto emailu případně osobě, kterou adresát zastupuje,
>> >předloženy nebo jejich existence je adresátovi či osobě jím
>zastoupené
>> >známá.
>> >
>> >This e-mail and any documents attached to it may be confidential and
>> >are intended only for its intended recipients.
>> >If you received this e-mail by mistake, please immediately inform
>its
>> >sender. Delete the contents of this e-mail with all attachments and
>its
>> >copies from your system.
>> >If you are not the intended recipient of this e-mail, you are not
>> >authorized to use, disseminate, copy or disclose this e-mail in any
>> >manner.
>> >The sender of this e-mail shall not be liable for any possible
>damage
>> >caused by modifications of the e-mail or by delay with transfer of
>the
>> >email.
>> >
>> >In case that this e-mail forms part of business dealings:
>> >- the sender reserves the right to end negotiations about entering
>into
>> >a contract in any time, for any reason, and without stating any
>> >reasoning.
>> >- if the e-mail contains an offer, the recipient is entitled to
>> >immediately accept such offer; The sender of this e-mail (offer)
>> >excludes any acceptance of the offer on the part of the recipient
>> >containing any amendment or variation.
>> >- the sender insists on that the respective contract is concluded
>only
>> >upon an express mutual agreement on all its aspects.
>> >- the sender of this e-mail informs that he/she is not authorized to
>> >enter into any contracts on behalf of the company except for cases
>in
>> >which he/she is expressly authorized to do so in writing, and such
>> >authorization or power of attorney is submitted to the recipient or
>the
>> >person represented by the recipient, or the existence of such
>> >authorization is known to the recipient of the person represented by
>> >the recipient.
>>
>> ______________________________________________
>> [hidden email] mailing list -- To UNSUBSCRIBE and more, see
>> https://stat.ethz.ch/mailman/listinfo/r-help
>> PLEASE do read the posting guide http://www.R-project.org/
>> posting-guide.html
>> and provide commented, minimal, self-contained, reproducible code.
>>

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Re: Replace NAs in split lists

Ek Esawi
In reply to this post by Jeff Newmiller
Thank you Jeff. Your code works, as usual , perfectly. I am just
wondering why if i put the whole code in one line, i get an error
message.
sdf2 <- lapply( sdf, function(z){z$Value
<-ifelse(is.na(z$Value),z$Value[!is.na(z$Value)][1],z$Value)z})
error. unexpected symbol in sdf2

Thanks again

EK


On Mon, Jan 8, 2018 at 3:12 AM, Jeff Newmiller <[hidden email]> wrote:

> Upon closer examination I see that you are not using the split version of
> df1 as I usually would, so here is a reproducible example:
>
> #----
> df1 <- read.table( text=
> "ID ID_2 Firist Value
> 1  a   aa   TRUE     2
> 2  a   ab  FALSE    NA
> 3  a   ac  FALSE    NA
> 4  b   aa   TRUE     5
> 5  b   ab  FALSE    NA
> ", header=TRUE, as.is=TRUE )
>
> sdf <- split( df1, df1$ID )
> # note the extra [ 1 ] in case you have more than one non-NA value # per ID
> sdf2 <- lapply( sdf
>               , function( z ) {
>                  z$Value <- ifelse( is.na( z$Value )
>                                   , z$Value[ !is.na( z$Value ) ][ 1 ]
>                                   , z$Value
>                                   )
>                  z
>                 }
>               )
> df2 <- do.call( rbind, sdf2 )
> df2
> #>     ID ID_2 Firist Value
> #> a.1  a   aa   TRUE     2
> #> a.2  a   ab  FALSE     2
> #> a.3  a   ac  FALSE     2
> #> b.4  b   aa   TRUE     5
> #> b.5  b   ab  FALSE     5
>
> # or using tidyverse methods
>
> library(dplyr)
> #>
> #> Attaching package: 'dplyr'
> #> The following objects are masked from 'package:stats':
> #>
> #>     filter, lag
> #> The following objects are masked from 'package:base':
> #>
> #>     intersect, setdiff, setequal, union
> df3 <- (   df1
>        %>% group_by( ID )
>        %>% do({
>               mutate( .
>                     , Value = ifelse( is.na( Value )
>                                     , Value[ !is.na( Value ) ][ 1 ]
>                                     , Value
>                                     )
>                     )
>            })
>        %>% ungroup
>        )
> df3
> #> # A tibble: 5 x 4
> #>   ID    ID_2  Firist Value
> #>   <chr> <chr> <lgl>  <int>
> #> 1 a     aa    T          2
> #> 2 a     ab    F          2
> #> 3 a     ac    F          2
> #> 4 b     aa    T          5
> #> 5 b     ab    F          5
> #----
>
>
> On Sun, 7 Jan 2018, Jeff Newmiller wrote:
>
>> Why do you want to modify df1?
>>
>> Why not just reassemble the parts as a new data frame and use that going
>> forward in your calculations? That is generally the preferred approach in R
>> so you can re-do your calculations easily if you find a mistake later.
>> --
>> Sent from my phone. Please excuse my brevity.
>>
>> On January 7, 2018 7:35:59 PM PST, Ek Esawi <[hidden email]> wrote:
>>>
>>> I just came up with a solution right after i posted the question, but
>>> i figured there must be a better and shorter one.than my solution
>>> sdf1[[1]][1,4]<-lapplyresults[[1]]
>>> sdf1[[2]][1,4]<-lapplyresults[[2]]
>>>
>>> EK
>>>
>>> On Sun, Jan 7, 2018 at 10:13 PM, Ek Esawi <[hidden email]> wrote:
>>>>
>>>> Hi all--
>>>>
>>>> I stumbled on this problem online. I did not like the solution given
>>>> there which was a long UDF. I thought why cannot split and l/s apply
>>>> work here. My aim is to split the data frame, use l/sapply, make
>>>> changes on the split lists and combine the split lists to new data
>>>> frame with the desired changes/output.
>>>>
>>>> The data frame shown below has a column named ID which has 2
>>>
>>> variables
>>>>
>>>> a and b; i want to replace the NAs on the Value column by 2, which is
>>>> the only numeric entry, for ID=a and by 5 for ID=b.
>>>>
>>>> I worked out the solution but could not replace the results in the
>>>
>>> split lists.
>>>>
>>>>
>>>> Original dataframe , df1
>>>>   ID ID_2 Firist Value
>>>> 1  a   aa   TRUE     2
>>>> 2  a   ab  FALSE    NA
>>>> 3  a   ac  FALSE    NA
>>>> 4  b   aa   TRUE     5
>>>> 5  b   ab  FALSE    NA
>>>> Sdf1
>>>> $a
>>>> ID ID_2 Firist Value
>>>> 1  a   aa   TRUE     2
>>>> 2  a   ab  FALSE    NA
>>>> 3  a   ac  FALSE    NA
>>>> $b
>>>>   ID ID_2 Firist Value
>>>> 4  b   aa   TRUE     5
>>>> 5  b   ab  FALSE    NA
>>>> Desired results
>>>> ID ID_2 Firist Value
>>>> 1  a   aa   TRUE    2
>>>> 2  a   ab  FALSE    2
>>>> 3  a   ac  FALSE    2
>>>>
>>>> $b
>>>>   ID ID_2 Firist Value
>>>> 4  b   aa   TRUE     5
>>>> 5  b   ab  FALSE     5
>>>>
>>>> My code
>>>>
>>>> sdf <- split(df1,df$ID)
>>>> lapply(sdf, function(z)
>>>
>>> ifelse(is.na(z$Value),z$Value[!is.na(z$Value)],z$Value))
>>>>
>>>> result:
>>>> $ a: num [1:3] 2 2 2
>>>> $ b: num [1:2] 5 5
>>>>
>>>> How could I put these two lists back in the split data frame, sdf1?
>>>> Then I could use do.call to reassemble a data frame from the split
>>>> lists,
>>>>
>>>> Thanks,
>>>> EK
>>>
>>>
>>> ______________________________________________
>>> [hidden email] mailing list -- To UNSUBSCRIBE and more, see
>>> https://stat.ethz.ch/mailman/listinfo/r-help
>>> PLEASE do read the posting guide
>>> http://www.R-project.org/posting-guide.html
>>> and provide commented, minimal, self-contained, reproducible code.
>>
>>
>> ______________________________________________
>> [hidden email] mailing list -- To UNSUBSCRIBE and more, see
>> https://stat.ethz.ch/mailman/listinfo/r-help
>> PLEASE do read the posting guide
>> http://www.R-project.org/posting-guide.html
>> and provide commented, minimal, self-contained, reproducible code.
>>
>
> ---------------------------------------------------------------------------
> Jeff Newmiller                        The     .....       .....  Go Live...
> DCN:<[hidden email]>        Basics: ##.#.       ##.#.  Live Go...
>                                       Live:   OO#.. Dead: OO#..  Playing
> Research Engineer (Solar/Batteries            O.O#.       #.O#.  with
> /Software/Embedded Controllers)               .OO#.       .OO#.  rocks...1k
> ---------------------------------------------------------------------------

