# [R] if else statement in loop

Kate Ignatius kate.ignatius at gmail.com
Mon Sep 29 12:54:08 CEST 2014

```Ooops,

I edited the code wrong to make it more easier for interpretation and
got X and Y's mixed up.  Try this:

for(i in length(1:(nrow(X)))){
Y\$IID1new <- ifelse((as.character(Y[,2]) == as.character(X[,i]) &
Y\$IID1new != ''), as.character(as.matrix(X[,(nrow(X)+i)])),'')
}

The second should be like this:

Y\$IID1new <- ifelse((as.character(Y[,2]) == as.character(X[,1])),
as.character(as.matrix(X[,(nrow(X)+1)])),'')

for(i in length(2:(nrow(X)))){
ifelse((as.character(Y[,i]) == as.character(X[,i])),
Y\$IID1new[is.na(Y\$IID1new)] <-
as.character(as.matrix(X[,(nrow(X)+i)])),'')
}

The reason why I'm selecting for number of rows seems a little odd
here I know but in real life this actually relies on a third data
frame, say Z, which for simplicity I didn't include here. But I only
want to start looking at the Nth column after twice as many rows in Z.
For instance, if Z has 4 rows, I want to  take values for IID1new
starting from column 9 in X to make IID1new in Y. Does that make
sense? Will this cause a problem?

So maybe it will probably be more like this if there were a Z

for(i in length(1:(2*nrow(Z)))){
Y\$IID1new <- ifelse((as.character(Y[,2]) == as.character(X[,i]) &
Y\$IID1new != ''), as.character(as.matrix(X[,(2*nrow(Z)+i)])),'')
}

But essentially what I would like is this:

FID IID IID1new
FAM01 samas4 samas4_father
FAM01 samas5 samas5_mother
FAM01 samas6 samas6_sibling

I hope this is a little clearer...

Let me know if there are more errors.

K.

On Mon, Sep 29, 2014 at 2:39 AM, PIKAL Petr <petr.pikal at precheza.cz> wrote:
> Hi
>
> Please, be more clear in what do you want. I get many errors trying your code and your explanation does not help much.
>
>> for(i in length(1:(2*nrow(X)))){
> +     Y\$IID1new <- ifelse((as.character(Y[,2]) == as.characterXl[,i]) & X\$IID1new != '') , as.character(as.matrix(X[,(2*nrow(X)+i)])),'')
> Error: unexpected ',' in:
> "for(i in length(1:(2*nrow(X)))){
>     Y\$IID1new <- ifelse((as.character(Y[,2]) == as.characterXl[,i]) & X\$IID1new != '') ,"
>> }
> Error: unexpected '}' in "}"
>> for(i in length(1:(2*nrow(X)))){
> +     Y\$IID1new <- ifelse((as.character(Y[,2]) == as.characterXl[,i]) &
> + X\$IID1new != '') , as.character(as.matrix(X[,(2*nrow(X)+i)])),'')
> Error: unexpected ',' in:
> "    Y\$IID1new <- ifelse((as.character(Y[,2]) == as.characterXl[,i]) &
> X\$IID1new != '') ,"
>> }
>
>
> Beside, this column X\$IID1new != '' does not exist in X
>
> Here you clearly ask for nonexistent column, and why the heck you want to select column by number of rows?
>
>> as.character(as.matrix(X[,(2*nrow(X)+1)]))
> Error in `[.data.frame`(X, , (2 * nrow(X) + 1)) :
>   undefined columns selected
>
> So based on your toy data frames, what shall be the result after your computation.
>
> Regards
> Petr
>
>
>> -----Original Message-----
>> From: r-help-bounces at r-project.org [mailto:r-help-bounces at r-
>> project.org] On Behalf Of Kate Ignatius
>> Sent: Sunday, September 28, 2014 9:14 PM
>> To: r-help
>> Subject: [R] if else statement in loop
>>
>> I have two data frames
>>
>> For simplicity:
>>
>> X=
>>
>> V1 V2 V3  V4 V5 V6
>> samas4 samas5 samas6 samas4_father samas5_mother samas6_sibling
>> samas4 samas5 samas6 samas4_father samas5_mother samas6_sibling
>> samas4 samas5 samas6 samas4_father samas5_mother samas6_sibling
>>
>> Y=
>>
>> FID IID
>> FAM01 samas4
>> FAM01 samas5
>> FAM01 samas6
>>
>> I want to set to create a new IID in Y using V4 V5 V6 in X using an
>> ifelse statement in a loop.  I've used something like the following
>> (after figuring out my factor problem):
>>
>> for(i in length(1:(2*nrow(X)))){
>>     Y\$IID1new <- ifelse((as.character(Y[,2]) == as.characterXl[,i]) &
>> X\$IID1new != '') , as.character(as.matrix(X[,(2*nrow(X)+i)])),'')
>> }
>>
>> But of course this tends to overwrite.
>>
>> Is there an easy way to set up a loop to replace missing values? This
>> didn't work either but not sure if its as easy as this:
>>
>> Y\$IID1new <- ifelse((as.character(Y[,2]) == as.characterXl[,i]) &
>> X\$IID1new != '') , as.character(as.matrix(X[,(2*nrow(X)+i)])),'')
>>
>> for(i in length(2:(2*nrow(X)))){
>> ifelse((as.character(Y[,i]) == as.character(Xl[,i])),
>> X[is.na(X\$IID1new)] <- as.character(as.matrix(X[(2*nrow(X)+i)])),'')
>> }
>>
>> Thanks!
>>
>> K.
>>
>> ______________________________________________
>> R-help at r-project.org mailing list
>> https://stat.ethz.ch/mailman/listinfo/r-help
>> guide.html
>> and provide commented, minimal, self-contained, reproducible code.
>
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