year and week to date - before 1/1 and after 12/31

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year and week to date - before 1/1 and after 12/31

peter salzman-2
hi,

to turn year and week into the date one can do the following:

as.Date('2018 05 Sun', "%Y %W %a")

however, when we want the Sunday (1st day of week) of the 1st week of 2018
we get NA because 1/1/2018 was on Monday

as.Date('2018 00 Mon',format="%Y %U %a")
## 2018-01-01
as.Date('2018 00 Sun',format="%Y %U %a")
## NA

btw the same goes for last week
as.Date('2017 53 Sun',format="%Y %U %a")
## 2017-12-31
as.Date('2017 53 Mon',format="%Y %U %a")
## NA

So my question is :
how do i get
from "2018 00 Sun" to 2018-12-31
and
from "2017 53 Mon" to 2018-01-01

i realize i can loop over days of week and do some if/then statements,
but is there a built in function?

thank you
peter





--
Peter Salzman, PhD
Department of Biostatistics and Computational Biology
University of Rochester

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Re: year and week to date - before 1/1 and after 12/31

peter salzman-2
it is simpler than i thought

first day of given week is the last day minus 6days

in other words:
d1 = as.Date('2018 00 Sat',format="%Y %U %a") - 6
d2 = as.Date('2018 00 Sun',format="%Y %U %a")
are the same as long both are not NA

therefore to get the one that is not NA one can do

max( c(d1,d2), na.rm=TRUE )

maybe there is some other trick

best,
peter






On Tue, Oct 16, 2018 at 10:22 AM peter salzman <[hidden email]>
wrote:

> hi,
>
> to turn year and week into the date one can do the following:
>
> as.Date('2018 05 Sun', "%Y %W %a")
>
> however, when we want the Sunday (1st day of week) of the 1st week of 2018
> we get NA because 1/1/2018 was on Monday
>
> as.Date('2018 00 Mon',format="%Y %U %a")
> ## 2018-01-01
> as.Date('2018 00 Sun',format="%Y %U %a")
> ## NA
>
> btw the same goes for last week
> as.Date('2017 53 Sun',format="%Y %U %a")
> ## 2017-12-31
> as.Date('2017 53 Mon',format="%Y %U %a")
> ## NA
>
> So my question is :
> how do i get
> from "2018 00 Sun" to 2018-12-31
> and
> from "2017 53 Mon" to 2018-01-01
>
> i realize i can loop over days of week and do some if/then statements,
> but is there a built in function?
>
> thank you
> peter
>
>
>
>
>
> --
> Peter Salzman, PhD
> Department of Biostatistics and Computational Biology
> University of Rochester
>


--
Peter Salzman, PhD
Department of Biostatistics and Computational Biology
University of Rochester

        [[alternative HTML version deleted]]

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Re: year and week to date - before 1/1 and after 12/31

Jeff Newmiller
If the date in your character representation does not exist then there is no requirement for a POSIX function to give any reliable answer... including NA. Using 00 as the week number won't always work.

The first week/weekday combination that is guaranteed to exist by POSIX is 1/1 (first Monday). If the corresponding mon/mday is 1/1 then no days exist in week zero for that year and the first Sunday is 6 days more than the mday of the first Monday, else the mday of the first Sunday is one day less than the mday of the first Monday.

You should if at all possible repair the computations that are creating the invalid string dates you mention.

On October 16, 2018 8:11:12 AM PDT, peter salzman <[hidden email]> wrote:

>it is simpler than i thought
>
>first day of given week is the last day minus 6days
>
>in other words:
>d1 = as.Date('2018 00 Sat',format="%Y %U %a") - 6
>d2 = as.Date('2018 00 Sun',format="%Y %U %a")
>are the same as long both are not NA
>
>therefore to get the one that is not NA one can do
>
>max( c(d1,d2), na.rm=TRUE )
>
>maybe there is some other trick
>
>best,
>peter
>
>
>
>
>
>
>On Tue, Oct 16, 2018 at 10:22 AM peter salzman
><[hidden email]>
>wrote:
>
>> hi,
>>
>> to turn year and week into the date one can do the following:
>>
>> as.Date('2018 05 Sun', "%Y %W %a")
>>
>> however, when we want the Sunday (1st day of week) of the 1st week of
>2018
>> we get NA because 1/1/2018 was on Monday
>>
>> as.Date('2018 00 Mon',format="%Y %U %a")
>> ## 2018-01-01
>> as.Date('2018 00 Sun',format="%Y %U %a")
>> ## NA
>>
>> btw the same goes for last week
>> as.Date('2017 53 Sun',format="%Y %U %a")
>> ## 2017-12-31
>> as.Date('2017 53 Mon',format="%Y %U %a")
>> ## NA
>>
>> So my question is :
>> how do i get
>> from "2018 00 Sun" to 2018-12-31
>> and
>> from "2017 53 Mon" to 2018-01-01
>>
>> i realize i can loop over days of week and do some if/then
>statements,
>> but is there a built in function?
>>
>> thank you
>> peter
>>
>>
>>
>>
>>
>> --
>> Peter Salzman, PhD
>> Department of Biostatistics and Computational Biology
>> University of Rochester
>>