______________________________________________
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Re: Replace NAs in split lists

Rui Barradas
In reply to this post by Ek Esawi
Because you need to separate the instructions with a ; (semi-colon).
Hope this helps
Rui Barradas


Enviado a partir do meu smartphone Samsung Galaxy.-------- Mensagem original --------De: Ek Esawi <[hidden email]> Data: 08/01/2018  16:03  (GMT+00:00) Para: Jeff Newmiller <[hidden email]>, [hidden email] Assunto: Re: [R] Replace NAs in split lists
Thank you Jeff. Your code works, as usual , perfectly. I am just
wondering why if i put the whole code in one line, i get an error
message.
sdf2 <- lapply( sdf, function(z){z$Value
<-ifelse(is.na(z$Value),z$Value[!is.na(z$Value)][1],z$Value)z})
error. unexpected symbol in sdf2

Thanks again

EK


On Mon, Jan 8, 2018 at 3:12 AM, Jeff Newmiller <[hidden email]> wrote:

> Upon closer examination I see that you are not using the split version of
> df1 as I usually would, so here is a reproducible example:
>
> #----
> df1 <- read.table( text=
> "ID ID_2 Firist Value
> 1  a   aa   TRUE     2
> 2  a   ab  FALSE    NA
> 3  a   ac  FALSE    NA
> 4  b   aa   TRUE     5
> 5  b   ab  FALSE    NA
> ", header=TRUE, as.is=TRUE )
>
> sdf <- split( df1, df1$ID )
> # note the extra [ 1 ] in case you have more than one non-NA value # per ID
> sdf2 <- lapply( sdf
>               , function( z ) {
>                  z$Value <- ifelse( is.na( z$Value )
>                                   , z$Value[ !is.na( z$Value ) ][ 1 ]
>                                   , z$Value
>                                   )
>                  z
>                 }
>               )
> df2 <- do.call( rbind, sdf2 )
> df2
> #>     ID ID_2 Firist Value
> #> a.1  a   aa   TRUE     2
> #> a.2  a   ab  FALSE     2
> #> a.3  a   ac  FALSE     2
> #> b.4  b   aa   TRUE     5
> #> b.5  b   ab  FALSE     5
>
> # or using tidyverse methods
>
> library(dplyr)
> #>
> #> Attaching package: 'dplyr'
> #> The following objects are masked from 'package:stats':
> #>
> #>     filter, lag
> #> The following objects are masked from 'package:base':
> #>
> #>     intersect, setdiff, setequal, union
> df3 <- (   df1
>        %>% group_by( ID )
>        %>% do({
>               mutate( .
>                     , Value = ifelse( is.na( Value )
>                                     , Value[ !is.na( Value ) ][ 1 ]
>                                     , Value
>                                     )
>                     )
>            })
>        %>% ungroup
>        )
> df3
> #> # A tibble: 5 x 4
> #>   ID    ID_2  Firist Value
> #>   <chr> <chr> <lgl>  <int>
> #> 1 a     aa    T          2
> #> 2 a     ab    F          2
> #> 3 a     ac    F          2
> #> 4 b     aa    T          5
> #> 5 b     ab    F          5
> #----
>
>
> On Sun, 7 Jan 2018, Jeff Newmiller wrote:
>
>> Why do you want to modify df1?
>>
>> Why not just reassemble the parts as a new data frame and use that going
>> forward in your calculations? That is generally the preferred approach in R
>> so you can re-do your calculations easily if you find a mistake later.
>> --
>> Sent from my phone. Please excuse my brevity.
>>
>> On January 7, 2018 7:35:59 PM PST, Ek Esawi <[hidden email]> wrote:
>>>
>>> I just came up with a solution right after i posted the question, but
>>> i figured there must be a better and shorter one.than my solution
>>> sdf1[[1]][1,4]<-lapplyresults[[1]]
>>> sdf1[[2]][1,4]<-lapplyresults[[2]]
>>>
>>> EK
>>>
>>> On Sun, Jan 7, 2018 at 10:13 PM, Ek Esawi <[hidden email]> wrote:
>>>>
>>>> Hi all--
>>>>
>>>> I stumbled on this problem online. I did not like the solution given
>>>> there which was a long UDF. I thought why cannot split and l/s apply
>>>> work here. My aim is to split the data frame, use l/sapply, make
>>>> changes on the split lists and combine the split lists to new data
>>>> frame with the desired changes/output.
>>>>
>>>> The data frame shown below has a column named ID which has 2
>>>
>>> variables
>>>>
>>>> a and b; i want to replace the NAs on the Value column by 2, which is
>>>> the only numeric entry, for ID=a and by 5 for ID=b.
>>>>
>>>> I worked out the solution but could not replace the results in the
>>>
>>> split lists.
>>>>
>>>>
>>>> Original dataframe , df1
>>>>   ID ID_2 Firist Value
>>>> 1  a   aa   TRUE     2
>>>> 2  a   ab  FALSE    NA
>>>> 3  a   ac  FALSE    NA
>>>> 4  b   aa   TRUE     5
>>>> 5  b   ab  FALSE    NA
>>>> Sdf1
>>>> $a
>>>> ID ID_2 Firist Value
>>>> 1  a   aa   TRUE     2
>>>> 2  a   ab  FALSE    NA
>>>> 3  a   ac  FALSE    NA
>>>> $b
>>>>   ID ID_2 Firist Value
>>>> 4  b   aa   TRUE     5
>>>> 5  b   ab  FALSE    NA
>>>> Desired results
>>>> ID ID_2 Firist Value
>>>> 1  a   aa   TRUE    2
>>>> 2  a   ab  FALSE    2
>>>> 3  a   ac  FALSE    2
>>>>
>>>> $b
>>>>   ID ID_2 Firist Value
>>>> 4  b   aa   TRUE     5
>>>> 5  b   ab  FALSE     5
>>>>
>>>> My code
>>>>
>>>> sdf <- split(df1,df$ID)
>>>> lapply(sdf, function(z)
>>>
>>> ifelse(is.na(z$Value),z$Value[!is.na(z$Value)],z$Value))
>>>>
>>>> result:
>>>> $ a: num [1:3] 2 2 2
>>>> $ b: num [1:2] 5 5
>>>>
>>>> How could I put these two lists back in the split data frame, sdf1?
>>>> Then I could use do.call to reassemble a data frame from the split
>>>> lists,
>>>>
>>>> Thanks,
>>>> EK
>>>
>>>
>>> ______________________________________________
>>> [hidden email] mailing list -- To UNSUBSCRIBE and more, see
>>> https://stat.ethz.ch/mailman/listinfo/r-help
>>> PLEASE do read the posting guide
>>> http://www.R-project.org/posting-guide.html
>>> and provide commented, minimal, self-contained, reproducible code.
>>
>>
>> ______________________________________________
>> [hidden email] mailing list -- To UNSUBSCRIBE and more, see
>> https://stat.ethz.ch/mailman/listinfo/r-help
>> PLEASE do read the posting guide
>> http://www.R-project.org/posting-guide.html
>> and provide commented, minimal, self-contained, reproducible code.
>>
>
> ---------------------------------------------------------------------------
> Jeff Newmiller                        The     .....       .....  Go Live...
> DCN:<[hidden email]>        Basics: ##.#.       ##.#.  Live Go...
>                                       Live:   OO#.. Dead: OO#..  Playing
> Research Engineer (Solar/Batteries            O.O#.       #.O#.  with
> /Software/Embedded Controllers)               .OO#.       .OO#.  rocks...1k
> ---------------------------------------------------------------------------

