Apply a function to each layer of a 3d array, returning an array

Question

Imagine you have a 3-dimensional array with rows, columns, and layers:

A <- array (1:27, c(3,3,3))

and imagine you have a function that takes a matrix as input and returns a matrix as output, like t.

How can you apply the function to each layer of the array, returning another array of the same size as the first?

I feel like I ought to be able to do it with apply somehow, but I can't.

Bonus question (I'd be very grateful if you answered this): is it faster to do this, or to make a list of each of the layer matrices and lapply the function to them?

--

Edit: please don't think that this question is answered - the answer below does not answer the question.

Answer 1

You have to think about which margin(s) over which you wish to extract the values.

you can transpose each of the 3rd dimension matrices by applying over dimensions 1 and 2 (rows and columns for want of a better word)

apply(A,1:2,t)
, , 1

     [,1] [,2] [,3]
[1,]    1    2    3
[2,]   10   11   12
[3,]   19   20   21

, , 2

     [,1] [,2] [,3]
[1,]    4    5    6
[2,]   13   14   15
[3,]   22   23   24

, , 3

     [,1] [,2] [,3]
[1,]    7    8    9
[2,]   16   17   18
[3,]   25   26   27

You can also use plyr and aaply which may act more intuitively

library(plyr)

aaply(A,3,t)
, ,  = 1


X1   1  2  3
  1  1  4  7
  2 10 13 16
  3 19 22 25

, ,  = 2


X1   1  2  3
  1  2  5  8
  2 11 14 17
  3 20 23 26

, ,  = 3


X1   1  2  3
  1  3  6  9
  2 12 15 18
  3 21 24 27

As to which is faster lapply or apply, I would think perhaps lapply would win, but you would still have to do the thinking about which margins you wanted to extract the matrices from.

I usually find it far easier to think in one dimension. Everything would be more straight forward if the earth were flat!

Answer 2

Another possibility for the transposition is function aperm.

A <- array (1:27, c(3,3,3))
# aperm permute the dimensions in the given order
# here we want to permute dimension 1 and 2
# so the new order is 2-1-3 instead of 1-2-3
aperm(A, c(2,1,3)) 
    , , 1

     [,1] [,2] [,3]
[1,]    1    2    3
[2,]    4    5    6
[3,]    7    8    9

, , 2

     [,1] [,2] [,3]
[1,]   10   11   12
[2,]   13   14   15
[3,]   16   17   18

, , 3

     [,1] [,2] [,3]
[1,]   19   20   21
[2,]   22   23   24
[3,]   25   26   27

As far as applying a function to each layer (say you want to multiply each layer by a different value) is concerned, here's another possibility:

vec <- c(1,5,3)
lapply(seq_along(vec), function(x)A[,,x]*vec[x])
[[1]]
     [,1] [,2] [,3]
[1,]    1    4    7
[2,]    2    5    8
[3,]    3    6    9

[[2]]
     [,1] [,2] [,3]
[1,]   50   65   80
[2,]   55   70   85
[3,]   60   75   90

[[3]]
     [,1] [,2] [,3]
[1,]   57   66   75
[2,]   60   69   78
[3,]   63   72   81

But as you see it creates a list, so it needs one more step using function simplify2array:

simplify2array(lapply(seq_along(vec),function(x)A[,,x]*vec[x]))
, , 1

     [,1] [,2] [,3]
[1,]    1    4    7
[2,]    2    5    8
[3,]    3    6    9

, , 2

     [,1] [,2] [,3]
[1,]   50   65   80
[2,]   55   70   85
[3,]   60   75   90

, , 3

     [,1] [,2] [,3]
[1,]   57   66   75
[2,]   60   69   78
[3,]   63   72   81

Or directly with sapply:

sapply(seq_along(vec), function(x)A[,,x]*vec[x], simplify="array")
, , 1

     [,1] [,2] [,3]
[1,]    1    4    7
[2,]    2    5    8
[3,]    3    6    9

, , 2

     [,1] [,2] [,3]
[1,]   50   65   80
[2,]   55   70   85
[3,]   60   75   90

, , 3

     [,1] [,2] [,3]
[1,]   57   66   75
[2,]   60   69   78
[3,]   63   72   81