Generate all combinations of items with two values in Julia? - julia

Question

I have m items. Each item is a pair of two values. For example, for m=4, I have the matrix:
julia> valid_pairs = [0 1;
1 2;
1 2;
2 3];
I would like to generate all combinations of the four items where each item i can take only the values in valid_pairs[i, :]. Based on the previous example, I would like to have:
julia> all_combs
4x16 Array{Int,2}
0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1
1 1 1 1 2 2 2 2 1 1 1 1 2 2 2 2
1 1 2 2 1 1 2 2 1 1 2 2 1 1 2 2
2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3
I feel like this can be done easily using Combinatorics.jl.
Though I used Combinatorics.jl, what I did was the following:
using Combinatorics
m = 4
combs = combinations(1:m) |> collect
L = length(combs)
all_combs = zeros(Int, m, L+1)
for j in 1:L
for i in 1:m
if !in(i, combs[j])
all_combs[i, j] = valid_pairs[i, 1]
else
all_combs[i, j] = valid_pairs[i, 2]
end
end
end
all_combs[:, end] = valid_pairs[:, 1]

Answer 1

Not the same order, but
julia> [collect(x) for x in Iterators.product(eachrow(valid_pairs)...)]
2×2×2×2 Array{Array{Int64,1},4}:
[:, :, 1, 1] =
[0, 1, 1, 2] [0, 2, 1, 2]
[1, 1, 1, 2] [1, 2, 1, 2]
[:, :, 2, 1] =
[0, 1, 2, 2] [0, 2, 2, 2]
[1, 1, 2, 2] [1, 2, 2, 2]
[:, :, 1, 2] =
[0, 1, 1, 3] [0, 2, 1, 3]
[1, 1, 1, 3] [1, 2, 1, 3]
[:, :, 2, 2] =
[0, 1, 2, 3] [0, 2, 2, 3]
[1, 1, 2, 3] [1, 2, 2, 3]
should do. If you really want a matrix (2D array), then you can hcat the previous answer, or directly do
julia> reduce(hcat, collect(x) for x in Iterators.product(eachrow(valid_pairs)...))
4×16 Array{Int64,2}:
0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1
1 1 2 2 1 1 2 2 1 1 2 2 1 1 2 2
1 1 1 1 2 2 2 2 1 1 1 1 2 2 2 2
2 2 2 2 2 2 2 2 3 3 3 3 3 3 3 3
EDIT: side note, I would define the pairs as tuples to clarify what's happening, so something like
valid_pairs = [(0,1), (1,2), (1,2), (2,3)]
and I would not create the 2D (or 4D, or m-D) array, but, instead, do
comb_pairs = Iterators.product(valid_pairs...)
which then gives you a lazy version of all the pair combinations, so that you can iterate on it without actually creating it first, which should be more efficient (and looks cleaner) I think.