return the elements of an array of unknown dimension at the ith index on the first dimension

geti

geti <- function (x, i) {
  
  # return the elements of an array `x` of unknown dimension
  # at the ith index of its first dimension
  
  # throw an error if x isn't an array
  stopifnot(is.array(x))
 
  # get the dimension of x
  k <- length(dim(x))
  
  # build an expression as text
  expr <- paste('x[i,',
                paste(rep(',', k - 2), collapse = ' '),
                ']')
  
  # evaluate it
  ans <- eval(parse(text = expr))  
  
  # return this
  return (ans)

}

# # test it
# # make an array
# arr <- array(rnorm(3 * 4 * 2),
#              dim = c(3, 4, 2))
# # print it
# arr
# 
# # get the elements at 2 on the first dimension
# geti(arr, 2)

Eew. Nasty problem. From the look of ?"[" there is no "right" way to do this, which is weird. But I guess arrays come in small enough numbers of dimensions you could probably just work with if/else blocks.

Here's marginally less hacky version that avoids eval(parse()) (see fortunes::fortune("parse")):

geti <- function(x, i, dim=1L) {
  k <- length(dim(x))
  idx <- rep(alist(,)[1], k) # dirty hack
  idx[dim] <- i
  eval(as.call(c(list(quote(`[`), x), idx)))
}

(this generalises your solution so passing a number other than 1 in to the dim argument subsets the dim'th dimension, so geti(x, 1, 2) gets the first slice through the second dimension).

Thanks!

@BenBondLamberty pointed me towards the abind package, wherein the asub function does the trick, in the same style as your answer. So the simplest solution seems to be:

library(abind)
asub(x, i, 1)

For an array x and index i on the first dimension