

Given this data frame (a simplified, essential reproducible example)
A<c(8,7,10,1,5)
A_flag<c(10,0,1,0,2)
B<c(5,6,2,1,0)
B_flag<c(12,9,0,5,0)
mydf<data.frame(A, A_flag, B, B_flag)
# this is my initial df
mydf
I want to get to this final situation
i<which(mydf$A_flag==0)
mydf$A[i]<NA
ii<which(mydf$B_flag==0)
mydf$B[ii]<NA
# this is my final df
mydf
By considering that I have to perform this task in a data frame with many columns I’m wondering if there is a compact and effective way to get the final result with just one ‘sweep’ of the dataframe?
I was thinking to the function apply or lapply but I can not properly conceive how to…
any hint for that?
[[alternative HTML version deleted]]
______________________________________________
[hidden email] mailing list  To UNSUBSCRIBE and more, see
https://stat.ethz.ch/mailman/listinfo/rhelpPLEASE do read the posting guide http://www.Rproject.org/postingguide.htmland provide commented, minimal, selfcontained, reproducible code.


Hello,
Try the following.
icol < which(grepl("flag", names(mydf)))
mydf[icol] < lapply(mydf[icol], function(x){
is.na(x) < x == 0
x
})
mydf
# A A_flag B B_flag
#1 8 10 5 12
#2 7 NA 6 9
#3 10 1 2 NA
#4 1 NA 1 5
#5 5 2 0 NA
Hope this helps,
Rui Barradas
On 11/22/2017 10:34 AM, Massimo Bressan wrote:
>
>
> Given this data frame (a simplified, essential reproducible example)
>
>
>
>
> A<c(8,7,10,1,5)
>
> A_flag<c(10,0,1,0,2)
>
> B<c(5,6,2,1,0)
>
> B_flag<c(12,9,0,5,0)
>
>
>
>
> mydf<data.frame(A, A_flag, B, B_flag)
>
>
>
>
> # this is my initial df
>
> mydf
>
>
>
>
> I want to get to this final situation
>
>
>
>
> i<which(mydf$A_flag==0)
>
> mydf$A[i]<NA
>
>
>
>
> ii<which(mydf$B_flag==0)
>
> mydf$B[ii]<NA
>
>
>
>
> # this is my final df
>
> mydf
>
>
>
>
> By considering that I have to perform this task in a data frame with many columns I’m wondering if there is a compact and effective way to get the final result with just one ‘sweep’ of the dataframe?
>
>
>
>
> I was thinking to the function apply or lapply but I can not properly conceive how to…
>
>
>
>
> any hint for that?
>
> [[alternative HTML version deleted]]
>
> ______________________________________________
> [hidden email] mailing list  To UNSUBSCRIBE and more, see
> https://stat.ethz.ch/mailman/listinfo/rhelp> PLEASE do read the posting guide http://www.Rproject.org/postingguide.html> and provide commented, minimal, selfcontained, reproducible code.
>
______________________________________________
[hidden email] mailing list  To UNSUBSCRIBE and more, see
https://stat.ethz.ch/mailman/listinfo/rhelpPLEASE do read the posting guide http://www.Rproject.org/postingguide.htmland provide commented, minimal, selfcontained, reproducible code.


...well, I don't think this is exactly the expected result (see my post)
to be noted that the columns affected should be "A" and "B"
thanks for the help
max
 Messaggio originale 
Da: "Rui Barradas" < [hidden email]>
A: "Massimo Bressan" < [hidden email]>, "rhelp" < [hidden email]>
Inviato: Mercoledì, 22 novembre 2017 11:49:08
Oggetto: Re: [R] assign NA to rows by test on multiple columns of a data frame
Hello,
Try the following.
icol < which(grepl("flag", names(mydf)))
mydf[icol] < lapply(mydf[icol], function(x){
is.na(x) < x == 0
x
})
mydf
# A A_flag B B_flag
#1 8 10 5 12
#2 7 NA 6 9
#3 10 1 2 NA
#4 1 NA 1 5
#5 5 2 0 NA
Hope this helps,
Rui Barradas
On 11/22/2017 10:34 AM, Massimo Bressan wrote:
>
>
> Given this data frame (a simplified, essential reproducible example)
>
>
>
>
> A<c(8,7,10,1,5)
>
> A_flag<c(10,0,1,0,2)
>
> B<c(5,6,2,1,0)
>
> B_flag<c(12,9,0,5,0)
>
>
>
>
> mydf<data.frame(A, A_flag, B, B_flag)
>
>
>
>
> # this is my initial df
>
> mydf
>
>
>
>
> I want to get to this final situation
>
>
>
>
> i<which(mydf$A_flag==0)
>
> mydf$A[i]<NA
>
>
>
>
> ii<which(mydf$B_flag==0)
>
> mydf$B[ii]<NA
>
>
>
>
> # this is my final df
>
> mydf
>
>
>
>
> By considering that I have to perform this task in a data frame with many columns I’m wondering if there is a compact and effective way to get the final result with just one ‘sweep’ of the dataframe?
>
>
>
>
> I was thinking to the function apply or lapply but I can not properly conceive how to…
>
>
>
>
> any hint for that?
>
> [[alternative HTML version deleted]]
>
> ______________________________________________
> [hidden email] mailing list  To UNSUBSCRIBE and more, see
> https://stat.ethz.ch/mailman/listinfo/rhelp> PLEASE do read the posting guide http://www.Rproject.org/postingguide.html> and provide commented, minimal, selfcontained, reproducible code.
>


