遍历 for 循环的所有列表的列表的所有组合

Question

I am working on a program and I am having a bit of difficulty trying to figure about to make the for loop do what I would like.

As of right now, the for statement goes through every single combination of all lists and makes 5 choices, that is right, I want 5 choices.

However, it creates a lot of overlap, since I am adding all the lists together and picking 5 items. I do get what I want, but along with many combinations I do not, since all I am doing is adding all 5 lists together, I want them to be picked independently.

I would like the for statement to only go through each list row by row, and pick all combinations, with no overlap.

from itertools import combinations

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def playerpick():
    P1 = [1,2,3,4,7]
    P2 = [1,8,13,14,17,29]
    P3 = [1,2,3]
    P4 = [1,7,8,12,15]
    P5 = [1,2,3,4]
    all_lists = P1 + P2 + P3 + P4 + P5
    for (a,b,c,d,e) in combinations(all_lists, 5):
        pick1 = a
        pick2 = b
        pick3 = c
        pick4 = d
        pick5 = e

playerpick()

The output I want is something like this, pull each item out from each list, one at a time:

output 1: [1,1,1,1,1]
output 2: [2,1,1,1,1]
output 3: [3,1,1,1,1]
output 4: [4,1,1,1,1]
output 5: [7,1,1,1,1]
output 6: [1,8,1,1,1](next combination)
output 7: [2,8,1,1,1]
output 8: [3,8,1,1,1]
output 9: [4,8,1,1,1]
output 10: [7,8,1,1,1]
output 11: [1,13,1,1,1](next combination)
output 12: [2,13,1,1,1]
...

Let me know if you have any questions, this is difficult to explain, I know.

Answer 1

Since its a iterator, you can use next :

In [240]: x = itertools.product(*a)

In [241]: next(x)
Out[241]: (1, 4, 7)

In [242]: next(x)
Out[242]: (1, 4, 8)

And so on.

You can have this next in your function which will yield one combination at a time.

Answer 2

This cartesian product shall do it in the order you want:

import itertools
import pprint
 
def cartesian_product(lists):
    return list(itertools.product(*lists))

for list in [(P1,P2,P3,P4,P5)]:
        print(list, '=>')
        pprint(cartesian_product(list), indent=2)

The cartesian product has no duplicates unless your input lists do, if it's the case maybe think of using set() .