最优雅的解决方案是使用itertools。python 2.6中的产品。

如果你使用的不是Python 2.6, itertools的文档。Product实际上显示了一个等效的功能，做产品的“手动”方式:

def product(*args, **kwds):
    # product('ABCD', 'xy') --> Ax Ay Bx By Cx Cy Dx Dy
    # product(range(2), repeat=3) --> 000 001 010 011 100 101 110 111
    pools = map(tuple, args) * kwds.get('repeat', 1)
    result = [[]]
    for pool in pools:
        result = [x+[y] for x in result for y in pool]
    for prod in result:
        yield tuple(prod)

最优雅的解决方案是使用itertools。python 2.6中的产品。

如果你使用的不是Python 2.6, itertools的文档。Product实际上显示了一个等效的功能，做产品的“手动”方式:

def product(*args, **kwds):
    # product('ABCD', 'xy') --> Ax Ay Bx By Cx Cy Dx Dy
    # product(range(2), repeat=3) --> 000 001 010 011 100 101 110 111
    pools = map(tuple, args) * kwds.get('repeat', 1)
    result = [[]]
    for pool in pools:
        result = [x+[y] for x in result for y in pool]
    for prod in result:
        yield tuple(prod)

你需要itertools.product:

>>> import itertools
>>> a = [[1,2,3],[4,5,6],[7,8,9,10]]
>>> list(itertools.product(*a))
[(1, 4, 7), (1, 4, 8), (1, 4, 9), (1, 4, 10), (1, 5, 7), (1, 5, 8), (1, 5, 9), (1, 5, 10), (1, 6, 7), (1, 6, 8), (1, 6, 9), (1, 6, 10), (2, 4, 7), (2, 4, 8), (2, 4, 9), (2, 4, 10), (2, 5, 7), (2, 5, 8), (2, 5, 9), (2, 5, 10), (2, 6, 7), (2, 6, 8), (2, 6, 9), (2, 6, 10), (3, 4, 7), (3, 4, 8), (3, 4, 9), (3, 4, 10), (3, 5, 7), (3, 5, 8), (3, 5, 9), (3, 5, 10), (3, 6, 7), (3, 6, 8), (3, 6, 9), (3, 6, 10)]

只需使用itertools.product:

listOLists = [[1,2,3],[4,5,6],[7,8,9,10]]
for l in itertools.product(*listOLists):
    print(l)

这主要是模仿Jarret Hardie使用itertools的Answer解决方案。产品，但有这些区别:

this passes parameters to itertools.product in-line, instead of via variable a - so no *args syntax needed on the inline parameters if your mypy type-linter acts like mine, and you can get your code to otherwise "work" with the *args syntax with inline product parameters (like product(*[[1,2,3],[4,5,6],[7,8,9,10]])), mypy might still fail it (with something like error: No overload variant of "product" matches argument type "List[object]") So solution to that mypy, is to not use *args syntax, like this:

    >>> import itertools
    >>> list(itertools.product([1,2,3],[4,5,6],[7,8,9,10]))
    [(1, 4, 7), (1, 4, 8), (1, 4, 9), (1, 4, 10), (1, 5, 7), (1, 5, 8), (1, 5, 9), (1, 5, 10), (1, 6, 7), (1, 6, 8), (1, 6, 9), (1, 6, 10), (2, 4, 7), (2, 4, 8), (2, 4, 9), (2, 4, 10), (2, 5, 7), (2, 5, 8), (2, 5, 9), (2, 5, 10), (2, 6, 7), (2, 6, 8), (2, 6, 9), (2, 6, 10), (3, 4, 7), (3, 4, 8), (3, 4, 9), (3, 4, 10), (3, 5, 7), (3, 5, 8), (3, 5, 9), (3, 5, 10), (3, 6, 7), (3, 6, 8), (3, 6, 9), (3, 6, 10)]

from itertools import product 
list_vals = [['Brand Acronym:CBIQ', 'Brand Acronym :KMEFIC'],['Brand Country:DXB','Brand Country:BH']]
list(product(*list_vals))

输出:

[('品牌首字母缩写:CBIQ'， '品牌国家:DXB')， (“品牌首字母缩写:CBIQ”，“品牌国家:BH”)， (“品牌首字母缩写:KMEFIC”，“品牌国家:DXB”)， (“品牌首字母缩写:KMEFIC”，“品牌国家:BH”)]