______________________________________________
[hidden email] mailing list -- To UNSUBSCRIBE and more, see
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.

        [[alternative HTML version deleted]]

______________________________________________
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Re: Replace NAs in split lists

Jeff Newmiller
In reply to this post by Ek Esawi
I don't know. You seem to be posting in HTML so your code is mangled. Can you post plain text and use the reprex package to make sure it produces the errorin a clean R session?
--
Sent from my phone. Please excuse my brevity.

On January 8, 2018 8:03:45 AM PST, Ek Esawi <[hidden email]> wrote:

>Thank you Jeff. Your code works, as usual , perfectly. I am just
>wondering why if i put the whole code in one line, i get an error
>message.
>sdf2 <- lapply( sdf, function(z){z$Value
><-ifelse(is.na(z$Value),z$Value[!is.na(z$Value)][1],z$Value)z})
>error. unexpected symbol in sdf2
>
>Thanks again
>
>EK
>
>
>On Mon, Jan 8, 2018 at 3:12 AM, Jeff Newmiller
><[hidden email]> wrote:
>> Upon closer examination I see that you are not using the split
>version of
>> df1 as I usually would, so here is a reproducible example:
>>
>> #----
>> df1 <- read.table( text=
>> "ID ID_2 Firist Value
>> 1  a   aa   TRUE     2
>> 2  a   ab  FALSE    NA
>> 3  a   ac  FALSE    NA
>> 4  b   aa   TRUE     5
>> 5  b   ab  FALSE    NA
>> ", header=TRUE, as.is=TRUE )
>>
>> sdf <- split( df1, df1$ID )
>> # note the extra [ 1 ] in case you have more than one non-NA value #
>per ID
>> sdf2 <- lapply( sdf
>>               , function( z ) {
>>                  z$Value <- ifelse( is.na( z$Value )
>>                                   , z$Value[ !is.na( z$Value ) ][ 1 ]
>>                                   , z$Value
>>                                   )
>>                  z
>>                 }
>>               )
>> df2 <- do.call( rbind, sdf2 )
>> df2
>> #>     ID ID_2 Firist Value
>> #> a.1  a   aa   TRUE     2
>> #> a.2  a   ab  FALSE     2
>> #> a.3  a   ac  FALSE     2
>> #> b.4  b   aa   TRUE     5
>> #> b.5  b   ab  FALSE     5
>>
>> # or using tidyverse methods
>>
>> library(dplyr)
>> #>
>> #> Attaching package: 'dplyr'
>> #> The following objects are masked from 'package:stats':
>> #>
>> #>     filter, lag
>> #> The following objects are masked from 'package:base':
>> #>
>> #>     intersect, setdiff, setequal, union
>> df3 <- (   df1
>>        %>% group_by( ID )
>>        %>% do({
>>               mutate( .
>>                     , Value = ifelse( is.na( Value )
>>                                     , Value[ !is.na( Value ) ][ 1 ]
>>                                     , Value
>>                                     )
>>                     )
>>            })
>>        %>% ungroup
>>        )
>> df3
>> #> # A tibble: 5 x 4
>> #>   ID    ID_2  Firist Value
>> #>   <chr> <chr> <lgl>  <int>
>> #> 1 a     aa    T          2
>> #> 2 a     ab    F          2
>> #> 3 a     ac    F          2
>> #> 4 b     aa    T          5
>> #> 5 b     ab    F          5
>> #----
>>
>>
>> On Sun, 7 Jan 2018, Jeff Newmiller wrote:
>>
>>> Why do you want to modify df1?
>>>
>>> Why not just reassemble the parts as a new data frame and use that
>going
>>> forward in your calculations? That is generally the preferred
>approach in R
>>> so you can re-do your calculations easily if you find a mistake
>later.
>>> --
>>> Sent from my phone. Please excuse my brevity.
>>>
>>> On January 7, 2018 7:35:59 PM PST, Ek Esawi <[hidden email]>
>wrote:
>>>>
>>>> I just came up with a solution right after i posted the question,
>but
>>>> i figured there must be a better and shorter one.than my solution
>>>> sdf1[[1]][1,4]<-lapplyresults[[1]]
>>>> sdf1[[2]][1,4]<-lapplyresults[[2]]
>>>>
>>>> EK
>>>>
>>>> On Sun, Jan 7, 2018 at 10:13 PM, Ek Esawi <[hidden email]>
>wrote:
>>>>>
>>>>> Hi all--
>>>>>
>>>>> I stumbled on this problem online. I did not like the solution
>given
>>>>> there which was a long UDF. I thought why cannot split and l/s
>apply
>>>>> work here. My aim is to split the data frame, use l/sapply, make
>>>>> changes on the split lists and combine the split lists to new data
>>>>> frame with the desired changes/output.
>>>>>
>>>>> The data frame shown below has a column named ID which has 2
>>>>
>>>> variables
>>>>>
>>>>> a and b; i want to replace the NAs on the Value column by 2, which
>is
>>>>> the only numeric entry, for ID=a and by 5 for ID=b.
>>>>>
>>>>> I worked out the solution but could not replace the results in the
>>>>
>>>> split lists.
>>>>>
>>>>>
>>>>> Original dataframe , df1
>>>>>   ID ID_2 Firist Value
>>>>> 1  a   aa   TRUE     2
>>>>> 2  a   ab  FALSE    NA
>>>>> 3  a   ac  FALSE    NA
>>>>> 4  b   aa   TRUE     5
>>>>> 5  b   ab  FALSE    NA
>>>>> Sdf1
>>>>> $a
>>>>> ID ID_2 Firist Value
>>>>> 1  a   aa   TRUE     2
>>>>> 2  a   ab  FALSE    NA
>>>>> 3  a   ac  FALSE    NA
>>>>> $b
>>>>>   ID ID_2 Firist Value
>>>>> 4  b   aa   TRUE     5
>>>>> 5  b   ab  FALSE    NA
>>>>> Desired results
>>>>> ID ID_2 Firist Value
>>>>> 1  a   aa   TRUE    2
>>>>> 2  a   ab  FALSE    2
>>>>> 3  a   ac  FALSE    2
>>>>>
>>>>> $b
>>>>>   ID ID_2 Firist Value
>>>>> 4  b   aa   TRUE     5
>>>>> 5  b   ab  FALSE     5
>>>>>
>>>>> My code
>>>>>
>>>>> sdf <- split(df1,df$ID)
>>>>> lapply(sdf, function(z)
>>>>
>>>> ifelse(is.na(z$Value),z$Value[!is.na(z$Value)],z$Value))
>>>>>
>>>>> result:
>>>>> $ a: num [1:3] 2 2 2
>>>>> $ b: num [1:2] 5 5
>>>>>
>>>>> How could I put these two lists back in the split data frame,
>sdf1?
>>>>> Then I could use do.call to reassemble a data frame from the split
>>>>> lists,
>>>>>
>>>>> Thanks,
>>>>> EK
>>>>
>>>>
>>>> ______________________________________________
>>>> [hidden email] mailing list -- To UNSUBSCRIBE and more, see
>>>> https://stat.ethz.ch/mailman/listinfo/r-help
>>>> PLEASE do read the posting guide
>>>> http://www.R-project.org/posting-guide.html
>>>> and provide commented, minimal, self-contained, reproducible code.
>>>
>>>
>>> ______________________________________________
>>> [hidden email] mailing list -- To UNSUBSCRIBE and more, see
>>> https://stat.ethz.ch/mailman/listinfo/r-help
>>> PLEASE do read the posting guide
>>> http://www.R-project.org/posting-guide.html
>>> and provide commented, minimal, self-contained, reproducible code.
>>>
>>
>>
>---------------------------------------------------------------------------
>> Jeff Newmiller                        The     .....       .....  Go
>Live...
>> DCN:<[hidden email]>        Basics: ##.#.       ##.#.  Live
>Go...
>>                                       Live:   OO#.. Dead: OO#..
>Playing
>> Research Engineer (Solar/Batteries            O.O#.       #.O#.  with
>> /Software/Embedded Controllers)               .OO#.       .OO#.
>rocks...1k
>>
>---------------------------------------------------------------------------

