Python生成器将另一个可迭代项分组为N个组 [英] Python generator that groups another iterable into groups of N

问题描述

我正在寻找一个函数，该函数采用可迭代的i和大小n，并产生长度为n的元组，这些元组是i的顺序值:

I'm looking for a function that takes an iterable i and a size n and yields tuples of length n that are sequential values from i:

x = [1,2,3,4,5,6,7,8,9,0]
[z for z in TheFunc(x,3)]

给予

[(1,2,3),(4,5,6),(7,8,9),(0)]

标准库中是否存在这样的功能?

Does such a function exist in the standard library?

如果它作为标准库的一部分存在，我似乎找不到它，而且我已经用完了所有的搜索条件.我可以自己写，但我不愿意.

If it exists as part of the standard library, I can't seem to find it and I've run out of terms to search for. I could write my own, but I'd rather not.

推荐答案

请参见

See the grouper recipe in the docs for the itertools package

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

(不过，这是有几个问题的副本.)

(However, this is a duplicate of quite a few questions.)

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