--
Sent from my phone. Please excuse my brevity.

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Re: year and week to date - before 1/1 and after 12/31

Jeff Newmiller
Er, my mistake, you are using %U not %W... but now I am really confused, because the first Sunday is trivial with %U/%u.

Can you clarify what your actual upstream input is? Is it an invalid date string as you say below, or is it year number?

On October 16, 2018 10:22:10 AM PDT, Jeff Newmiller <[hidden email]> wrote:

>If the date in your character representation does not exist then there
>is no requirement for a POSIX function to give any reliable answer...
>including NA. Using 00 as the week number won't always work.
>
>The first week/weekday combination that is guaranteed to exist by POSIX
>is 1/1 (first Monday). If the corresponding mon/mday is 1/1 then no
>days exist in week zero for that year and the first Sunday is 6 days
>more than the mday of the first Monday, else the mday of the first
>Sunday is one day less than the mday of the first Monday.
>
>You should if at all possible repair the computations that are creating
>the invalid string dates you mention.
>
>On October 16, 2018 8:11:12 AM PDT, peter salzman
><[hidden email]> wrote:
>>it is simpler than i thought
>>
>>first day of given week is the last day minus 6days
>>
>>in other words:
>>d1 = as.Date('2018 00 Sat',format="%Y %U %a") - 6
>>d2 = as.Date('2018 00 Sun',format="%Y %U %a")
>>are the same as long both are not NA
>>
>>therefore to get the one that is not NA one can do
>>
>>max( c(d1,d2), na.rm=TRUE )
>>
>>maybe there is some other trick
>>
>>best,
>>peter
>>
>>
>>
>>
>>
>>
>>On Tue, Oct 16, 2018 at 10:22 AM peter salzman
>><[hidden email]>
>>wrote:
>>
>>> hi,
>>>
>>> to turn year and week into the date one can do the following:
>>>
>>> as.Date('2018 05 Sun', "%Y %W %a")
>>>
>>> however, when we want the Sunday (1st day of week) of the 1st week
>of
>>2018
>>> we get NA because 1/1/2018 was on Monday
>>>
>>> as.Date('2018 00 Mon',format="%Y %U %a")
>>> ## 2018-01-01
>>> as.Date('2018 00 Sun',format="%Y %U %a")
>>> ## NA
>>>
>>> btw the same goes for last week
>>> as.Date('2017 53 Sun',format="%Y %U %a")
>>> ## 2017-12-31
>>> as.Date('2017 53 Mon',format="%Y %U %a")
>>> ## NA
>>>
>>> So my question is :
>>> how do i get
>>> from "2018 00 Sun" to 2018-12-31
>>> and
>>> from "2017 53 Mon" to 2018-01-01
>>>
>>> i realize i can loop over days of week and do some if/then
>>statements,
>>> but is there a built in function?
>>>
>>> thank you
>>> peter
>>>
>>>
>>>
>>>
>>>
>>> --
>>> Peter Salzman, PhD
>>> Department of Biostatistics and Computational Biology
>>> University of Rochester
>>>

--
Sent from my phone. Please excuse my brevity.