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Re: Replace NAs in split lists

Ek Esawi
OPS!  Sorry i did indeed posted the code in HTML; should have known better.

ifelse(is.na(z$Value),z$Value[!is.na(z$Value)][1],z$Value)z})
error. unexpected symbol in sdf2

On Mon, Jan 8, 2018 at 11:44 AM, Jeff Newmiller
<[hidden email]> wrote:

> I don't know. You seem to be posting in HTML so your code is mangled. Can you post plain text and use the reprex package to make sure it produces the errorin a clean R session?
> --
> Sent from my phone. Please excuse my brevity.
>
> On January 8, 2018 8:03:45 AM PST, Ek Esawi <[hidden email]> wrote:
>>Thank you Jeff. Your code works, as usual , perfectly. I am just
>>wondering why if i put the whole code in one line, i get an error
>>message.
>>sdf2 <- lapply( sdf, function(z){z$Value
>><-ifelse(is.na(z$Value),z$Value[!is.na(z$Value)][1],z$Value)z})
>>error. unexpected symbol in sdf2
>>
>>Thanks again
>>
>>EK
>>
>>
>>On Mon, Jan 8, 2018 at 3:12 AM, Jeff Newmiller
>><[hidden email]> wrote:
>>> Upon closer examination I see that you are not using the split
>>version of
>>> df1 as I usually would, so here is a reproducible example:
>>>
>>> #----
>>> df1 <- read.table( text=
>>> "ID ID_2 Firist Value
>>> 1  a   aa   TRUE     2
>>> 2  a   ab  FALSE    NA
>>> 3  a   ac  FALSE    NA
>>> 4  b   aa   TRUE     5
>>> 5  b   ab  FALSE    NA
>>> ", header=TRUE, as.is=TRUE )
>>>
>>> sdf <- split( df1, df1$ID )
>>> # note the extra [ 1 ] in case you have more than one non-NA value #
>>per ID
>>> sdf2 <- lapply( sdf
>>>               , function( z ) {
>>>                  z$Value <- ifelse( is.na( z$Value )
>>>                                   , z$Value[ !is.na( z$Value ) ][ 1 ]
>>>                                   , z$Value
>>>                                   )
>>>                  z
>>>                 }
>>>               )
>>> df2 <- do.call( rbind, sdf2 )
>>> df2
>>> #>     ID ID_2 Firist Value
>>> #> a.1  a   aa   TRUE     2
>>> #> a.2  a   ab  FALSE     2
>>> #> a.3  a   ac  FALSE     2
>>> #> b.4  b   aa   TRUE     5
>>> #> b.5  b   ab  FALSE     5
>>>
>>> # or using tidyverse methods
>>>
>>> library(dplyr)
>>> #>
>>> #> Attaching package: 'dplyr'
>>> #> The following objects are masked from 'package:stats':
>>> #>
>>> #>     filter, lag
>>> #> The following objects are masked from 'package:base':
>>> #>
>>> #>     intersect, setdiff, setequal, union
>>> df3 <- (   df1
>>>        %>% group_by( ID )
>>>        %>% do({
>>>               mutate( .
>>>                     , Value = ifelse( is.na( Value )
>>>                                     , Value[ !is.na( Value ) ][ 1 ]
>>>                                     , Value
>>>                                     )
>>>                     )
>>>            })
>>>        %>% ungroup
>>>        )
>>> df3
>>> #> # A tibble: 5 x 4
>>> #>   ID    ID_2  Firist Value
>>> #>   <chr> <chr> <lgl>  <int>
>>> #> 1 a     aa    T          2
>>> #> 2 a     ab    F          2
>>> #> 3 a     ac    F          2
>>> #> 4 b     aa    T          5
>>> #> 5 b     ab    F          5
>>> #----
>>>
>>>
>>> On Sun, 7 Jan 2018, Jeff Newmiller wrote:
>>>
>>>> Why do you want to modify df1?
>>>>
>>>> Why not just reassemble the parts as a new data frame and use that
>>going
>>>> forward in your calculations? That is generally the preferred
>>approach in R
>>>> so you can re-do your calculations easily if you find a mistake
>>later.
>>>> --
>>>> Sent from my phone. Please excuse my brevity.
>>>>
>>>> On January 7, 2018 7:35:59 PM PST, Ek Esawi <[hidden email]>
>>wrote:
>>>>>
>>>>> I just came up with a solution right after i posted the question,
>>but
>>>>> i figured there must be a better and shorter one.than my solution
>>>>> sdf1[[1]][1,4]<-lapplyresults[[1]]
>>>>> sdf1[[2]][1,4]<-lapplyresults[[2]]
>>>>>
>>>>> EK
>>>>>
>>>>> On Sun, Jan 7, 2018 at 10:13 PM, Ek Esawi <[hidden email]>
>>wrote:
>>>>>>
>>>>>> Hi all--
>>>>>>
>>>>>> I stumbled on this problem online. I did not like the solution
>>given
>>>>>> there which was a long UDF. I thought why cannot split and l/s
>>apply
>>>>>> work here. My aim is to split the data frame, use l/sapply, make
>>>>>> changes on the split lists and combine the split lists to new data
>>>>>> frame with the desired changes/output.
>>>>>>
>>>>>> The data frame shown below has a column named ID which has 2
>>>>>
>>>>> variables
>>>>>>
>>>>>> a and b; i want to replace the NAs on the Value column by 2, which
>>is
>>>>>> the only numeric entry, for ID=a and by 5 for ID=b.
>>>>>>
>>>>>> I worked out the solution but could not replace the results in the
>>>>>
>>>>> split lists.
>>>>>>
>>>>>>
>>>>>> Original dataframe , df1
>>>>>>   ID ID_2 Firist Value
>>>>>> 1  a   aa   TRUE     2
>>>>>> 2  a   ab  FALSE    NA
>>>>>> 3  a   ac  FALSE    NA
>>>>>> 4  b   aa   TRUE     5
>>>>>> 5  b   ab  FALSE    NA
>>>>>> Sdf1
>>>>>> $a
>>>>>> ID ID_2 Firist Value
>>>>>> 1  a   aa   TRUE     2
>>>>>> 2  a   ab  FALSE    NA
>>>>>> 3  a   ac  FALSE    NA
>>>>>> $b
>>>>>>   ID ID_2 Firist Value
>>>>>> 4  b   aa   TRUE     5
>>>>>> 5  b   ab  FALSE    NA
>>>>>> Desired results
>>>>>> ID ID_2 Firist Value
>>>>>> 1  a   aa   TRUE    2
>>>>>> 2  a   ab  FALSE    2
>>>>>> 3  a   ac  FALSE    2
>>>>>>
>>>>>> $b
>>>>>>   ID ID_2 Firist Value
>>>>>> 4  b   aa   TRUE     5
>>>>>> 5  b   ab  FALSE     5
>>>>>>
>>>>>> My code
>>>>>>
>>>>>> sdf <- split(df1,df$ID)
>>>>>> lapply(sdf, function(z)
>>>>>
>>>>> ifelse(is.na(z$Value),z$Value[!is.na(z$Value)],z$Value))
>>>>>>
>>>>>> result:
>>>>>> $ a: num [1:3] 2 2 2
>>>>>> $ b: num [1:2] 5 5
>>>>>>
>>>>>> How could I put these two lists back in the split data frame,
>>sdf1?
>>>>>> Then I could use do.call to reassemble a data frame from the split
>>>>>> lists,
>>>>>>
>>>>>> Thanks,
>>>>>> EK
>>>>>
>>>>>
>>>>> ______________________________________________
>>>>> [hidden email] mailing list -- To UNSUBSCRIBE and more, see
>>>>> https://stat.ethz.ch/mailman/listinfo/r-help
>>>>> PLEASE do read the posting guide
>>>>> http://www.R-project.org/posting-guide.html
>>>>> and provide commented, minimal, self-contained, reproducible code.
>>>>
>>>>
>>>> ______________________________________________
>>>> [hidden email] mailing list -- To UNSUBSCRIBE and more, see
>>>> https://stat.ethz.ch/mailman/listinfo/r-help
>>>> PLEASE do read the posting guide
>>>> http://www.R-project.org/posting-guide.html
>>>> and provide commented, minimal, self-contained, reproducible code.
>>>>
>>>
>>>
>>---------------------------------------------------------------------------
>>> Jeff Newmiller                        The     .....       .....  Go
>>Live...
>>> DCN:<[hidden email]>        Basics: ##.#.       ##.#.  Live
>>Go...
>>>                                       Live:   OO#.. Dead: OO#..
>>Playing
>>> Research Engineer (Solar/Batteries            O.O#.       #.O#.  with
>>> /Software/Embedded Controllers)               .OO#.       .OO#.
>>rocks...1k
>>>
>>---------------------------------------------------------------------------

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Re: Replace NAs in split lists

R help mailing list-2
I don't get exactly that error message,
  > ifelse(is.na(z$Value),z$Value[!is.na(z$Value)][1],z$Value)z})
  Error: unexpected symbol in "ifelse(is.na(z$Value),z$Value[!is.na
(z$Value)][1],z$Value)z"
The 'symbol' in "unexpected symbol" refers to a "name" ('z' in this case).
The problem is usually at the end of the code snippet shown in the error
message as in
  > f(a)b # missing newline or semicolon?
  Error: unexpected symbol in "f(a)b"
or
  > f(a b) # missing comma?
  Error: unexpected symbol in "f(a b"

You need a newline or semicolon between the end of 'ifelse(...)' and 'z' or
maybe
you don't want the z there at all.

After you fix that it will complain about the unmatched right brace and
parenthesis.



Bill Dunlap
TIBCO Software
wdunlap tibco.com

On Mon, Jan 8, 2018 at 8:55 AM, Ek Esawi <[hidden email]> wrote:

