两个数组的组合,在Julia中排序

Question

如果我有

a=[1,3,5,7,9]
b=[2,4,6,8,10]

我想用排序创建两个列表中长度为5的每个组合.

到目前为止,我可以通过以下各种方式获得:

ab=hcat(a,b)
collect(combinations(ab,5))

但我想只收到32个(在这种情况下)有序组合.

类似于我正在寻找的函数将是Mathematica中的元组[Transpose @ {a,b}]函数.

编辑:Mathematica输出如下

a = {1, 3, 5, 7, 9};
b = {2, 4, 6, 8, 10};
combin = Tuples[Transpose@{a, b}]
Length[combin]

Out[1]:= {{1, 3, 5, 7, 9}, {1, 3, 5, 7, 10}, {1, 3, 5, 8, 9}, {1, 3, 5, 8,
10}, {1, 3, 6, 7, 9}, {1, 3, 6, 7, 10}, {1, 3, 6, 8, 9}, {1, 3, 6,
8, 10}, {1, 4, 5, 7, 9}, {1, 4, 5, 7, 10}, {1, 4, 5, 8, 9}, {1, 4,
5, 8, 10}, {1, 4, 6, 7, 9}, {1, 4, 6, 7, 10}, {1, 4, 6, 8, 9}, {1,
4, 6, 8, 10}, {2, 3, 5, 7, 9}, {2, 3, 5, 7, 10}, {2, 3, 5, 8,
9}, {2, 3, 5, 8, 10}, {2, 3, 6, 7, 9}, {2, 3, 6, 7, 10}, {2, 3, 6,
8, 9}, {2, 3, 6, 8, 10}, {2, 4, 5, 7, 9}, {2, 4, 5, 7, 10}, {2, 4,
5, 8, 9}, {2, 4, 5, 8, 10}, {2, 4, 6, 7, 9}, {2, 4, 6, 7, 10}, {2,
4, 6, 8, 9}, {2, 4, 6, 8, 10}}

Out[2]:= 32

Answer 1

这是使用的v0.5解决方案Base.product.

同

a = [1,3,5,7,9]
b = [2,4,6,8,10]

创建元组数组

julia> vec(collect(Base.product(zip(a, b)...)))
32-element Array{Tuple{Int64,Int64,Int64,Int64,Int64},1}:
 (1,3,5,7,9) 
 (2,3,5,7,9) 
 (1,4,5,7,9) 
 (2,4,5,7,9) 
 (1,3,6,7,9) 
 (2,3,6,7,9) 
 (1,4,6,7,9) 
 (2,4,6,7,9) 
 (1,3,5,8,9) 
 (2,3,5,8,9) 
 ?           
 (2,4,6,7,10)
 (1,3,5,8,10)
 (2,3,5,8,10)
 (1,4,5,8,10)
 (2,4,5,8,10)
 (1,3,6,8,10)
 (2,3,6,8,10)
 (1,4,6,8,10)
 (2,4,6,8,10)

并将该结果收集到矩阵中

julia> hcat((collect(row) for row in ans)...)
5×32 Array{Int64,2}:
 1  2  1  2  1  2  1  2  1  2  1  2  1  …   2   1   2   1   2   1   2   1   2
 3  3  4  4  3  3  4  4  3  3  4  4  3      4   3   3   4   4   3   3   4   4
 5  5  5  5  6  6  6  6  5  5  5  5  6      6   5   5   5   5   6   6   6   6
 7  7  7  7  7  7  7  7  8  8  8  8  8      7   8   8   8   8   8   8   8   8
 9  9  9  9  9  9  9  9  9  9  9  9  9     10  10  10  10  10  10  10  10  10