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
,则可以例如更改如下代码:
>>> 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]]