# Functional way to stack list of 2d matrices into 3d matrix

After a clever lapply, I'm left with a list of 2-dimensional matrices.

For example:

set.seed(1) test <- replicate( 5, matrix(runif(25),ncol=5), simplify=FALSE ) > test [[1]] [,1] [,2] [,3] [,4] [,5] [1,] 0.8357088 0.29589546 0.9994045 0.2862853 0.6973738 [2,] 0.2377494 0.14704832 0.0348748 0.7377974 0.6414624 [3,] 0.3539861 0.70399206 0.3383913 0.8340543 0.6439229 [4,] 0.8568854 0.10380669 0.9150638 0.3142708 0.9778534 [5,] 0.8537634 0.03372777 0.6172353 0.4925665 0.4147353 [[2]] [,1] [,2] [,3] [,4] [,5] [1,] 0.1194048 0.9833502 0.9674695 0.6687715 0.1928159 [2,] 0.5260297 0.3883191 0.5150718 0.4189159 0.8967387 [3,] 0.2250734 0.2292448 0.1630703 0.3233450 0.3081196 [4,] 0.4864118 0.6232975 0.6219023 0.8352553 0.3633005 [5,] 0.3702148 0.1365402 0.9859542 0.1438170 0.7839465 [[3]] ...

I'd like to turn that into a 3-dimensional array:

set.seed(1) replicate( 5, matrix(runif(25),ncol=5) )

Obviously, if I'm using replicate I can just turn on simplify, but sapply does not simplify the result properly, and stack fails utterly. do.call(rbind,mylist) turns it into a 2d matrix rather than 3d array.

I can do this with a loop, but I'm looking for a neat and functional way to handle it.

The closest way I've come up with is:

array( do.call( c, test ), dim=c(dim(test[[1]]),length(test)) )

But I feel like that's inelegant (because it disassembles and then reassembles the array attributes of the vectors, and needs a lot of testing to make safe (e.g. that the dimensions of each element are the same).

## Answers

You can use the abind package and then use do.call(abind, c(test, along = 3))

library(abind) testArray <- do.call(abind, c(test, along = 3))

Or you could use simplify = 'array' in a call to sapply, (instead of lapply). simplify = 'array' is not the same as simplify = TRUE, as it will change the argument higher in simplify2array

eg

foo <- function(x) matrix(1:10, ncol = 5) # the default is simplify = TRUE sapply(1:5, foo) [,1] [,2] [,3] [,4] [,5] [1,] 1 1 1 1 1 [2,] 2 2 2 2 2 [3,] 3 3 3 3 3 [4,] 4 4 4 4 4 [5,] 5 5 5 5 5 [6,] 6 6 6 6 6 [7,] 7 7 7 7 7 [8,] 8 8 8 8 8 [9,] 9 9 9 9 9 [10,] 10 10 10 10 10 # which is *not* what you want # so set `simplify = 'array' sapply(1:5, foo, simplify = 'array') , , 1 [,1] [,2] [,3] [,4] [,5] [1,] 1 3 5 7 9 [2,] 2 4 6 8 10 , , 2 [,1] [,2] [,3] [,4] [,5] [1,] 1 3 5 7 9 [2,] 2 4 6 8 10 , , 3 [,1] [,2] [,3] [,4] [,5] [1,] 1 3 5 7 9 [2,] 2 4 6 8 10 , , 4 [,1] [,2] [,3] [,4] [,5] [1,] 1 3 5 7 9 [2,] 2 4 6 8 10 , , 5 [,1] [,2] [,3] [,4] [,5] [1,] 1 3 5 7 9 [2,] 2 4 6 8 10

Try this:

simplify2array(test)

test2 <- unlist(test) dim(test2) <- c(dim(test[[1]]),5)

or if you do not know the expected size ahead of time:

dim3 <- c(dim(test[[1]]), length(test2)/prod(dim(test[[1]]))) dim(test2) <- dim3

An array is simply an atomic vector with dimensions. Each of the matrix components of test is really just a vector with dimensions too. Hence the simplest solution I can think of is to unroll the list test into a vector and convert that to an array using array and suitably supplied dimensions.

set.seed(1) foo <- replicate( 5, matrix(runif(25),ncol=5) ) tmp <- array(unlist(test), dim = c(5,5,5)) > all.equal(foo, tmp) [1] TRUE > is.array(tmp) [1] TRUE > dim(tmp) [1] 5 5 5

If you don't want to hardcode the dimensions, we have to make some assumptions but can easily fill in the dimension from test, e.g.

tmp2 <- array(unlist(test), dim = c(dim(test[[1]]), length(test))) > all.equal(foo, tmp2) [1] TRUE

This assumes that the dimensions of each component are all the same, but then I don't see how you could put sub-matrices into a 3-d array if that condition doesn't hold.

This may seem hacky, to unroll the list, but this is simply exploiting how R handles matrices and arrays as vectors with dimensions.