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Re: year and week to date - before 1/1 and after 12/31

peter salzman-2
hi,

thanks for replying,

it has taken some time to understand

i have year+week and i need to find the 1st day and the last day of that
week
i can decide when week starts

for example these 3 examples:
df <- data.frame(id = 1:3, year = c(2018, 2018, 2018), week=c(0,1,52))

## now run for all 3 rows:
for (kk in 1:3) {
  print(df[kk,])
  print('## version 1')
  print(as.Date(paste(df$year[kk],df$week[kk],'Sun',sep=' '), format = "%Y
%U %a")       )
  print(as.Date(paste(df$year[kk],df$week[kk],'Mon',sep=' '), format = "%Y
%U %a")       )
  print(as.Date(paste(df$year[kk],df$week[kk],'Tue',sep=' '), format = "%Y
%U %a")       )
  print(as.Date(paste(df$year[kk],df$week[kk],'Wed',sep=' '), format = "%Y
%U %a")       )
  print(as.Date(paste(df$year[kk],df$week[kk],'Thu',sep=' '), format = "%Y
%U %a")       )
  print(as.Date(paste(df$year[kk],df$week[kk],'Fri',sep=' '), format = "%Y
%U %a")       )
  print(as.Date(paste(df$year[kk],df$week[kk],'Sat',sep=' '), format = "%Y
%U %a")       )

  print('## version 2')
  print(as.Date(paste(df$year[kk],df$week[kk],'7',sep=' '), format = "%Y %U
%u")        )
  print(as.Date(paste(df$year[kk],df$week[kk],'1',sep=' '), format = "%Y %U
%u")         )
  print(as.Date(paste(df$year[kk],df$week[kk],'2',sep=' '), format = "%Y %U
%u")         )
  print(as.Date(paste(df$year[kk],df$week[kk],'3',sep=' '), format = "%Y %U
%u")         )
  print(as.Date(paste(df$year[kk],df$week[kk],'4',sep=' '), format = "%Y %U
%u")         )
  print(as.Date(paste(df$year[kk],df$week[kk],'5',sep=' '), format = "%Y %U
%u")         )
  print(as.Date(paste(df$year[kk],df$week[kk],'6',sep=' '), format = "%Y %U
%u")         )
}

for week 0 we get NA for Sunday because it was Dec 31, 2017
similarly for week 52 we get NA for Tue,Wed, ... because these are in
January of 2019

my hope was to write
as.Date(paste(year,week,1), format "%Y %month %weekday")
as.Date(paste(year,week,7), format "%Y %month %weekday")
and get the first and last day of the given week even at the beginning and
end of year
for example
as.Date("2018 0 Sun","%Y %U %a")   = '2017-12-31'

i hope this makes sense.

thanks for replying
peter










On Tue, Oct 16, 2018 at 2:11 PM Jeff Newmiller <[hidden email]>
wrote:

> Er, my mistake, you are using %U not %W... but now I am really confused,
> because the first Sunday is trivial with %U/%u.
>
> Can you clarify what your actual upstream input is? Is it an invalid date
> string as you say below, or is it year number?
>
> On October 16, 2018 10:22:10 AM PDT, Jeff Newmiller <
> [hidden email]> wrote:
> >If the date in your character representation does not exist then there
> >is no requirement for a POSIX function to give any reliable answer...
> >including NA. Using 00 as the week number won't always work.
> >
> >The first week/weekday combination that is guaranteed to exist by POSIX
> >is 1/1 (first Monday). If the corresponding mon/mday is 1/1 then no
> >days exist in week zero for that year and the first Sunday is 6 days
> >more than the mday of the first Monday, else the mday of the first
> >Sunday is one day less than the mday of the first Monday.
> >
> >You should if at all possible repair the computations that are creating
> >the invalid string dates you mention.
> >
> >On October 16, 2018 8:11:12 AM PDT, peter salzman
> ><[hidden email]> wrote:
> >>it is simpler than i thought
> >>
> >>first day of given week is the last day minus 6days
> >>
> >>in other words:
> >>d1 = as.Date('2018 00 Sat',format="%Y %U %a") - 6
> >>d2 = as.Date('2018 00 Sun',format="%Y %U %a")
> >>are the same as long both are not NA
> >>
> >>therefore to get the one that is not NA one can do
> >>
> >>max( c(d1,d2), na.rm=TRUE )
> >>
> >>maybe there is some other trick
> >>
> >>best,
> >>peter
> >>
> >>
> >>
> >>
> >>
> >>
> >>On Tue, Oct 16, 2018 at 10:22 AM peter salzman
> >><[hidden email]>
> >>wrote:
> >>
> >>> hi,
> >>>
> >>> to turn year and week into the date one can do the following:
> >>>
> >>> as.Date('2018 05 Sun', "%Y %W %a")
> >>>
> >>> however, when we want the Sunday (1st day of week) of the 1st week
> >of
> >>2018
> >>> we get NA because 1/1/2018 was on Monday
> >>>
> >>> as.Date('2018 00 Mon',format="%Y %U %a")
> >>> ## 2018-01-01
> >>> as.Date('2018 00 Sun',format="%Y %U %a")
> >>> ## NA
> >>>
> >>> btw the same goes for last week
> >>> as.Date('2017 53 Sun',format="%Y %U %a")
> >>> ## 2017-12-31
> >>> as.Date('2017 53 Mon',format="%Y %U %a")
> >>> ## NA
> >>>
> >>> So my question is :
> >>> how do i get
> >>> from "2018 00 Sun" to 2018-12-31
> >>> and
> >>> from "2017 53 Mon" to 2018-01-01
> >>>
> >>> i realize i can loop over days of week and do some if/then
> >>statements,
> >>> but is there a built in function?
> >>>
> >>> thank you
> >>> peter
> >>>
> >>>
> >>>
> >>>
> >>>
> >>> --
> >>> Peter Salzman, PhD
> >>> Department of Biostatistics and Computational Biology
> >>> University of Rochester
> >>>
>
> --
> Sent from my phone. Please excuse my brevity.
>


