在Python中按(n)块迭代迭代器?

问题描述

你能想出一个很好的方法(也许使用 itertools)来将迭代器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] 与 chunks(l,3) 成为迭代器 [1,2,3], [4,5,6], [7]

Therefore l=[1,2,3,4,5,6,7] with chunks(l,3) becomes an iterator [1,2,3], [4,5,6], [7]

我可以想到一个小程序来做到这一点，但可能不是使用 itertools 的好方法.

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

推荐答案

itertools 文档的 grouper() 配方.python.org/3/library/itertools.html#itertools-recipes" rel="nofollow noreferrer">recipes 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)

不过，它会用一个填充值填充最后一个块.

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

一种不太通用的解决方案，它只适用于序列，但可以根据需要处理最后一个块

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，一个适用于通用迭代器并按需要运行的解决方案是

Finally, a solution that works on general iterators and 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

这篇关于在Python中按(n)块迭代迭代器?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持html5模板网！