Massimo Bressan
ARPAV
Agenzia Regionale per la Prevenzione e
Protezione Ambientale del Veneto
Dipartimento Provinciale di Treviso
Via Santa Barbara, 5/a
31100 Treviso, Italy
tel: +39 0422 558545
fax: +39 0422 558516
email: [hidden email]
______________________________________________
[hidden email] mailing list  To UNSUBSCRIBE and more, see
https://stat.ethz.ch/mailman/listinfo/rhelpPLEASE do read the posting guide http://www.Rproject.org/postingguide.htmland provide commented, minimal, selfcontained, reproducible code.


Hello,
Sorry, I obviously read in a hurry.
icol < grepl("flag", names(mydf))
is.na(mydf[!icol]) < mydf[icol] == 0
mydf
# A A_flag B B_flag
#1 8 10 5 12
#2 NA 0 6 9
#3 10 1 NA 0
#4 NA 0 1 5
#5 5 2 NA 0
Hope this helps,
Rui Barradas
On 11/22/2017 11:19 AM, Massimo Bressan wrote:
> ...well, I don't think this is exactly the expected result (see my post)
>
> to be noted that the columns affected should be "A" and "B"
>
> thanks for the help
>
> max
>
>  Messaggio originale 
> Da: "Rui Barradas" < [hidden email]>
> A: "Massimo Bressan" < [hidden email]>, "rhelp" < [hidden email]>
> Inviato: Mercoledì, 22 novembre 2017 11:49:08
> Oggetto: Re: [R] assign NA to rows by test on multiple columns of a data frame
>
> Hello,
>
> Try the following.
>
>
> icol < which(grepl("flag", names(mydf)))
> mydf[icol] < lapply(mydf[icol], function(x){
> is.na(x) < x == 0
> x
> })
>
> mydf
> # A A_flag B B_flag
> #1 8 10 5 12
> #2 7 NA 6 9
> #3 10 1 2 NA
> #4 1 NA 1 5
> #5 5 2 0 NA
>
>
> Hope this helps,
>
> Rui Barradas
>
> On 11/22/2017 10:34 AM, Massimo Bressan wrote:
>>
>>
>> Given this data frame (a simplified, essential reproducible example)
>>
>>
>>
>>
>> A<c(8,7,10,1,5)
>>
>> A_flag<c(10,0,1,0,2)
>>
>> B<c(5,6,2,1,0)
>>
>> B_flag<c(12,9,0,5,0)
>>
>>
>>
>>
>> mydf<data.frame(A, A_flag, B, B_flag)
>>
>>
>>
>>
>> # this is my initial df
>>
>> mydf
>>
>>
>>
>>
>> I want to get to this final situation
>>
>>
>>
>>
>> i<which(mydf$A_flag==0)
>>
>> mydf$A[i]<NA
>>
>>
>>
>>
>> ii<which(mydf$B_flag==0)
>>
>> mydf$B[ii]<NA
>>
>>
>>
>>
>> # this is my final df
>>
>> mydf
>>
>>
>>
>>
>> By considering that I have to perform this task in a data frame with many columns I’m wondering if there is a compact and effective way to get the final result with just one ‘sweep’ of the dataframe?
>>
>>
>>
>>
>> I was thinking to the function apply or lapply but I can not properly conceive how to…
>>
>>
>>
>>
>> any hint for that?
>>
>> [[alternative HTML version deleted]]
>>
>> ______________________________________________
>> [hidden email] mailing list  To UNSUBSCRIBE and more, see
>> https://stat.ethz.ch/mailman/listinfo/rhelp>> PLEASE do read the posting guide http://www.Rproject.org/postingguide.html>> and provide commented, minimal, selfcontained, reproducible code.
>>
______________________________________________
[hidden email] mailing list  To UNSUBSCRIBE and more, see
https://stat.ethz.ch/mailman/listinfo/rhelpPLEASE do read the posting guide http://www.Rproject.org/postingguide.htmland provide commented, minimal, selfcontained, reproducible code.