--
Peter Salzman, PhD
BMS
Greater Boston Area, MA

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Re: year and week to date - before 1/1 and after 12/31

Jeff Newmiller
You cannot obtain a predictable result by sending invalid time
representation data to strptime... you have to work with valid time
representations.
See sample approach below:

############################
weekEnds <- function( DF ) {
    d1_1 <- as.Date( sprintf( "%04d 1 1"
                            , DF[[ "Y" ]]
                            )
                   , format = "%Y %U %u"
                   )
    d52_7 <- as.Date( sprintf( "%04d 52 7"
                             , DF[[ "Y" ]]
                             )
                    , format = "%Y %U %u"
                    )
    week <- as.difftime( 7, units = "days" )
    day <- as.difftime( 1, units = "days" )
    d <- as.Date( sprintf( "%04d %d 1"
                         , DF[[ "Y" ]]
                         , DF[[ "wn" ]]
                         )
                , format = "%Y %U %u"
                )
    before <- 0 == DF[[ "wn" ]]
    after <- 53 == DF[[ "wn" ]]
    d[ before ] <- d1_1[ before ] - week
    d[ after ] <- d52_7[ after ] + day
    DF[[ "weekBegin" ]] <- d
    DF[[ "weekEnd" ]] <- d + week
    DF
}

tst <- expand.grid( Y = 2000:2028
                   , wn = c( 0, 1, 53 )
                   )

result <- weekEnds( tst )
set.seed( 42 )
result[ sample( nrow( result ), 5 ), ]
#>       Y wn  weekBegin    weekEnd
#> 80 2021 53 2021-12-27 2022-01-03
#> 81 2022 53 2022-12-26 2023-01-02
#> 25 2024  0 2024-01-01 2024-01-08
#> 70 2011 53 2011-12-26 2012-01-02
#> 54 2024  1 2024-01-08 2024-01-15
sum( is.na( result$weekBegin ) )
#> [1] 0
############################

On Tue, 16 Oct 2018, peter salzman wrote:

