Python乘法数组串联

Question

If I have a array of an inconsistent size, lists that consists of an (inconsistent) amount of lists:

 lists = [List1, List2, List3,...ListN]

where each contained list is of inconsistent size.

How can I concatenate the contents multiplicatively of each contained array.

Example of target output:

A = ["A","B","C","D"]
B = ["1","2","3","4"]
C = ["D","E","F","G"]
A[0-N] + B[0-N] + C[0-N]

Giving ["A1D","B1D","C1D","D1D",
        "A2D","B2D","C2D","D2D"
        "A3D","B3D","C3D","D3D"
        "A4D","B4D","C4D","D4D"
        "A1E","B1E","C1E","D1E"

         ... "C4G","D4G"  ]

For this specific example it should yield a list length of 4^3. (List length to the power of the number of lists)

However list length is not constant so it is really

List1 Length * List2 Length * List3 Length * ... * ListN Length

For an inconsistent list length:

  A = ["A","B"]
  B = ["1","2","3","4"]

 = ["A1","A2","A3","A4","B1","B2","B3","B4"]

I have tried python maps and zips, but I am having trouble doing for example:

zip(list1, list2, list3)

when:

Amount of lists is not consistent

The lists are not stored separately but collated in one large list,

and the list size is not consistent

Methods described on other SO Question only address consistent size, 2 list, situations. I am having trouble applying these techniques in this situation.

Answer 1

import itertools
A = ["A","B"]
B = ["1","2","3","4"]
list(itertools.product(A, B))

Generic

lists_var = [List1, List2, List3,...ListN]
list(itertools.product(*lists_var))

Answer 2

Use itertools to get the results, and then format as you want:

import itertools

A = ["A","B","C","D"]
B = ["1","2","3","4"]
C = ["D","E","F","G"]

lists = [A, B, C]

results = [''.join(t) for t initertools.product(*lists)]

print(results)

prints:

['A1D', 'A1E', 'A1F', 'A1G', 'A2D', 'A2E', 'A2F', 'A2G', 'A3D', 'A3E', 'A3F', 'A3G', 'A4D', 'A4E', 'A4F', 'A4G', 'B1D', 'B1E', 'B1F', 'B1G', 'B2D', 'B2E', 'B2F', 'B2G', 'B3D', 'B3E', 'B3F', 'B3G', 'B4D', 'B4E', 'B4F', 'B4G', 'C1D', 'C1E', 'C1F', 'C1G', 'C2D', 'C2E', 'C2F', 'C2G', 'C3D', 'C3E', 'C3F', 'C3G', 'C4D', 'C4E', 'C4F', 'C4G', 'D1D', 'D1E', 'D1F', 'D1G', 'D2D', 'D2E', 'D2F', 'D2G', 'D3D', 'D3E', 'D3F', 'D3G', 'D4D', 'D4E', 'D4F', 'D4G']

Answer 3

itertools.product Is what you are looking for, I think:

>>> import itertools
>>> A = ["A","B"]
>>> B = ["1","2","3","4"]
>>> C = [A,B]
>>> [''.join(i) for i in itertools.product(*C)]
['A1', 'A2', 'A3', 'A4', 'B1', 'B2', 'B3', 'B4']