> OPS!  Sorry i did indeed posted the code in HTML; should have known better.
>
> ifelse(is.na(z$Value),z$Value[!is.na(z$Value)][1],z$Value)z})
> error. unexpected symbol in sdf2
>
> On Mon, Jan 8, 2018 at 11:44 AM, Jeff Newmiller
> <[hidden email]> wrote:
> > I don't know. You seem to be posting in HTML so your code is mangled.
> Can you post plain text and use the reprex package to make sure it produces
> the errorin a clean R session?
> > --
> > Sent from my phone. Please excuse my brevity.
> >
> > On January 8, 2018 8:03:45 AM PST, Ek Esawi <[hidden email]> wrote:
> >>Thank you Jeff. Your code works, as usual , perfectly. I am just
> >>wondering why if i put the whole code in one line, i get an error
> >>message.
> >>sdf2 <- lapply( sdf, function(z){z$Value
> >><-ifelse(is.na(z$Value),z$Value[!is.na(z$Value)][1],z$Value)z})
> >>error. unexpected symbol in sdf2
> >>
> >>Thanks again
> >>
> >>EK
> >>
> >>
> >>On Mon, Jan 8, 2018 at 3:12 AM, Jeff Newmiller
> >><[hidden email]> wrote:
> >>> Upon closer examination I see that you are not using the split
> >>version of
> >>> df1 as I usually would, so here is a reproducible example:
> >>>
> >>> #----
> >>> df1 <- read.table( text=
> >>> "ID ID_2 Firist Value
> >>> 1  a   aa   TRUE     2
> >>> 2  a   ab  FALSE    NA
> >>> 3  a   ac  FALSE    NA
> >>> 4  b   aa   TRUE     5
> >>> 5  b   ab  FALSE    NA
> >>> ", header=TRUE, as.is=TRUE )
> >>>
> >>> sdf <- split( df1, df1$ID )
> >>> # note the extra [ 1 ] in case you have more than one non-NA value #
> >>per ID
> >>> sdf2 <- lapply( sdf
> >>>               , function( z ) {
> >>>                  z$Value <- ifelse( is.na( z$Value )
> >>>                                   , z$Value[ !is.na( z$Value ) ][ 1 ]
> >>>                                   , z$Value
> >>>                                   )
> >>>                  z
> >>>                 }
> >>>               )
> >>> df2 <- do.call( rbind, sdf2 )
> >>> df2
> >>> #>     ID ID_2 Firist Value
> >>> #> a.1  a   aa   TRUE     2
> >>> #> a.2  a   ab  FALSE     2
> >>> #> a.3  a   ac  FALSE     2
> >>> #> b.4  b   aa   TRUE     5
> >>> #> b.5  b   ab  FALSE     5
> >>>
> >>> # or using tidyverse methods
> >>>
> >>> library(dplyr)
> >>> #>
> >>> #> Attaching package: 'dplyr'
> >>> #> The following objects are masked from 'package:stats':
> >>> #>
> >>> #>     filter, lag
> >>> #> The following objects are masked from 'package:base':
> >>> #>
> >>> #>     intersect, setdiff, setequal, union
> >>> df3 <- (   df1
> >>>        %>% group_by( ID )
> >>>        %>% do({
> >>>               mutate( .
> >>>                     , Value = ifelse( is.na( Value )
> >>>                                     , Value[ !is.na( Value ) ][ 1 ]
> >>>                                     , Value
> >>>                                     )
> >>>                     )
> >>>            })
> >>>        %>% ungroup
> >>>        )
> >>> df3
> >>> #> # A tibble: 5 x 4
> >>> #>   ID    ID_2  Firist Value
> >>> #>   <chr> <chr> <lgl>  <int>
> >>> #> 1 a     aa    T          2
> >>> #> 2 a     ab    F          2
> >>> #> 3 a     ac    F          2
> >>> #> 4 b     aa    T          5
> >>> #> 5 b     ab    F          5
> >>> #----
> >>>
> >>>
> >>> On Sun, 7 Jan 2018, Jeff Newmiller wrote:
> >>>
> >>>> Why do you want to modify df1?
> >>>>
> >>>> Why not just reassemble the parts as a new data frame and use that
> >>going
> >>>> forward in your calculations? That is generally the preferred
> >>approach in R
> >>>> so you can re-do your calculations easily if you find a mistake
> >>later.
> >>>> --
> >>>> Sent from my phone. Please excuse my brevity.
> >>>>
> >>>> On January 7, 2018 7:35:59 PM PST, Ek Esawi <[hidden email]>
> >>wrote:
> >>>>>
> >>>>> I just came up with a solution right after i posted the question,
> >>but
> >>>>> i figured there must be a better and shorter one.than my solution
> >>>>> sdf1[[1]][1,4]<-lapplyresults[[1]]
> >>>>> sdf1[[2]][1,4]<-lapplyresults[[2]]
> >>>>>
> >>>>> EK
> >>>>>
> >>>>> On Sun, Jan 7, 2018 at 10:13 PM, Ek Esawi <[hidden email]>
> >>wrote:
> >>>>>>
> >>>>>> Hi all--
> >>>>>>
> >>>>>> I stumbled on this problem online. I did not like the solution
> >>given
> >>>>>> there which was a long UDF. I thought why cannot split and l/s
> >>apply
> >>>>>> work here. My aim is to split the data frame, use l/sapply, make
> >>>>>> changes on the split lists and combine the split lists to new data
> >>>>>> frame with the desired changes/output.
> >>>>>>
> >>>>>> The data frame shown below has a column named ID which has 2
> >>>>>
> >>>>> variables
> >>>>>>
> >>>>>> a and b; i want to replace the NAs on the Value column by 2, which
> >>is
> >>>>>> the only numeric entry, for ID=a and by 5 for ID=b.
> >>>>>>
> >>>>>> I worked out the solution but could not replace the results in the
> >>>>>
> >>>>> split lists.
> >>>>>>
> >>>>>>
> >>>>>> Original dataframe , df1
> >>>>>>   ID ID_2 Firist Value
> >>>>>> 1  a   aa   TRUE     2
> >>>>>> 2  a   ab  FALSE    NA
> >>>>>> 3  a   ac  FALSE    NA
> >>>>>> 4  b   aa   TRUE     5
> >>>>>> 5  b   ab  FALSE    NA
> >>>>>> Sdf1
> >>>>>> $a
> >>>>>> ID ID_2 Firist Value
> >>>>>> 1  a   aa   TRUE     2
> >>>>>> 2  a   ab  FALSE    NA
> >>>>>> 3  a   ac  FALSE    NA
> >>>>>> $b
> >>>>>>   ID ID_2 Firist Value
> >>>>>> 4  b   aa   TRUE     5
> >>>>>> 5  b   ab  FALSE    NA
> >>>>>> Desired results
> >>>>>> ID ID_2 Firist Value
> >>>>>> 1  a   aa   TRUE    2
> >>>>>> 2  a   ab  FALSE    2
> >>>>>> 3  a   ac  FALSE    2
> >>>>>>
> >>>>>> $b
> >>>>>>   ID ID_2 Firist Value
> >>>>>> 4  b   aa   TRUE     5
> >>>>>> 5  b   ab  FALSE     5
> >>>>>>
> >>>>>> My code
> >>>>>>
> >>>>>> sdf <- split(df1,df$ID)
> >>>>>> lapply(sdf, function(z)
> >>>>>
> >>>>> ifelse(is.na(z$Value),z$Value[!is.na(z$Value)],z$Value))
> >>>>>>
> >>>>>> result:
> >>>>>> $ a: num [1:3] 2 2 2
> >>>>>> $ b: num [1:2] 5 5
> >>>>>>
> >>>>>> How could I put these two lists back in the split data frame,
> >>sdf1?
> >>>>>> Then I could use do.call to reassemble a data frame from the split
> >>>>>> lists,
> >>>>>>
> >>>>>> Thanks,
> >>>>>> EK
> >>>>>
> >>>>>
> >>>>> ______________________________________________
> >>>>> [hidden email] mailing list -- To UNSUBSCRIBE and more, see
> >>>>> https://stat.ethz.ch/mailman/listinfo/r-help
> >>>>> PLEASE do read the posting guide
> >>>>> http://www.R-project.org/posting-guide.html
> >>>>> and provide commented, minimal, self-contained, reproducible code.
> >>>>
> >>>>
> >>>> ______________________________________________
> >>>> [hidden email] mailing list -- To UNSUBSCRIBE and more, see
> >>>> https://stat.ethz.ch/mailman/listinfo/r-help
> >>>> PLEASE do read the posting guide
> >>>> http://www.R-project.org/posting-guide.html
> >>>> and provide commented, minimal, self-contained, reproducible code.
> >>>>
> >>>
> >>>
> >>----------------------------------------------------------
> -----------------
> >>> Jeff Newmiller                        The     .....       .....  Go
> >>Live...
> >>> DCN:<[hidden email]>        Basics: ##.#.       ##.#.  Live
> >>Go...
> >>>                                       Live:   OO#.. Dead: OO#..
> >>Playing
> >>> Research Engineer (Solar/Batteries            O.O#.       #.O#.  with
> >>> /Software/Embedded Controllers)               .OO#.       .OO#.
> >>rocks...1k
> >>>
> >>----------------------------------------------------------
> -----------------
>
> ______________________________________________
> [hidden email] mailing list -- To UNSUBSCRIBE and more, see
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/
> posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>

