在Python中用块（n）迭代迭代器？ [英] Iterate an iterator by chunks (of n) in Python?

问题描述

你能想出一个很好的方法（可能用itertools）将迭代器拆分成给定大小的块吗？

Can you think of a nice way (maybe with itertools) to split an iterator into chunks of given size?

因此 l = [ 1,2,3,4,5,6,7] 带块（l，3）成为迭代器 [1,2,3]，[4,5,6]，[7]

我能想到一个小程序要做这可能不是一个很好的方式，可能是itertools。

I can think of a small program to do that but not a nice way with maybe itertools.

推荐答案

石斑鱼（）来自 itertools 文档的配方食谱接近你想要的：

The grouper() recipe from the itertools documentation's recipes comes close to what you want:

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

它会填充最后一个带有填充值的块。

It will fill up the last chunk with a fill value, though.

一种不常用的解决方案序列，但确实处理最后一个块是

A less general solution that only works on sequences but does handle the last chunk as desired is

[my_list[i:i + chunk_size] for i in range(0, len(my_list), chunk_size)]

最后，适用于一般迭代器的解决方案的行为符合要求是

Finally, a solution that works on general iterators an behaves as desired is

def grouper(n, iterable):
    it = iter(iterable)
    while True:
       chunk = tuple(itertools.islice(it, n))
       if not chunk:
           return
       yield chunk

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