How to get column index matrix of an array in R?

Question

Imagine I have a simple 4x3x2 array in R.

> x <- array(1:24,c(4,3,2), dimnames=list(c('a','b','c','d'),c('x','y','z'),1:2))
>  x
, , 1

  x y  z
a 1 5  9
b 2 6 10
c 3 7 11
d 4 8 12

, , 2

   x  y  z
a 13 17 21
b 14 18 22
c 15 19 23
d 16 20 24

What I'd like, is a simple function on the array that gives me back the name of the index of each element for an arbitrary dimension. In this case, dimension 2.

The function would behave like this:

> arraydims(x,2)  #Where 2 is dimension I want names for.

, , 1

     [,1] [,2] [,3]
[1,] "x"  "y"  "z" 
[2,] "x"  "y"  "z" 
[3,] "x"  "y"  "z" 
[4,] "x"  "y"  "z" 

, , 2

     [,1] [,2] [,3]
[1,] "x"  "y"  "z" 
[2,] "x"  "y"  "z" 
[3,] "x"  "y"  "z" 
[4,] "x"  "y"  "z"

Does it have to be returned as an array? `library(reshape)` `y <- subset(melt(x), X3 == 2)` — Brandon Bertelsen, Jul 21 '11 at 01:28

IRTFM · Accepted Answer · 2011-07-21T01:40:07.643

4

The function is just

colmtx <- function(x, n) {  return( array( 
                   rep(dimnames(x)[[n]], each=prod(dim(x)[0:(n-1)])), 
                   dim=dim(x) ) ) }

edited Jul 21 '11 at 01:40

answered Jul 21 '11 at 01:17

IRTFM

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+1 - Yup, that's the version of my answer that's actually correct! ;) – joran Jul 21 '11 at 01:52
I had an incorrect answer up for a few minutes, too, but deleted it when I saw I had goofed. – IRTFM Jul 21 '11 at 02:28
Thanks - was trying to avoid the route of relying on knowledge of the array's implementation, but this works as intended and extends to n-dimensional arrays. – evanrsparks Jul 21 '11 at 13:42
It shouldn't be a problem to rely on knowledge of the implementation, but if you want to avoid it for some reason, see my answer. – Aaron left Stack Overflow Jul 21 '11 at 13:49

score 0 · Answer 2 · answered Jul 21 '11 at 01:24

I think this works as intended. My coding is probably a bit dodgy at best and there may well be a more straightforward solution.

arraydims <- function(arrname,arrdim) {

#rows

if(arrdim==1) {
arrname[,,] <- dimnames(arrname)[[1]]
print(arrname)
         }

#columns

if(arrdim==2) {
arrname[,,] <- rep(dimnames(arrname)[[2]],each=dim(arrname)[1])
print(arrname)
         }

#tables   

if(arrdim==3) {
arrname[,,] <- rep(dimnames(arrname)[[3]],each=dim(arrname)[1]*dim(arrname)[2])
print(arrname)
         }

}

score 0 · Answer 3 · answered Jul 21 '11 at 11:48

Here's another way to do it. It gets the names for all dimensions at once (though that could be changed; I did it that way because the which returned all the indices anyway.

arraydims <- function(x) {
  x[] <- 1
  idx <- which(x==1, arr.ind=TRUE)
  lapply(1:ncol(x), function(k) { 
    array(dimnames(x)[[k]][idx[,k]], dim(x), dimnames=dimnames(x))
  })
}

How to get column index matrix of an array in R?

3 Answers3

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