        [[alternative HTML version deleted]]

______________________________________________
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Re: Replace NAs in split lists

Ek Esawi
In reply to this post by Ek Esawi
Thank you all. Now everything works. Happy 2018 and beyond
EK

On Sun, Jan 7, 2018 at 10:13 PM, Ek Esawi <[hidden email]> wrote:

> Hi all--
>
> I stumbled on this problem online. I did not like the solution given
> there which was a long UDF. I thought why cannot split and l/s apply
> work here. My aim is to split the data frame, use l/sapply, make
> changes on the split lists and combine the split lists to new data
> frame with the desired changes/output.
>
> The data frame shown below has a column named ID which has 2 variables
> a and b; i want to replace the NAs on the Value column by 2, which is
> the only numeric entry, for ID=a and by 5 for ID=b.
>
> I worked out the solution but could not replace the results in the split lists.
>
> Original dataframe , df1
>   ID ID_2 Firist Value
> 1  a   aa   TRUE     2
> 2  a   ab  FALSE    NA
> 3  a   ac  FALSE    NA
> 4  b   aa   TRUE     5
> 5  b   ab  FALSE    NA
> Sdf1
> $a
> ID ID_2 Firist Value
> 1  a   aa   TRUE     2
> 2  a   ab  FALSE    NA
> 3  a   ac  FALSE    NA
> $b
>   ID ID_2 Firist Value
> 4  b   aa   TRUE     5
> 5  b   ab  FALSE    NA
> Desired results
> ID ID_2 Firist Value
> 1  a   aa   TRUE    2
> 2  a   ab  FALSE    2
> 3  a   ac  FALSE    2
>
> $b
>   ID ID_2 Firist Value
> 4  b   aa   TRUE     5
> 5  b   ab  FALSE     5
>
> My code
>
> sdf <- split(df1,df$ID)
> lapply(sdf, function(z) ifelse(is.na(z$Value),z$Value[!is.na(z$Value)],z$Value))
> result:
> $ a: num [1:3] 2 2 2
> $ b: num [1:2] 5 5
>
> How could I put these two lists back in the split data frame, sdf1?
> Then I could use do.call to reassemble a data frame from the split
> lists,
>
> Thanks,
> EK

______________________________________________
[hidden email] mailing list -- To UNSUBSCRIBE and more, see
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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Re: Replace NAs in split lists

PIKAL Petr
In reply to this post by Jeff Newmiller
Hi

Yes, that is why I mentioned "the example". Of course if there are missing values in other columns or data layout is different, na.locf could give undesired result.

However it is simple and works with data similar to the example.

