itertools.combinations,如果两个列表都像这里一样.
或者在一般情况下itertools.product,然后进行一些过滤:
In [7]: a = ["1", "2", "3"]
...: b = ["a", "b", "c"]
In [8]: list(filter(lambda t: t[0] < t[1], product(a,b)))
Out[8]:
[('1', 'a'),
('1', 'b'),
('1', 'c'),
('2', 'a'),
('2', 'b'),
('2', 'c'),
('3', 'a'),
('3', 'b'),
('3', 'c')]
此外,我认为术语组合已经意味着结果中元素的顺序无关紧要.
好的,Theodros是对的.为了补偿,这里的版本应该适用于任何列表列表:
l = [['1','2','3'], ['a','b'], ['x','y']]
set(tuple(sorted(p)) for p in product(*l) if len(set(p)) > 1)
给出(适当排序)
set([('1', 'a', 'x'),
('3', 'a', 'y'),
('2', 'b', 'y'),
('2', 'a', 'y'),
('1', 'a', 'y'),
('1', 'b', 'y'),
('2', 'a', 'x'),
('3', 'b', 'y'),
('1', 'b', 'x'),
('2', 'b', 'x'),
('3', 'a', 'x'),
('3', 'b', 'x')])
它也适用于前面的反例l = [[1,2,3],[1,3,4,5]]:
set([(1, 2), (1, 3), (1, 4), (1, 5), (2, 3), (2, 5), (3, 4), (2, 4), (3, 5)])