> hi,
> thanks for replying,
>
> it has taken some time to understand
>
> i have year+week and i need to find the 1st day and the last day of that week
> i can decide when week starts
>
> for example these 3 examples:
> df <- data.frame(id = 1:3, year = c(2018, 2018, 2018), week=c(0,1,52))
>
> ## now run for all 3 rows:
> for (kk in 1:3) {
>   print(df[kk,])
>   print('## version 1')
>   print(as.Date(paste(df$year[kk],df$week[kk],'Sun',sep=' '), format = "%Y %U %a")       ) 
>   print(as.Date(paste(df$year[kk],df$week[kk],'Mon',sep=' '), format = "%Y %U %a")       ) 
>   print(as.Date(paste(df$year[kk],df$week[kk],'Tue',sep=' '), format = "%Y %U %a")       ) 
>   print(as.Date(paste(df$year[kk],df$week[kk],'Wed',sep=' '), format = "%Y %U %a")       ) 
>   print(as.Date(paste(df$year[kk],df$week[kk],'Thu',sep=' '), format = "%Y %U %a")       ) 
>   print(as.Date(paste(df$year[kk],df$week[kk],'Fri',sep=' '), format = "%Y %U %a")       ) 
>   print(as.Date(paste(df$year[kk],df$week[kk],'Sat',sep=' '), format = "%Y %U %a")       )
>   
>   print('## version 2')
>   print(as.Date(paste(df$year[kk],df$week[kk],'7',sep=' '), format = "%Y %U %u")        ) 
>   print(as.Date(paste(df$year[kk],df$week[kk],'1',sep=' '), format = "%Y %U %u")         ) 
>   print(as.Date(paste(df$year[kk],df$week[kk],'2',sep=' '), format = "%Y %U %u")         ) 
>   print(as.Date(paste(df$year[kk],df$week[kk],'3',sep=' '), format = "%Y %U %u")         ) 
>   print(as.Date(paste(df$year[kk],df$week[kk],'4',sep=' '), format = "%Y %U %u")         ) 
>   print(as.Date(paste(df$year[kk],df$week[kk],'5',sep=' '), format = "%Y %U %u")         ) 
>   print(as.Date(paste(df$year[kk],df$week[kk],'6',sep=' '), format = "%Y %U %u")         )
> }
>
> for week 0 we get NA for Sunday because it was Dec 31, 2017
> similarly for week 52 we get NA for Tue,Wed, ... because these are in January of 2019
>
> my hope was to write 
> as.Date(paste(year,week,1), format "%Y %month %weekday")
> as.Date(paste(year,week,7), format "%Y %month %weekday")
> and get the first and last day of the given week even at the beginning and end of year
> for example
> as.Date("2018 0 Sun","%Y %U %a")   = '2017-12-31'
>
> i hope this makes sense.
>
> thanks for replying
> peter
>
>
>
>
>
>
>
>
>
>
> On Tue, Oct 16, 2018 at 2:11 PM Jeff Newmiller <[hidden email]> wrote:
>       Er, my mistake, you are using %U not %W... but now I am really confused, because the first Sunday is trivial with %U/%u.
>
>       Can you clarify what your actual upstream input is? Is it an invalid date string as you say below, or is it year number?
>
>       On October 16, 2018 10:22:10 AM PDT, Jeff Newmiller <[hidden email]> wrote:
>       >If the date in your character representation does not exist then there
>       >is no requirement for a POSIX function to give any reliable answer...
>       >including NA. Using 00 as the week number won't always work.
>       >
>       >The first week/weekday combination that is guaranteed to exist by POSIX
>       >is 1/1 (first Monday). If the corresponding mon/mday is 1/1 then no
>       >days exist in week zero for that year and the first Sunday is 6 days
>       >more than the mday of the first Monday, else the mday of the first
>       >Sunday is one day less than the mday of the first Monday.
>       >
>       >You should if at all possible repair the computations that are creating
>       >the invalid string dates you mention.
>       >
>       >On October 16, 2018 8:11:12 AM PDT, peter salzman
>       ><[hidden email]> wrote:
>       >>it is simpler than i thought
>       >>
>       >>first day of given week is the last day minus 6days
>       >>
>       >>in other words:
>       >>d1 = as.Date('2018 00 Sat',format="%Y %U %a") - 6
>       >>d2 = as.Date('2018 00 Sun',format="%Y %U %a")
>       >>are the same as long both are not NA
>       >>
>       >>therefore to get the one that is not NA one can do
>       >>
>       >>max( c(d1,d2), na.rm=TRUE )
>       >>
>       >>maybe there is some other trick
>       >>
>       >>best,
>       >>peter
>       >>
>       >>
>       >>
>       >>
>       >>
>       >>
>       >>On Tue, Oct 16, 2018 at 10:22 AM peter salzman
>       >><[hidden email]>
>       >>wrote:
>       >>
>       >>> hi,
>       >>>
>       >>> to turn year and week into the date one can do the following:
>       >>>
>       >>> as.Date('2018 05 Sun', "%Y %W %a")
>       >>>
>       >>> however, when we want the Sunday (1st day of week) of the 1st week
>       >of
>       >>2018
>       >>> we get NA because 1/1/2018 was on Monday
>       >>>
>       >>> as.Date('2018 00 Mon',format="%Y %U %a")
>       >>> ## 2018-01-01
>       >>> as.Date('2018 00 Sun',format="%Y %U %a")
>       >>> ## NA
>       >>>
>       >>> btw the same goes for last week
>       >>> as.Date('2017 53 Sun',format="%Y %U %a")
>       >>> ## 2017-12-31
>       >>> as.Date('2017 53 Mon',format="%Y %U %a")
>       >>> ## NA
>       >>>
>       >>> So my question is :
>       >>> how do i get
>       >>> from "2018 00 Sun" to 2018-12-31
>       >>> and
>       >>> from "2017 53 Mon" to 2018-01-01
>       >>>
>       >>> i realize i can loop over days of week and do some if/then
>       >>statements,
>       >>> but is there a built in function?
>       >>>
>       >>> thank you
>       >>> peter
>       >>>
>       >>>
>       >>>
>       >>>
>       >>>
>       >>> --
>       >>> Peter Salzman, PhD
>       >>> Department of Biostatistics and Computational Biology
>       >>> University of Rochester
>       >>>
>
>       --
>       Sent from my phone. Please excuse my brevity.
>
>
>
> --
> Peter Salzman, PhD
> BMS
> Greater Boston Area, MA
>
>

