对于Python 2.6x，请使用itertools模块的izip_longest。

对于Python 3，使用zip_longest代替(没有i前导)。

>>> list(itertools.izip_longest(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

为了补充已经给出的答案，以下内容适用于任何可迭代对象，不使用itertools，回答@ProdIssue的问题:

def zip_longest(*iterables, default_value):
    iterators = tuple(iter(i) for i in iterables)
    sentinel = object()
    while True:
        new = tuple(next(i, sentinel) for i in iterators)
        if all(n is sentinel for n in new):
            return
        yield tuple(default_value if n is sentinel else n for n in new)

需要使用sentinel，以便产生default_value的迭代器不会被错误地标识为空。

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.

我使用一个2d数组，但概念是类似的使用python 2.x:

if len(set([len(p) for p in printer])) > 1:
    printer = [column+['']*(max([len(p) for p in printer])-len(column)) for column in printer]

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)))