从子列表中获取所有对,其中没有一对单独来自一个子列表

以下是一种方法:将两个列表组合在一起。然后做一个组合的乘积。

from itertools import product, combinations, chain
res = list(chain.from_iterable(product(a, b) for a, b in combinations(lists, 2)))

print(res)
[('a', 7),
 ('a', 1),
 ('a', 2),
 ('a', 3),
 ('a', 5),
 ('b', 7),
 ('b', 1),
 ('b', 2),
 ('b', 3),
 ('b', 5),
 ('c', 7),
 ('c', 1),
 ('c', 2),
 ('c', 3),
 ('c', 5),
 ('a', 'alpha'),
 ('a', 'gamma'),
 ('b', 'alpha'),
 ('b', 'gamma'),
 ('c', 'alpha'),
 ('c', 'gamma'),
 (7, 'alpha'),
 (7, 'gamma'),
 (1, 'alpha'),
 (1, 'gamma'),
 (2, 'alpha'),
 (2, 'gamma'),
 (3, 'alpha'),
 (3, 'gamma'),
 (5, 'alpha'),
 (5, 'gamma')]

itertools.combinations + itertools.product 和 set 理解删除重复项

{item for z in (itertools.combinations(x, 2) for x in itertools.product(*lists)) for item in z}


{('a', 'alpha'),
 ('a', 'gamma'),
 ('a', 1),
 ('a', 2),
 ('a', 3),
 ('a', 5),
 ('a', 7),
 ('b', 'alpha'),
 ('b', 'gamma'),
 ('b', 1),
 ('b', 2),
 ('b', 3),
 ('b', 5),
 ('b', 7),
 ('c', 'alpha'),
 ('c', 'gamma'),
 ('c', 1),
 ('c', 2),
 ('c', 3),
 ('c', 5),
 ('c', 7),
 (1, 'alpha'),
 (1, 'gamma'),
 (2, 'alpha'),
 (2, 'gamma'),
 (3, 'alpha'),
 (3, 'gamma'),
 (5, 'alpha'),
 (5, 'gamma'),
 (7, 'alpha'),
 (7, 'gamma')}

多亏了@domochevksi的评论:

n = len(lists)
combinations = [
    [item0, item1]
    for i in range(n)
    for j in range(i+1, n)
    for item0 in lists[i]
    for item1 in lists[j]
    ]