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Re: year and week to date - before 1/1 and after 12/31

Gabor Grothendieck
In reply to this post by peter salzman-2
Replace the week in the date with week 2, say -- a week in which
nothing will go wrong and then add or subtract the appropriate number
of weeks.

  d <- c('2016 00 Sun', '2017 53 Sun', '2017 53 Mon') # test data

  as.Date(sub(" .. ", "02", d), "%Y %U %a") +
     7 * (as.numeric(sub(".... (..) ...", "\\1", d)) - 2)
  ## [1] "2015-12-27" "2017-12-31" "2018-01-01"

On Tue, Oct 16, 2018 at 10:23 AM peter salzman
<[hidden email]> wrote:

>
> hi,
>
> to turn year and week into the date one can do the following:
>
> as.Date('2018 05 Sun', "%Y %W %a")
>
> however, when we want the Sunday (1st day of week) of the 1st week of 2018
> we get NA because 1/1/2018 was on Monday
>
> as.Date('2018 00 Mon',format="%Y %U %a")
> ## 2018-01-01
> as.Date('2018 00 Sun',format="%Y %U %a")
> ## NA
>
> btw the same goes for last week
> as.Date('2017 53 Sun',format="%Y %U %a")
> ## 2017-12-31
> as.Date('2017 53 Mon',format="%Y %U %a")
> ## NA
>
> So my question is :
> how do i get
> from "2018 00 Sun" to 2018-12-31
> and
> from "2017 53 Mon" to 2018-01-01
>
> i realize i can loop over days of week and do some if/then statements,
> but is there a built in function?
>
> thank you
> peter
>
>
>
>
>
> --
> Peter Salzman, PhD
> Department of Biostatistics and Computational Biology
> University of Rochester
>
>         [[alternative HTML version deleted]]
>
> ______________________________________________
> [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
> and provide commented, minimal, self-contained, reproducible code.



--
Statistics & Software Consulting
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email: ggrothendieck at gmail.com

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Re: year and week to date - before 1/1 and after 12/31

peter salzman-2
In reply to this post by Jeff Newmiller
thanks Jeff and Gabor,
appreciate you spending time on this,

you both use similar ideas - add/subtract days/weeks from a day that exists

combining all i learned i would go with something like this:

## step 1) find 1st sunday of year Y
d11 <- as.Date( sprintf( "%04d 1 1"
                            , DF[[ "Y" ]]
                            )
                   , format = "%Y %U %u"
                   )
## btw note that when Jan 1 is on sunday then d11 will be Jan 8th
## step 2) for start of week just add the number of weeks that is needed
start <- d11 + (w-1)*week
## step 3) for end of week add number of weeks and subtract 1 day
start <- d11 + w*week - day

thanks
peter






On Wed, Oct 17, 2018 at 11:15 PM Jeff Newmiller
<[hidden email]> wrote:

