Stack with new axis last

General Usage
1

This makes a 5×2×3 array. If I want instead a 2×3×5 array, can I do it by passing ;dims= or do I need to permutedims afterwards?

julia> a = map(collect∘join, (Iterators.product((["AB", "CD"], ["rst", "uvw", "xyz"])...)))
2×3 Matrix{Vector{Char}}:
 ['A', 'B', 'r', 's', 't']  ['A', 'B', 'u', 'v', 'w']  ['A', 'B', 'x', 'y', 'z']
 ['C', 'D', 'r', 's', 't']  ['C', 'D', 'u', 'v', 'w']  ['C', 'D', 'x', 'y', 'z']

julia> stack(a)
5×2×3 Array{Char, 3}:
[:, :, 1] =
 'A'  'C'
 'B'  'D'
 'r'  'r'
 's'  's'
 't'  't'

[:, :, 2] =
 'A'  'C'
 'B'  'D'
 'u'  'u'
 'v'  'v'
 'w'  'w'

[:, :, 3] =
 'A'  'C'
 'B'  'D'
 'x'  'x'
 'y'  'y'
 'z'  'z'
2

You can use stack(a; dims=1) and then reshape for this, or a comprehension:

julia> c = stack(a; dims=1); summary(c)  # outer axes combined to 1st, like vec(a)
"6×5 Matrix{Char}"

julia> d = reshape(c, size(a)..., :); summary(d)
"2×3×5 Array{Char, 3}"

julia> d[1,1,:] == a[1,1]
true

julia> e = [a[i,j][k] for i in axes(a,1), j in axes(a,2), k in axes(a[1],1)];

julia> e == d
true

This operation would be stack(a; dims=(1,2)) if that existed. But right now, keeping multiple outer axes only works when they are last, as in stack(a).

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3

@mcabbott, what are the advantages of your proposed solution over OP’s original idea of using: permutedims(stack(a),(2,3,1)) ?

Thanks.

NB:
This post answers the question.

1 Like