1

I am trying to use the loop over the column names of the existing dataframe and then create new columns based on one of the old column.Here is my sample data: 

 sample<-list(c(10,12,17,7,9,10),c(NA,NA,NA,10,12,13),c(1,1,1,0,0,0))
    sample<-as.data.frame(sample)
    colnames(sample)<-c("x1","x2","D")

>sample
x1  x2  D
10  NA  1
12  NA  1
17  NA  1
7   10  0
9   20  0
10  13  0

Now, I am trying to use for loop to generate two variables x1.imp and x2.imp that have values related to D=0 when D=1 and values related to D=1 when D=0(Here I actually don't need for loop but for my original dataset with large cols (variables), I really need the loop) based on the following condition:

for (i in names(sample[,1:2])){
sample$i.imp<-with (sample, ifelse (D==1, i[D==0],i[D==1]))
i=i+1
return(sample)
}


Error in i + 1 : non-numeric argument to binary operator

However, the following works, but it doesn't give the names of new cols as imp.x2 and imp.x3

for(i in sample[,1:2]){
impt.i<-with(sample,ifelse(D==1,i[D==0],i[D==1]))
i=i+1
print(as.data.frame(impt.i))
 }

impt.i
1      7
2      9
3     10
4     10
5     12
6     17
  impt.i
1     10
2     12
3     13
4     NA
5     NA
6     NA

Note that I already know the solution without loop [here]. I want with loop.

Expected output:

x1  x2  D   x1.impt x2.imp 
10  NA  1   7       10      
12  NA  1   9       20
17  NA  1   10      13
7   10  0   10      NA
9   20  0   12      NA
10  13  0   17      NA

I would greatly appreciate your valuable input in this regard.

Community
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2 Answers2

3

This is nuts, but since you are asking for it... Your code with minimum changes would be:

for (i in colnames(sample)[1:2]){
  sample[[paste0(i, '.impt')]] <- with(sample, ifelse(D==1, get(i)[D==0],get(i)[D==1]))
}

A few comments:

  1. replaced names(sample[,1:2]) with the more elegant colnames(sample)[1:2]
  2. the $ is for interactive usage. Instead, when programming, i.e. when the column name is to be interpreted, you need to use [ or [[, hence I replaced sample$i.imp with sample[[paste0(i, '.impt')]]
  3. inside with, i[D==0] will not give you x1[D==0] when i is "x1", hence the need to dereference it using get.
  4. you should not name your data.frame sample as it is also the name of a pretty common function
flodel
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1

This should work,

test <- sample[,"D"] == 1
for (.name in names(sample)[1:2]){
  newvar <- paste(.name, "impt", sep=".")
  sample[[newvar]] <- ifelse(test, sample[!test, .name], 
                                   sample[test, .name]) 
}

sample
baptiste
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