将列表分成n组的另一种方法[重复]

甲的Python配方（在Python
2.6，使用

itertools.izip_longest

def grouper(n, iterable, fillvalue=None):    "grouper(3, 'ABCDEFG', 'x') --> ABC DEF Gxx"    args = [iter(iterable)] * n    return itertools.zip_longest(*args, fillvalue=fillvalue)

用法示例：

>>> list(grouper(3, range(9)))[(0, 1, 2), (3, 4, 5), (6, 7, 8)]>>> list(grouper(3, range(10)))[(0, 1, 2), (3, 4, 5), (6, 7, 8), (9, None, None)]

如果您希望最后一组短于其他组而不是用填充

fillvalue

>>> def mygrouper(n, iterable):...     args = [iter(iterable)] * n...     return ([e for e in t if e != None] for t in itertools.zip_longest(*args))... >>> list(mygrouper(3, range(9)))[[0, 1, 2], [3, 4, 5], [6, 7, 8]]>>> list(mygrouper(3, range(10)))[[0, 1, 2], [3, 4, 5], [6, 7, 8], [9]]