Hi *Massimo,*
*Try this.*
*a < mydf==0mydf[a] < NAHTHEK*
On Wed, Nov 22, 2017 at 5:34 AM, Massimo Bressan <
[hidden email]> wrote:
>
>
> Given this data frame (a simplified, essential reproducible example)
>
>
>
>
> A<c(8,7,10,1,5)
>
> A_flag<c(10,0,1,0,2)
>
> B<c(5,6,2,1,0)
>
> B_flag<c(12,9,0,5,0)
>
>
>
>
> mydf<data.frame(A, A_flag, B, B_flag)
>
>
>
>
> # this is my initial df
>
> mydf
>
>
>
>
> I want to get to this final situation
>
>
>
>
> i<which(mydf$A_flag==0)
>
> mydf$A[i]<NA
>
>
>
>
> ii<which(mydf$B_flag==0)
>
> mydf$B[ii]<NA
>
>
>
>
> # this is my final df
>
> mydf
>
>
>
>
> By considering that I have to perform this task in a data frame with many
> columns I’m wondering if there is a compact and effective way to get the
> final result with just one ‘sweep’ of the dataframe?
>
>
>
>
> I was thinking to the function apply or lapply but I can not properly
> conceive how to…
>
>
>
>
> any hint for that?
>
> [[alternative HTML version deleted]]
>
> ______________________________________________
> [hidden email] mailing list  To UNSUBSCRIBE and more, see
> https://stat.ethz.ch/mailman/listinfo/rhelp> PLEASE do read the posting guide http://www.Rproject.org/> postingguide.html
> and provide commented, minimal, selfcontained, reproducible code.
[[alternative HTML version deleted]]
______________________________________________
[hidden email] mailing list  To UNSUBSCRIBE and more, see
https://stat.ethz.ch/mailman/listinfo/rhelpPLEASE do read the posting guide http://www.Rproject.org/postingguide.htmland provide commented, minimal, selfcontained, reproducible code.


OPS,
Sorry i did not read the post carfully. Mine will not work if you have
zeros on columns A and B.. But you could modify it to work for specific
columns i believe.
EK
On Wed, Nov 22, 2017 at 8:37 AM, Ek Esawi < [hidden email]> wrote:
> Hi *Massimo,*
>
> *Try this.*
>
> *a < mydf==0mydf[a] < NAHTHEK*
>
> On Wed, Nov 22, 2017 at 5:34 AM, Massimo Bressan <
> [hidden email]> wrote:
>
>>
>>
>> Given this data frame (a simplified, essential reproducible example)
>>
>>
>>
>>
>> A<c(8,7,10,1,5)
>>
>> A_flag<c(10,0,1,0,2)
>>
>> B<c(5,6,2,1,0)
>>
>> B_flag<c(12,9,0,5,0)
>>
>>
>>
>>
>> mydf<data.frame(A, A_flag, B, B_flag)
>>
>>
>>
>>
>> # this is my initial df
>>
>> mydf
>>
>>
>>
>>
>> I want to get to this final situation
>>
>>
>>
>>
>> i<which(mydf$A_flag==0)
>>
>> mydf$A[i]<NA
>>
>>
>>
>>
>> ii<which(mydf$B_flag==0)
>>
>> mydf$B[ii]<NA
>>
>>
>>
>>
>> # this is my final df
>>
>> mydf
>>
>>
>>
>>
>> By considering that I have to perform this task in a data frame with many
>> columns I’m wondering if there is a compact and effective way to get the
>> final result with just one ‘sweep’ of the dataframe?
>>
>>
>>
>>
>> I was thinking to the function apply or lapply but I can not properly
>> conceive how to…
>>
>>
>>
>>
>> any hint for that?
>>
>> [[alternative HTML version deleted]]
>>
>> ______________________________________________
>> [hidden email] mailing list  To UNSUBSCRIBE and more, see
>> https://stat.ethz.ch/mailman/listinfo/rhelp>> PLEASE do read the posting guide http://www.Rproject.org/posti>> ngguide.html
>> and provide commented, minimal, selfcontained, reproducible code.
>
>
>
[[alternative HTML version deleted]]
______________________________________________
[hidden email] mailing list  To UNSUBSCRIBE and more, see
https://stat.ethz.ch/mailman/listinfo/rhelpPLEASE do read the posting guide http://www.Rproject.org/postingguide.htmland provide commented, minimal, selfcontained, reproducible code.


