Python：如何获取列表集的所有有序组合

Question

let say I have the following lists:

a = [1, 2, 3]
b = [11, 12, 13]
c = [111, 112, 113]

I want to have the following output

[1,2,3]
[1,2,13]
[1,2,113]
[1,12,3]
[1, 12, 13]
[1, 12, 113]
[1,112,3]
[1,112,13]
[1, 112, 113]
[11,2,3]
[11, 2, 13]
[11, 2, 113]
[11, 12, 3]
[11, 12, 13]
[11, 12, 113]
[11, 112, 3]
[11, 112, 13]
[11, 112, 113]
...

Hence, I want to have a function that will give me all the listing combinations, which are selections of some members of a set where order is disregarded - a set of lists when each element in them keep his index from the original list.

I went over all the options in itertools and didn't find any solution.

Answer 1

Just zip your lists before using product :

from itertools import product

a = [1, 2, 3]
b = [11, 12, 13]
c = [111, 112, 113]

for p in product(*(zip(a, b, c))):
    print(p)

Output:

(1, 2, 3)
(1, 2, 13)
(1, 2, 113)
(1, 12, 3)
(1, 12, 13)
(1, 12, 113)
(1, 112, 3)
(1, 112, 13)
(1, 112, 113)
(11, 2, 3)
(11, 2, 13)
(11, 2, 113)
(11, 12, 3)
....

To answer your comment: the output of zip is:

print(list(zip(a, b, c)))
#  [(1, 11, 111), (2, 12, 112), (3, 13, 113)]

Then product will generate its output by taking one value in each of these 3 tuples.

Answer 2

 a = [1, 2, 3]
 b = [11, 12, 13]
 c = [111, 112, 113]
 d=[]
 for i in range(3):
     d.append((a[i],b[i],c[i]))
 print(d)
 for i in range(len(d)):
     for k in range(3):
         for j in range(3):
             print(d[0][i],d[1][k],d[2][j])

This will help i guess