Why does sapply return a matrix that I need to transpose, and then the transposed matrix will not attach to a dataframe?

Question

I would appreciate insight into why this happens and how I might do this more eloquently.

When I use sapply, I would like it to return a 3x2 matrix, but it returns a 2x3 matrix. Why is this? And why is it difficult to attach this to another data frame?

a <- data.frame(id=c('a','b','c'), var1 = c(1,2,3), var2 = c(3,2,1))
out <- sapply(a$id, function(x) out = a[x, c('var1', 'var2')])
#out is 3x2, but I would like it to be 2x3
#I then want to append t(out) (out as a 2x3 matrix) to b, a 1x3 dataframe
b <- data.frame(var3=c(0,0,0))

when I try to attach these,

b[,c('col2','col3')] <- t(out)

The error that I get is:

Warning message:
In `[<-.data.frame`(`*tmp*`, , c("col2", "col3"), value = list(1,  :
  provided 6 variables to replace 2 variables

although the following appears to give the desired result:

rownames(out) <- c('col1', 'col2')
b <- cbind(b, t(out))

I can not operate on the variables:

b$var1/b$var2

returns

Error in b$var1/b$var2 : non-numeric argument to binary operator

Thanks!

Answer 1

To expand on DWin's answer: it would help to look at the structure of your out object. It explains why b$var1/b$var2 doesn't do what you expect.

> out <- sapply(a$id, function(x) out = a[x, c('var1', 'var2')])
> str(out)  # this isn't a data.frame or a matrix...
List of 6
 $ : num 1
 $ : num 3
 $ : num 2
 $ : num 2
 $ : num 3
 $ : num 1
 - attr(*, "dim")= int [1:2] 2 3
 - attr(*, "dimnames")=List of 2
  ..$ : chr [1:2] "var1" "var2"
  ..$ : NULL

The apply family of functions are designed to work on vectors and arrays, so you need to take care when using them with data.frames (which are usually lists of vectors). You can use the fact that data.frames are lists to your advantage with lapply.

> out <- lapply(a$id, function(x) a[x, c('var1', 'var2')])  # list of data.frames
> out <- do.call(rbind, out) # data.frame
> b <- cbind(b,out)
> str(b)
'data.frame':   3 obs. of  4 variables:
 $ var3: num  0 0 0
 $ var1: num  1 2 3
 $ var2: num  3 2 1
 $ var3: num  0 0 0
> b$var1/b$var2
[1] 0.3333333 1.0000000 3.0000000