Hi
I too misread the question twice and i may have mistakenly posted nontext
answer earlier. Below is a step by step solution that works provided that
your real data frame has the same structure (alternative columns as in your
example). You could combine all the steps in 2 statements.
Best of luck
EK
a < grep("_flag",colnames(mydf))
b < mydf[,a]==0
c < mydf[,a1]
c[b] < NA
mydf[,a1] < c
mydf
A A_flag B B_flag
1 8 10 5 12
2 NA 0 6 9
3 10 1 NA 0
4 NA 0 1 5
5 5 2 NA 0
On Wed, Nov 22, 2017 at 8:44 AM, Ek Esawi < [hidden email]> wrote:
> OPS,
>
> Sorry i did not read the post carfully. Mine will not work if you have
> zeros on columns A and B.. But you could modify it to work for specific
> columns i believe.
>
> EK
>
> On Wed, Nov 22, 2017 at 8:37 AM, Ek Esawi < [hidden email]> wrote:
>
>> Hi *Massimo,*
>>
>> *Try this.*
>>
>> *a < mydf==0mydf[a] < NAHTHEK*
>>
>> On Wed, Nov 22, 2017 at 5:34 AM, Massimo Bressan <
>> [hidden email]> wrote:
>>
>>>
>>>
>>> Given this data frame (a simplified, essential reproducible example)
>>>
>>>
>>>
>>>
>>> A<c(8,7,10,1,5)
>>>
>>> A_flag<c(10,0,1,0,2)
>>>
>>> B<c(5,6,2,1,0)
>>>
>>> B_flag<c(12,9,0,5,0)
>>>
>>>
>>>
>>>
>>> mydf<data.frame(A, A_flag, B, B_flag)
>>>
>>>
>>>
>>>
>>> # this is my initial df
>>>
>>> mydf
>>>
>>>
>>>
>>>
>>> I want to get to this final situation
>>>
>>>
>>>
>>>
>>> i<which(mydf$A_flag==0)
>>>
>>> mydf$A[i]<NA
>>>
>>>
>>>
>>>
>>> ii<which(mydf$B_flag==0)
>>>
>>> mydf$B[ii]<NA
>>>
>>>
>>>
>>>
>>> # this is my final df
>>>
>>> mydf
>>>
>>>
>>>
>>>
>>> By considering that I have to perform this task in a data frame with
>>> many columns I’m wondering if there is a compact and effective way to get
>>> the final result with just one ‘sweep’ of the dataframe?
>>>
>>>
>>>
>>>
>>> I was thinking to the function apply or lapply but I can not properly
>>> conceive how to…
>>>
>>>
>>>
>>>
>>> any hint for that?
>>>
>>> [[alternative HTML version deleted]]
>>>
>>> ______________________________________________
>>> [hidden email] mailing list  To UNSUBSCRIBE and more, see
>>> https://stat.ethz.ch/mailman/listinfo/rhelp>>> PLEASE do read the posting guide http://www.Rproject.org/posti>>> ngguide.html
>>> and provide commented, minimal, selfcontained, reproducible code.
>>
>>
>>
>
[[alternative HTML version deleted]]
______________________________________________
[hidden email] mailing list  To UNSUBSCRIBE and more, see
https://stat.ethz.ch/mailman/listinfo/rhelpPLEASE do read the posting guide http://www.Rproject.org/postingguide.htmland provide commented, minimal, selfcontained, reproducible code.