>
> You cannot obtain a predictable result by sending invalid time
> representation data to strptime... you have to work with valid time
> representations.
> See sample approach below:
>
> ############################
> weekEnds <- function( DF ) {
>     d1_1 <- as.Date( sprintf( "%04d 1 1"
>                             , DF[[ "Y" ]]
>                             )
>                    , format = "%Y %U %u"
>                    )
>     d52_7 <- as.Date( sprintf( "%04d 52 7"
>                              , DF[[ "Y" ]]
>                              )
>                     , format = "%Y %U %u"
>                     )
>     week <- as.difftime( 7, units = "days" )
>     day <- as.difftime( 1, units = "days" )
>     d <- as.Date( sprintf( "%04d %d 1"
>                          , DF[[ "Y" ]]
>                          , DF[[ "wn" ]]
>                          )
>                 , format = "%Y %U %u"
>                 )
>     before <- 0 == DF[[ "wn" ]]
>     after <- 53 == DF[[ "wn" ]]
>     d[ before ] <- d1_1[ before ] - week
>     d[ after ] <- d52_7[ after ] + day
>     DF[[ "weekBegin" ]] <- d
>     DF[[ "weekEnd" ]] <- d + week
>     DF
> }
>
> tst <- expand.grid( Y = 2000:2028
>                    , wn = c( 0, 1, 53 )
>                    )
>
> result <- weekEnds( tst )
> set.seed( 42 )
> result[ sample( nrow( result ), 5 ), ]
> #>       Y wn  weekBegin    weekEnd
> #> 80 2021 53 2021-12-27 2022-01-03
> #> 81 2022 53 2022-12-26 2023-01-02
> #> 25 2024  0 2024-01-01 2024-01-08
> #> 70 2011 53 2011-12-26 2012-01-02
> #> 54 2024  1 2024-01-08 2024-01-15
> sum( is.na( result$weekBegin ) )
> #> [1] 0
> ############################
>
> On Tue, 16 Oct 2018, peter salzman wrote:
>
> > hi,
> > thanks for replying,
> >
> > it has taken some time to understand
> >
> > i have year+week and i need to find the 1st day and the last day of that week
> > i can decide when week starts
> >
> > for example these 3 examples:
> > df <- data.frame(id = 1:3, year = c(2018, 2018, 2018), week=c(0,1,52))
> >
> > ## now run for all 3 rows:
> > for (kk in 1:3) {
> >   print(df[kk,])
> >   print('## version 1')
> >   print(as.Date(paste(df$year[kk],df$week[kk],'Sun',sep=' '), format = "%Y %U %a")       )
> >   print(as.Date(paste(df$year[kk],df$week[kk],'Mon',sep=' '), format = "%Y %U %a")       )
> >   print(as.Date(paste(df$year[kk],df$week[kk],'Tue',sep=' '), format = "%Y %U %a")       )
> >   print(as.Date(paste(df$year[kk],df$week[kk],'Wed',sep=' '), format = "%Y %U %a")       )
> >   print(as.Date(paste(df$year[kk],df$week[kk],'Thu',sep=' '), format = "%Y %U %a")       )
> >   print(as.Date(paste(df$year[kk],df$week[kk],'Fri',sep=' '), format = "%Y %U %a")       )
> >   print(as.Date(paste(df$year[kk],df$week[kk],'Sat',sep=' '), format = "%Y %U %a")       )
> >
> >   print('## version 2')
> >   print(as.Date(paste(df$year[kk],df$week[kk],'7',sep=' '), format = "%Y %U %u")        )
> >   print(as.Date(paste(df$year[kk],df$week[kk],'1',sep=' '), format = "%Y %U %u")         )
> >   print(as.Date(paste(df$year[kk],df$week[kk],'2',sep=' '), format = "%Y %U %u")         )
> >   print(as.Date(paste(df$year[kk],df$week[kk],'3',sep=' '), format = "%Y %U %u")         )
> >   print(as.Date(paste(df$year[kk],df$week[kk],'4',sep=' '), format = "%Y %U %u")         )
> >   print(as.Date(paste(df$year[kk],df$week[kk],'5',sep=' '), format = "%Y %U %u")         )
> >   print(as.Date(paste(df$year[kk],df$week[kk],'6',sep=' '), format = "%Y %U %u")         )
> > }
> >
> > for week 0 we get NA for Sunday because it was Dec 31, 2017
> > similarly for week 52 we get NA for Tue,Wed, ... because these are in January of 2019
> >
> > my hope was to write
> > as.Date(paste(year,week,1), format "%Y %month %weekday")
> > as.Date(paste(year,week,7), format "%Y %month %weekday")
> > and get the first and last day of the given week even at the beginning and end of year
> > for example
> > as.Date("2018 0 Sun","%Y %U %a")   = '2017-12-31'
> >
> > i hope this makes sense.
> >
> > thanks for replying
