Easiest way to replace non-NA values by column index - r

Question

I have a data frame like the following, with some NAs:
mydf=data.frame(ID=LETTERS[1:10], aaa=runif(10), bbb=runif(10), ccc=runif(10), ddd=runif(10))
mydf[c(1,4,5,7:10),2]=NA
mydf[c(1,2,4:8),3]=NA
mydf[c(3,4,6:10),4]=NA
mydf[c(1,3,4,6,9,10),5]=NA
> mydf
ID aaa bbb ccc ddd
1 A NA NA 0.08844614 NA
2 B 0.4912790 NA 0.88925139 0.1233173
3 C 0.1325188 0.1389260 NA NA
4 D NA NA NA NA
5 E NA NA 0.60750723 0.6357998
6 F 0.8218579 NA NA NA
7 G NA NA NA 0.5988206
8 H NA NA NA 0.4008338
9 I NA 0.8784563 NA NA
10 J NA 0.2959320 NA NA
What I want to accomplish here is the following:
1- replace non-NA values by column index -1, so that the output looks like this:
> mydf
ID aaa bbb ccc ddd
1 A NA NA 3 NA
2 B 1 NA 3 4
3 C 1 2 NA NA
4 D NA NA NA NA
5 E NA NA 3 4
6 F 1 NA NA NA
7 G NA NA NA 4
8 H NA NA NA 4
9 I NA 2 NA NA
10 J NA 2 NA NA
2- Then I would like to add an extra column that shows the following:
0 for all NAs in a row
0 for a row with more than 1 non-NA value
the actual value when it is the only non-NA value in a row
The final result should look like this:
> mydf
ID aaa bbb ccc ddd final
1 A NA NA 3 NA 3
2 B 1 NA 3 4 0
3 C 1 2 NA NA 0
4 D NA NA NA NA 0
5 E NA NA 3 4 0
6 F 1 NA NA NA 1
7 G NA NA NA 4 4
8 H NA NA NA 4 4
9 I NA 2 NA NA 2
10 J NA 2 NA NA 2
I could probably do all this with an ugly for loop, then aggregate for the final column, and substitute by 0 where appropriate...
But I was wondering if there would be a clean way to do this with some apply calls in just a few lines...
Thanks!

Answer 1

You could do:
mydf[-1] <- sapply(1:4, \(x) x * mydf[x+1]/mydf[x+1])
mydf$final <- apply(mydf[-1], 1, function(x) {
if(all(is.na(x)) | sum(!is.na(x)) > 1) 0 else na.omit(x)
})
Result:
mydf
#> ID aaa bbb ccc ddd final
#> 1 A NA NA 3 NA 3
#> 2 B 1 NA 3 4 0
#> 3 C 1 2 NA NA 0
#> 4 D NA NA NA NA 0
#> 5 E NA NA 3 4 0
#> 6 F 1 NA NA NA 1
#> 7 G NA NA NA 4 4
#> 8 H NA NA NA 4 4
#> 9 I NA 2 NA NA 2
#> 10 J NA 2 NA NA 2
Created on 2022-12-16 with reprex v2.0.2

Answer 2

Here is an idea,
mydf1 <- cbind.data.frame(ID = mydf$ID, mapply(function(x, y) replace(x, !is.na(x), y),
mydf,
seq(ncol(mydf)) - 1)[,-1])
mydf1$final <- apply(mydf1[-1], 1, \(i)
ifelse(sum(is.na(i)) == (ncol(mydf) - 1) | sum(!is.na(i)) > 1, 0, i[!is.na(i)]))
mydf1
ID aaa bbb ccc ddd final
1 A <NA> <NA> 3 <NA> 3
2 B 1 <NA> 3 4 0
3 C 1 2 <NA> <NA> 0
4 D <NA> <NA> <NA> <NA> 0
5 E <NA> <NA> 3 4 0
6 F 1 <NA> <NA> <NA> 1
7 G <NA> <NA> <NA> 4 4
8 H <NA> <NA> <NA> 4 4
9 I <NA> 2 <NA> <NA> 2
10 J <NA> 2 <NA> <NA> 2

Answer 3

A third option could be
tmp <- mydf[,-1]
tmp[!is.na(tmp)] <- 1
(mydf[,-1] <- tmp * as.list(1:4))
# aaa bbb ccc ddd
#1 NA NA 3 NA
#2 1 NA 3 4
#3 1 2 NA NA
#4 NA NA NA NA
#5 NA NA 3 4
#6 1 NA NA NA
#7 NA NA NA 4
#8 NA NA NA 4
#9 NA 2 NA NA
#10 NA 2 NA NA
The final column can be generated like this
idx <- rowSums(tmp, na.rm = TRUE) == 1
mydf$final <- idx * max.col(replace(tmp, is.na(tmp), -Inf))
Result
mydf
# ID aaa bbb ccc ddd final
#1 A NA NA 3 NA 3
#2 B 1 NA 3 4 0
#3 C 1 2 NA NA 0
#4 D NA NA NA NA 0
#5 E NA NA 3 4 0
#6 F 1 NA NA NA 1
#7 G NA NA NA 4 4
#8 H NA NA NA 4 4
#9 I NA 2 NA NA 2
#10 J NA 2 NA NA 2