tuples - 如何在 Julia 中编写一组元素的外积？

Question

一组元素的排列

我有一个 Julia 片段，它生成所有可能的 4 元组，每个槽中有 3 个可能的元素:

julia> collect(Iterators.product(ntuple(_ -> 1:3, 4)...))
3×3×3×3 Array{NTuple{4, Int64}, 4}:
[:, :, 1, 1] =
 (1, 1, 1, 1)  (1, 2, 1, 1)  (1, 3, 1, 1)
 (2, 1, 1, 1)  (2, 2, 1, 1)  (2, 3, 1, 1)
 (3, 1, 1, 1)  (3, 2, 1, 1)  (3, 3, 1, 1)

[:, :, 2, 1] =
 (1, 1, 2, 1)  (1, 2, 2, 1)  (1, 3, 2, 1)
 (2, 1, 2, 1)  (2, 2, 2, 1)  (2, 3, 2, 1)
 (3, 1, 2, 1)  (3, 2, 2, 1)  (3, 3, 2, 1)

[:, :, 3, 1] =
 (1, 1, 3, 1)  (1, 2, 3, 1)  (1, 3, 3, 1)
 (2, 1, 3, 1)  (2, 2, 3, 1)  (2, 3, 3, 1)
 (3, 1, 3, 1)  (3, 2, 3, 1)  (3, 3, 3, 1)

[:, :, 1, 2] =
 (1, 1, 1, 2)  (1, 2, 1, 2)  (1, 3, 1, 2)
 (2, 1, 1, 2)  (2, 2, 1, 2)  (2, 3, 1, 2)
 (3, 1, 1, 2)  (3, 2, 1, 2)  (3, 3, 1, 2)
...

问题

如何修改此代码，而不是对元素使用范围 1:3，而是从数组中选择元素，比如 [1,5]，或者一组 Set([1,5])，只有 2 种可能性？

Answer 1

我想这就是您要找的:

julia> original = collect(Iterators.product(ntuple(_ -> [1,5], 4)...))
2×2×2×2 Array{NTuple{4, Int64}, 4}:
[:, :, 1, 1] =
 (1, 1, 1, 1)  (1, 5, 1, 1)
 (5, 1, 1, 1)  (5, 5, 1, 1)

[:, :, 2, 1] =
 (1, 1, 5, 1)  (1, 5, 5, 1)
 (5, 1, 5, 1)  (5, 5, 5, 1)

[:, :, 1, 2] =
 (1, 1, 1, 5)  (1, 5, 1, 5)
 (5, 1, 1, 5)  (5, 5, 1, 5)

[:, :, 2, 2] =
 (1, 1, 5, 5)  (1, 5, 5, 5)
 (5, 1, 5, 5)  (5, 5, 5, 5)

julia> flattened = [original...]
16-element Vector{NTuple{4, Int64}}:
 (1, 1, 1, 1)
 (5, 1, 1, 1)
 (1, 5, 1, 1)
 (5, 5, 1, 1)
 (1, 1, 5, 1)
 (5, 1, 5, 1)
 (1, 5, 5, 1)
 (5, 5, 5, 1)
 (1, 1, 1, 5)
 (5, 1, 1, 5)
 (1, 5, 1, 5)
 (5, 5, 1, 5)
 (1, 1, 5, 5)
 (5, 1, 5, 5)
 (1, 5, 5, 5)
 (5, 5, 5, 5)

因此，可以简单地用数组或集合代替迭代器。之后，您可以flatten the resulting matrix与 splat operator .

此外，这里是如何使用 list comprehension 以 A[i,j] 形式访问元素的:

julia> original  =  collect(Iterators.product(ntuple(_ -> 1:3, 4)...));
julia> newMatrix = [original[:,:,a,b] for a in 1:3, b in 1:3];

julia> newMatrix[2,3]
3×3 Matrix{NTuple{4, Int64}}:
 (1, 1, 2, 3)  (1, 2, 2, 3)  (1, 3, 2, 3)
 (2, 1, 2, 3)  (2, 2, 2, 3)  (2, 3, 2, 3)
 (3, 1, 2, 3)  (3, 2, 2, 3)  (3, 3, 2, 3)

julia> typeof(newMatrix)
Matrix{Matrix{NTuple{4, Int64}}} (alias for Array{Array{NTuple{4, Int64}, 2}, 2})