除了接受的答案外，如果你使用的可迭代对象可能长度不同，但不应该是，建议将strict=True传递给zip()(自Python 3.10起支持)。

引用文件:

zip() is often used in cases where the iterables are assumed to be of equal length. In such cases, it’s recommended to use the strict=True option. Its output is the same as regular zip(): >>> list(zip(('a', 'b', 'c'), (1, 2, 3), strict=True)) [('a', 1), ('b', 2), ('c', 3)] Unlike the default behavior, it checks that the lengths of iterables are identical, raising a ValueError if they aren’t: >>> list(zip(range(3), ['fee', 'fi', 'fo', 'fum'], strict=True)) Traceback (most recent call last): ... ValueError: zip() argument 2 is longer than argument 1 Without the strict=True argument, any bug that results in iterables of different lengths will be silenced, possibly manifesting as a hard-to-find bug in another part of the program.

non itertools Python 3 解决方案：

def zip_longest(*lists):
    def g(l):
        for item in l:
            yield item
        while True:
            yield None
    gens = [g(l) for l in lists]    
    for _ in range(max(map(len, lists))):
        yield tuple(next(g) for g in gens)

我的Python 2解决方案:

if len(list1) < len(list2):
    list1.extend([None] * (len(list2) - len(list1)))
else:
    list2.extend([None] * (len(list1) - len(list2)))

除了接受的答案外，如果你使用的可迭代对象可能长度不同，但不应该是，建议将strict=True传递给zip()(自Python 3.10起支持)。

引用文件:

zip() is often used in cases where the iterables are assumed to be of equal length. In such cases, it’s recommended to use the strict=True option. Its output is the same as regular zip(): >>> list(zip(('a', 'b', 'c'), (1, 2, 3), strict=True)) [('a', 1), ('b', 2), ('c', 3)] Unlike the default behavior, it checks that the lengths of iterables are identical, raising a ValueError if they aren’t: >>> list(zip(range(3), ['fee', 'fi', 'fo', 'fum'], strict=True)) Traceback (most recent call last): ... ValueError: zip() argument 2 is longer than argument 1 Without the strict=True argument, any bug that results in iterables of different lengths will be silenced, possibly manifesting as a hard-to-find bug in another part of the program.

在Python 3中，您可以使用itertools.zip_longest

>>> list(itertools.zip_longest(a, b, c))
[('a1', 'b1', 'c1'), (None, 'b2', 'c2'), (None, 'b3', None)]

你可以使用fillvalue参数填充不同于None的值:

>>> list(itertools.zip_longest(a, b, c, fillvalue='foo'))
[('a1', 'b1', 'c1'), ('foo', 'b2', 'c2'), ('foo', 'b3', 'foo')]

在Python 2中，你可以使用itertools。izip_longest (Python 2.6+)，或者可以使用map和None。这是map的一个鲜为人知的特性(但是map在Python 3中改变了。x，所以这只适用于Python 2.x)。

>>> map(None, a, b, c)
[('a1', 'b1', 'c1'), (None, 'b2', 'c2'), (None, 'b3', None)]

只是使用迭代器，没什么特别的。

def zip_longest(*iterables):
    items = 0
    for iterable in iterables:
        items = max(items, len(iterable))

    iters = [iter(iterable) for iterable in iterables]
    while items:
        yield (*[next(i, None) for i in iters],)
        items -= 1