Access (and fill) different rows in array - r

Question

Let's say I have a 3-dimensional array:
a <- array(dim = c(3, 2, 3))
and a vector with indices for the 1st dimension:
ind <- c(1,2,3)
Now I want to put a number (e.g. 1) into the rows (ind) across 2nd and 2rd dimension.
a[ind,,] <- matrix(1, ncol = 2, nrow = length(ind))
clearly does not work! A loop would work, however is there a better solution?
The result should be:
, , 1
[,1] [,2]
[1,] 1 1
[2,] NA NA
[3,] NA NA
, , 2
[,1] [,2]
[1,] NA NA
[2,] 1 1
[3,] NA NA
, , 3
[,1] [,2]
[1,] NA NA
[2,] NA NA
[3,] 1 1

Answer 1

Using mapply.
mapply(function(x, y) {a[x,,y] <<- 1;a}, ind, ind)
a
# , , 1
#
# [,1] [,2]
# [1,] 1 1
# [2,] NA NA
# [3,] NA NA
#
# , , 2
#
# [,1] [,2]
# [1,] NA NA
# [2,] 1 1
# [3,] NA NA
#
# , , 3
#
# [,1] [,2]
# [1,] NA NA
# [2,] NA NA
# [3,] 1 1
Or, for this specific case where x and y is the same, just:
sapply(ind, function(x) {a[x,,x] <<- 1;a})
Note: If you find the console output unnecessary wrap an invisible() around the code.

Answer 2

Another alternative that replaces everything in one step. It takes advantage of ['s operations when its argument is a matrix, each column indexing on a dimension of the array.
To confirm that it is putting values where we intend, I'll modify the replacement matrix to be 1:6.
# original data
a <- array(dim = c(3, 2, 3))
# the replacement matrix
r <- matrix(1:6, ncol = 2, nrow = length(ind))
# this will be what we index `a` on
m <- expand.grid(ind, seq_len(dim(a)[2]))
m$Var3 <- m$Var1 # repeat `ind` in the third dimension
m
# Var1 Var2 Var3
# 1 1 1 1
# 2 2 1 2
# 3 3 1 3
# 4 1 2 1
# 5 2 2 2
# 6 3 2 3
And the replacement:
a[as.matrix(m)] <- r
a
# , , 1
# [,1] [,2]
# [1,] 1 4
# [2,] NA NA
# [3,] NA NA
# , , 2
# [,1] [,2]
# [1,] NA NA
# [2,] 2 5
# [3,] NA NA
# , , 3
# [,1] [,2]
# [1,] NA NA
# [2,] NA NA
# [3,] 3 6