Do you mean like this:
mydf < within(mydf, {
is.na(A)< !A_flag
is.na(B)< !B_flag
}
)
> mydf
A A_flag B B_flag
1 8 10 5 12
2 NA 0 6 9
3 10 1 NA 0
4 NA 0 1 5
5 5 2 NA 0
Cheers,
Bert
Bert Gunter
"The trouble with having an open mind is that people keep coming along and
sticking things into it."
 Opus (aka Berkeley Breathed in his "Bloom County" comic strip )
On Wed, Nov 22, 2017 at 2:34 AM, Massimo Bressan <
[hidden email]> wrote:
>
>
> Given this data frame (a simplified, essential reproducible example)
>
>
>
>
> A<c(8,7,10,1,5)
>
> A_flag<c(10,0,1,0,2)
>
> B<c(5,6,2,1,0)
>
> B_flag<c(12,9,0,5,0)
>
>
>
>
> mydf<data.frame(A, A_flag, B, B_flag)
>
>
>
>
> # this is my initial df
>
> mydf
>
>
>
>
> I want to get to this final situation
>
>
>
>
> i<which(mydf$A_flag==0)
>
> mydf$A[i]<NA
>
>
>
>
> ii<which(mydf$B_flag==0)
>
> mydf$B[ii]<NA
>
>
>
>
> # this is my final df
>
> mydf
>
>
>
>
> By considering that I have to perform this task in a data frame with many
> columns I’m wondering if there is a compact and effective way to get the
> final result with just one ‘sweep’ of the dataframe?
>
>
>
>
> I was thinking to the function apply or lapply but I can not properly
> conceive how to…
>
>
>
>
> any hint for that?
>
> [[alternative HTML version deleted]]
>
> ______________________________________________
> [hidden email] mailing list  To UNSUBSCRIBE and more, see
> https://stat.ethz.ch/mailman/listinfo/rhelp> PLEASE do read the posting guide http://www.Rproject.org/> postingguide.html
> and provide commented, minimal, selfcontained, reproducible code.
[[alternative HTML version deleted]]
______________________________________________
[hidden email] mailing list  To UNSUBSCRIBE and more, see
https://stat.ethz.ch/mailman/listinfo/rhelpPLEASE do read the posting guide http://www.Rproject.org/postingguide.htmland provide commented, minimal, selfcontained, reproducible code.


yes, it works, even if I do not really get how and why it's working the combination of logical results (could you provide some insights for that?)
moreover, and most of all, I was hoping for a compact solution because I need to deal with MANY columns (more than 40) in data frame with the same basic structure as the simplified example I posted
thanks
m
 Messaggio originale 
Da: "Bert Gunter" < [hidden email]>
A: "Massimo Bressan" < [hidden email]>
Cc: "rhelp" < [hidden email]>
Inviato: Mercoledì, 22 novembre 2017 17:32:33
Oggetto: Re: [R] assign NA to rows by test on multiple columns of a data frame
Do you mean like this:
mydf < within(mydf, {
is.na(A)< !A_flag
is.na(B)< !B_flag
}
)
> mydf
A A_flag B B_flag
1 8 10 5 12
2 NA 0 6 9
3 10 1 NA 0
4 NA 0 1 5
5 5 2 NA 0
Cheers,
Bert
Bert Gunter
"The trouble with having an open mind is that people keep coming along and
sticking things into it."
 Opus (aka Berkeley Breathed in his "Bloom County" comic strip )
On Wed, Nov 22, 2017 at 2:34 AM, Massimo Bressan <
[hidden email]> wrote:
>
>
> Given this data frame (a simplified, essential reproducible example)
>
>
>
>
> A<c(8,7,10,1,5)
>
> A_flag<c(10,0,1,0,2)
>
> B<c(5,6,2,1,0)
>
> B_flag<c(12,9,0,5,0)
>
>
>
>
> mydf<data.frame(A, A_flag, B, B_flag)
>
>
>
>
> # this is my initial df
>
> mydf
>
>
>
>
> I want to get to this final situation
>
>
>
>
> i<which(mydf$A_flag==0)
>
> mydf$A[i]<NA
>
>
>
>
> ii<which(mydf$B_flag==0)
>
> mydf$B[ii]<NA
>
>
>
>
> # this is my final df
>
> mydf
>
>
>
>
> By considering that I have to perform this task in a data frame with many
> columns I’m wondering if there is a compact and effective way to get the
> final result with just one ‘sweep’ of the dataframe?
>
>
>
>
> I was thinking to the function apply or lapply but I can not properly
> conceive how to…
>
>
>
>
> any hint for that?
>
> [[alternative HTML version deleted]]
>
> ______________________________________________
> [hidden email] mailing list  To UNSUBSCRIBE and more, see
> https://stat.ethz.ch/mailman/listinfo/rhelp> PLEASE do read the posting guide http://www.Rproject.org/> postingguide.html
> and provide commented, minimal, selfcontained, reproducible code.