Cheers
Petr

> -----Original Message-----
> From: Jeff Newmiller [mailto:[hidden email]]
> Sent: Monday, January 8, 2018 4:45 PM
> To: Eric Berger <[hidden email]>
> Cc: PIKAL Petr <[hidden email]>; [hidden email]; Ek Esawi
> <[hidden email]>
> Subject: Re: [R] Replace NAs in split lists
>
> "Enforce" is overstating it... results will differ if there are no non-NA values for
> a given ID, and there is a potential further discrepancy if there are multiple
> non-NA values. But these issues were not identified by the OP, so may not be
> relevant in their case.
> --
> Sent from my phone. Please excuse my brevity.
>
> On January 8, 2018 6:41:33 AM PST, Eric Berger <[hidden email]>
> wrote:
> >You can enforce these assumptions by sorting on multiple columns, which
> >leads to
> >
> >na.locf(df1[ order(df1$ID,df1$Value), ])
> >
> >
> >
> >On Mon, Jan 8, 2018 at 4:19 PM, Jeff Newmiller
> ><[hidden email]>
> >wrote:
> >
> >> Yes, you are right if the IDs are always sequentially-adjacent and
> >the
> >> first non-NA value appears in the first record for each ID.
> >> --
> >> Sent from my phone. Please excuse my brevity.
> >>
> >> On January 8, 2018 2:29:40 AM PST, PIKAL Petr
> ><[hidden email]>
> >> wrote:
> >> >Hi
> >> >
> >> >With the example, na.locf seems to be the easiest way.
> >> >> library(zoo)
> >> >
> >> >> na.locf(df1)
> >> >  ID ID_2 Firist Value
> >> >1  a   aa   TRUE     2
> >> >2  a   ab  FALSE     2
> >> >3  a   ac  FALSE     2
> >> >4  b   aa   TRUE     5
> >> >5  b   ab  FALSE     5
> >> >
> >> >Cheers
> >> >Petr
> >> >
> >> >> -----Original Message-----
> >> >> From: R-help [mailto:[hidden email]] On Behalf Of
> >Jeff
> >> >> Newmiller
> >> >> Sent: Monday, January 8, 2018 9:13 AM
> >> >> To: [hidden email]; Ek Esawi <[hidden email]>
> >> >> Subject: Re: [R] Replace NAs in split lists
> >> >>
> >> >> Upon closer examination I see that you are not using the split
> >> >version of
> >> >> df1 as I usually would, so here is a reproducible example:
> >> >>
> >> >> #----
> >> >> df1 <- read.table( text=
> >> >> "ID ID_2 Firist Value
> >> >> 1  a   aa   TRUE     2
> >> >> 2  a   ab  FALSE    NA
> >> >> 3  a   ac  FALSE    NA
> >> >> 4  b   aa   TRUE     5
> >> >> 5  b   ab  FALSE    NA
> >> >> ", header=TRUE, as.is=TRUE )
> >> >>
> >> >> sdf <- split( df1, df1$ID )
> >> >> # note the extra [ 1 ] in case you have more than one non-NA value
> >#
> >> >per ID
> >> >> sdf2 <- lapply( sdf
> >> >>                , function( z ) {
> >> >>                   z$Value <- ifelse( is.na( z$Value )
> >> >>                                    , z$Value[ !is.na( z$Value ) ][
> >1
> >> >]
> >> >>                                    , z$Value
> >> >>                                    )
> >> >>                   z
> >> >>                  }
> >> >>                )
> >> >> df2 <- do.call( rbind, sdf2 )
> >> >> df2
> >> >> #>     ID ID_2 Firist Value
> >> >> #> a.1  a   aa   TRUE     2
> >> >> #> a.2  a   ab  FALSE     2
> >> >> #> a.3  a   ac  FALSE     2
> >> >> #> b.4  b   aa   TRUE     5
> >> >> #> b.5  b   ab  FALSE     5
> >> >>
> >> >> # or using tidyverse methods
> >> >>
> >> >> library(dplyr)
> >> >> #>
> >> >> #> Attaching package: 'dplyr'
> >> >> #> The following objects are masked from 'package:stats':
> >> >> #>
> >> >> #>     filter, lag
> >> >> #> The following objects are masked from 'package:base':
> >> >> #>
> >> >> #>     intersect, setdiff, setequal, union
> >> >> df3 <- (   df1
> >> >>         %>% group_by( ID )
> >> >>         %>% do({
> >> >>                mutate( .
> >> >>                      , Value = ifelse( is.na( Value )
> >> >>                                      , Value[ !is.na( Value ) ][ 1
> >]
> >> >>                                      , Value
> >> >>                                      )
> >> >>                      )
> >> >>             })
> >> >>         %>% ungroup
> >> >>         )
> >> >> df3
> >> >> #> # A tibble: 5 x 4
> >> >> #>   ID    ID_2  Firist Value
> >> >> #>   <chr> <chr> <lgl>  <int>
> >> >> #> 1 a     aa    T          2
> >> >> #> 2 a     ab    F          2
> >> >> #> 3 a     ac    F          2
> >> >> #> 4 b     aa    T          5
> >> >> #> 5 b     ab    F          5
> >> >> #----
> >> >>
> >> >> On Sun, 7 Jan 2018, Jeff Newmiller wrote:
> >> >>
> >> >> > Why do you want to modify df1?
> >> >> >
> >> >> > Why not just reassemble the parts as a new data frame and use
> >that
> >> >> > going forward in your calculations? That is generally the
> >preferred
> >> >> > approach in R so you can re-do your calculations easily if you
> >find
> >> >a
> >> >> > mistake later.
> >> >> > --
> >> >> > Sent from my phone. Please excuse my brevity.
> >> >> >