> > peter
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> > On Tue, Oct 16, 2018 at 2:11 PM Jeff Newmiller <[hidden email]> wrote:
> >       Er, my mistake, you are using %U not %W... but now I am really confused, because the first Sunday is trivial with %U/%u.
> >
> >       Can you clarify what your actual upstream input is? Is it an invalid date string as you say below, or is it year number?
> >
> >       On October 16, 2018 10:22:10 AM PDT, Jeff Newmiller <[hidden email]> wrote:
> >       >If the date in your character representation does not exist then there
> >       >is no requirement for a POSIX function to give any reliable answer...
> >       >including NA. Using 00 as the week number won't always work.
> >       >
> >       >The first week/weekday combination that is guaranteed to exist by POSIX
> >       >is 1/1 (first Monday). If the corresponding mon/mday is 1/1 then no
> >       >days exist in week zero for that year and the first Sunday is 6 days
> >       >more than the mday of the first Monday, else the mday of the first
> >       >Sunday is one day less than the mday of the first Monday.
> >       >
> >       >You should if at all possible repair the computations that are creating
> >       >the invalid string dates you mention.
> >       >
> >       >On October 16, 2018 8:11:12 AM PDT, peter salzman
> >       ><[hidden email]> wrote:
> >       >>it is simpler than i thought
> >       >>
> >       >>first day of given week is the last day minus 6days
> >       >>
> >       >>in other words:
> >       >>d1 = as.Date('2018 00 Sat',format="%Y %U %a") - 6
> >       >>d2 = as.Date('2018 00 Sun',format="%Y %U %a")
> >       >>are the same as long both are not NA
> >       >>
> >       >>therefore to get the one that is not NA one can do
> >       >>
> >       >>max( c(d1,d2), na.rm=TRUE )
> >       >>
> >       >>maybe there is some other trick
> >       >>
> >       >>best,
> >       >>peter
> >       >>
> >       >>
> >       >>
> >       >>
> >       >>
> >       >>
> >       >>On Tue, Oct 16, 2018 at 10:22 AM peter salzman
> >       >><[hidden email]>
> >       >>wrote:
> >       >>
> >       >>> hi,
> >       >>>
> >       >>> to turn year and week into the date one can do the following:
> >       >>>
> >       >>> as.Date('2018 05 Sun', "%Y %W %a")
> >       >>>
> >       >>> however, when we want the Sunday (1st day of week) of the 1st week
> >       >of
> >       >>2018
> >       >>> we get NA because 1/1/2018 was on Monday
> >       >>>
> >       >>> as.Date('2018 00 Mon',format="%Y %U %a")
> >       >>> ## 2018-01-01
> >       >>> as.Date('2018 00 Sun',format="%Y %U %a")
> >       >>> ## NA
> >       >>>
> >       >>> btw the same goes for last week
> >       >>> as.Date('2017 53 Sun',format="%Y %U %a")
> >       >>> ## 2017-12-31
> >       >>> as.Date('2017 53 Mon',format="%Y %U %a")
> >       >>> ## NA
> >       >>>
> >       >>> So my question is :
> >       >>> how do i get
> >       >>> from "2018 00 Sun" to 2018-12-31
> >       >>> and
> >       >>> from "2017 53 Mon" to 2018-01-01
> >       >>>
> >       >>> i realize i can loop over days of week and do some if/then
> >       >>statements,
> >       >>> but is there a built in function?
> >       >>>
> >       >>> thank you
> >       >>> peter
> >       >>>
> >       >>>
> >       >>>
> >       >>>
> >       >>>
> >       >>> --
> >       >>> Peter Salzman, PhD
> >       >>> Department of Biostatistics and Computational Biology
> >       >>> University of Rochester
> >       >>>
> >
> >       --
> >       Sent from my phone. Please excuse my brevity.
> >
> >
> >
> > --
> > Peter Salzman, PhD
> > BMS
> > Greater Boston Area, MA
> >
> >
>
> ---------------------------------------------------------------------------
> 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
> ---------------------------------------------------------------------------



--
Peter Salzman, PhD
BMS
Greater Boston Area, MA

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