______________________________________________
[hidden email] mailing list  To UNSUBSCRIBE and more, see
https://stat.ethz.ch/mailman/listinfo/rhelpPLEASE do read the posting guide http://www.Rproject.org/postingguide.htmland provide commented, minimal, selfcontained, reproducible code.


Inline.
 Bert
On Wed, Nov 22, 2017 at 11:52 PM, Massimo Bressan <
[hidden email]> wrote:
> yes, it works, even if I do not really get how and why it's working the
> combination of logical results (could you provide some insights for that?)
>
Read the docs!
?"!"
?NA
explain
Then a more "compact" (but probably less efficient) solution just loops
over the indices:
##UNTESTED
choices < ***column numbers of _flag columns***
chng < ***column numbers of columns that will have values changed to NA***
for(i in seq_along choices) mydf[, ] < is.na(mydf[, chng[i]) < !mydf[,
choices[i]]
>
> moreover, and most of all, I was hoping for a compact solution because I
> need to deal with MANY columns (more than 40) in data frame with the same
> basic structure as the simplified example I posted
>
> thanks
>
> m
>
>
>  Messaggio originale 
> Da: "Bert Gunter" < [hidden email]>
> A: "Massimo Bressan" < [hidden email]>
> Cc: "rhelp" < [hidden email]>
> Inviato: Mercoledì, 22 novembre 2017 17:32:33
> Oggetto: Re: [R] assign NA to rows by test on multiple columns of a data
> frame
>
> Do you mean like this:
>
> mydf < within(mydf, {
> is.na(A)< !A_flag
> is.na(B)< !B_flag
> }
> )
>
> > mydf
> A A_flag B B_flag
> 1 8 10 5 12
> 2 NA 0 6 9
> 3 10 1 NA 0
> 4 NA 0 1 5
> 5 5 2 NA 0
>
>
> Cheers,
> Bert
>
>
> Bert Gunter
>
> "The trouble with having an open mind is that people keep coming along and
> sticking things into it."
>  Opus (aka Berkeley Breathed in his "Bloom County" comic strip )
>
> On Wed, Nov 22, 2017 at 2:34 AM, Massimo Bressan <
> [hidden email]> wrote:
>
> >
> >
> > Given this data frame (a simplified, essential reproducible example)
> >
> >
> >
> >
> > A<c(8,7,10,1,5)
> >
> > A_flag<c(10,0,1,0,2)
> >
> > B<c(5,6,2,1,0)
> >
> > B_flag<c(12,9,0,5,0)
> >
> >
> >
> >
> > mydf<data.frame(A, A_flag, B, B_flag)
> >
> >
> >
> >
> > # this is my initial df
> >
> > mydf
> >
> >
> >
> >
> > I want to get to this final situation
> >
> >
> >
> >
> > i<which(mydf$A_flag==0)
> >
> > mydf$A[i]<NA
> >
> >
> >
> >
> > ii<which(mydf$B_flag==0)
> >
> > mydf$B[ii]<NA
> >
> >
> >
> >
> > # this is my final df
> >
> > mydf
> >
> >
> >
> >
> > By considering that I have to perform this task in a data frame with many
> > columns I’m wondering if there is a compact and effective way to get the
> > final result with just one ‘sweep’ of the dataframe?
> >
> >
> >
> >
> > I was thinking to the function apply or lapply but I can not properly
> > conceive how to…
> >
> >
> >
> >
> > any hint for that?
> >
> > [[alternative HTML version deleted]]
> >
> > ______________________________________________
> > [hidden email] mailing list  To UNSUBSCRIBE and more, see
> > https://stat.ethz.ch/mailman/listinfo/rhelp> > PLEASE do read the posting guide http://www.Rproject.org/> > postingguide.html
> > and provide commented, minimal, selfcontained, reproducible code.
> 
>
[[alternative HTML version deleted]]
______________________________________________
[hidden email] mailing list  To UNSUBSCRIBE and more, see
https://stat.ethz.ch/mailman/listinfo/rhelpPLEASE do read the posting guide http://www.Rproject.org/postingguide.htmland provide commented, minimal, selfcontained, reproducible code.