> >> >> > On January 7, 2018 7:35:59 PM PST, Ek Esawi <[hidden email]>
> >> >wrote:
> >> >> >> I just came up with a solution right after i posted the
> >question,
> >> >but
> >> >> >> i figured there must be a better and shorter one.than my
> >solution
> >> >> >> sdf1[[1]][1,4]<-lapplyresults[[1]]
> >> >> >> sdf1[[2]][1,4]<-lapplyresults[[2]]
> >> >> >>
> >> >> >> EK
> >> >> >>
> >> >> >> On Sun, Jan 7, 2018 at 10:13 PM, Ek Esawi <[hidden email]>
> >> >wrote:
> >> >> >>> Hi all--
> >> >> >>>
> >> >> >>> I stumbled on this problem online. I did not like the solution
> >> >given
> >> >> >>> there which was a long UDF. I thought why cannot split and l/s
> >> >apply
> >> >> >>> work here. My aim is to split the data frame, use l/sapply,
> >make
> >> >> >>> changes on the split lists and combine the split lists to new
> >> >data
> >> >> >>> frame with the desired changes/output.
> >> >> >>>
> >> >> >>> The data frame shown below has a column named ID which has 2
> >> >> >> variables
> >> >> >>> a and b; i want to replace the NAs on the Value column by 2,
> >> >which
> >> >> >>> is the only numeric entry, for ID=a and by 5 for ID=b.
> >> >> >>>
> >> >> >>> I worked out the solution but could not replace the results in
> >> >the
> >> >> >> split lists.
> >> >> >>>
> >> >> >>> Original dataframe , df1
> >> >> >>>   ID ID_2 Firist Value
> >> >> >>> 1  a   aa   TRUE     2
> >> >> >>> 2  a   ab  FALSE    NA
> >> >> >>> 3  a   ac  FALSE    NA
> >> >> >>> 4  b   aa   TRUE     5
> >> >> >>> 5  b   ab  FALSE    NA
> >> >> >>> Sdf1
> >> >> >>> $a
> >> >> >>> ID ID_2 Firist Value
> >> >> >>> 1  a   aa   TRUE     2
> >> >> >>> 2  a   ab  FALSE    NA
> >> >> >>> 3  a   ac  FALSE    NA
> >> >> >>> $b
> >> >> >>>   ID ID_2 Firist Value
> >> >> >>> 4  b   aa   TRUE     5
> >> >> >>> 5  b   ab  FALSE    NA
> >> >> >>> Desired results
> >> >> >>> ID ID_2 Firist Value
> >> >> >>> 1  a   aa   TRUE    2
> >> >> >>> 2  a   ab  FALSE    2
> >> >> >>> 3  a   ac  FALSE    2
> >> >> >>>
> >> >> >>> $b
> >> >> >>>   ID ID_2 Firist Value
> >> >> >>> 4  b   aa   TRUE     5
> >> >> >>> 5  b   ab  FALSE     5
> >> >> >>>
> >> >> >>> My code
> >> >> >>>
> >> >> >>> sdf <- split(df1,df$ID)
> >> >> >>> lapply(sdf, function(z)
> >> >> >> ifelse(is.na(z$Value),z$Value[!is.na(z$Value)],z$Value))
> >> >> >>> result:
> >> >> >>> $ a: num [1:3] 2 2 2
> >> >> >>> $ b: num [1:2] 5 5
> >> >> >>>
> >> >> >>> How could I put these two lists back in the split data frame,
> >> >sdf1?
> >> >> >>> Then I could use do.call to reassemble a data frame from the
> >> >split
> >> >> >>> lists,
> >> >> >>>
> >> >> >>> Thanks,
> >> >> >>> EK
> >> >> >>
> >> >> >> ______________________________________________
> >> >> >> [hidden email] mailing list -- To UNSUBSCRIBE and more,
> >see
> >> >> >> https://stat.ethz.ch/mailman/listinfo/r-help
> >> >> >> PLEASE do read the posting guide
> >> >> >> http://www.R-project.org/posting-guide.html
> >> >> >> and provide commented, minimal, self-contained, reproducible
> >code.
> >> >> >
> >> >> > ______________________________________________
> >> >> > [hidden email] mailing list -- To UNSUBSCRIBE and more,
> >see
> >> >> > https://stat.ethz.ch/mailman/listinfo/r-help
> >> >> > PLEASE do read the posting guide
> >> >> > http://www.R-project.org/posting-guide.html
> >> >> > and provide commented, minimal, self-contained, reproducible
> >code.
> >> >> >
> >> >>
> >> >>
> >> >-----------------------------------------------------------
> >> ----------------
> >> >> Jeff Newmiller                        The     .....       .....
> >Go
> >> >Live...
> >> >> DCN:<[hidden email]>        Basics: ##.#.       ##.#.
> >Live
> >> >Go...
> >> >>                                        Live:   OO#.. Dead: OO#..
> >> >Playing
> >> >> Research Engineer (Solar/Batteries            O.O#.       #.O#.
> >with
> >> >> /Software/Embedded Controllers)               .OO#.       .OO#.
> >> >rocks...1k
> >> >>
> >> >> ______________________________________________
> >> >> [hidden email] mailing list -- To UNSUBSCRIBE and more, see
> >> >> https://stat.ethz.ch/mailman/listinfo/r-help
> >> >> PLEASE do read the posting guide
> >> >http://www.R-project.org/posting-guide.html
> >> >> and provide commented, minimal, self-contained, reproducible code.
> >> >
> >> >________________________________
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________________________________
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- a obsahuje-li nabídku, je adresát oprávněn nabídku bezodkladně přijmout; Odesílatel tohoto e-mailu (nabídky) vylučuje přijetí nabídky ze strany příjemce s dodatkem či odchylkou.
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