Cha*_*ang 5 combinations julia
有一个数组[1, 2, ..., m],有一个整数n.
如果m=2和n=3,我想获得
[1, 1, 1]
[1, 1, 2]
[1, 2, 1]
[1, 2, 2]
[2, 1, 1]
[2, 1, 2]
[2, 2, 1]
[2, 2, 2]
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就像我一样
for i=1:m
for j=1:m
for k=1:m
\\ get [i, j, k]
end
end
end
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但是,无论是m和n是变量.我怎样才能做到这一点?我正在使用Julia,但任何一般建议都没问题.
我不清楚"我想要获得"是什么意思\\ get [i, j, k],但你可能会觉得这很有用/有趣.
julia> using Iterators
julia> collect(product(repeated(1:2,3)...))
8-element Array{(Int32,Int32,Int32),1}:
(1,1,1)
(2,1,1)
(1,2,1)
(2,2,1)
(1,1,2)
(2,1,2)
(1,2,2)
(2,2,2)
julia> A=reshape(1:8,(2,2,2))
2x2x2 Array{Int32,3}:
[:, :, 1] =
1 3
2 4
[:, :, 2] =
5 7
6 8
julia> for i in product(repeated(1:2,3)...)
@show A[i...]
end
A[i...] => 1
A[i...] => 2
A[i...] => 3
A[i...] => 4
A[i...] => 5
A[i...] => 6
A[i...] => 7
A[i...] => 8
julia> cartesianmap((k...)->println(A[k...]^2+1),tuple(repeated(2,3)...))
2
5
10
17
26
37
50
65
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甚至没有Iterators包裹......
julia> cartesianmap((k...)->println(A[k...]),tuple(repmat([2],3)...))
1
2
3
4
5
6
7
8
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