将列表分成 n 组的另一种方法 [英] Alternative way to split a list into groups of n

问题描述

假设我有一个任意长度的列表，L:

L = list(range(1000))

将该列表分成 n 组的最佳方法是什么?这是我能想出的最好的结构，但出于某种原因，这并不是完成任务的最佳方式:

n = 25对于范围内的 i (0, len(L), n):块 = L[i:i+25]

是否有内置工具可以执行此操作?

早期的答案是将我的 for 循环改造成 listcomp，这不是想法；你基本上是在以不同的形式给我我的确切答案.我正在查看是否有替代方法来完成此操作，例如列表中的假设 .split 或其他内容.我也在昨晚写的一些代码中使用它作为生成器:

def split_list(L, n):assert type(L) 是列表，L 不是列表"对于范围内的 i (0, len(L), n):产量 L[i:i+n]

解决方案

A Python配方(在 Python 2.6 中，使用 itertools.izip_longest):

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

示例用法:

<预><代码>>>>列表(石斑鱼(3，范围(9)))[(0, 1, 2), (3, 4, 5), (6, 7, 8)]>>>列表(石斑鱼(3，范围(10)))[(0, 1, 2), (3, 4, 5), (6, 7, 8), (9, None, None)]

如果您希望最后一组比其他组短而不是用 fillvalue 填充，那么您可以例如像这样更改代码:

<预><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))...>>>列表(mygrouper(3，范围(9)))[[0, 1, 2], [3, 4, 5], [6, 7, 8]]>>>列表(mygrouper(3，范围(10)))[[0, 1, 2], [3, 4, 5], [6, 7, 8], [9]]

Let's say I have a list of arbitrary length, L:

L = list(range(1000))

What is the best way to split that list into groups of n? This is the best structure that I have been able to come up with, and for some reason it does not feel like it is the best way of accomplishing the task:

n = 25
for i in range(0, len(L), n):
    chunk = L[i:i+25]

Is there a built-in to do this I'm missing?

Edit: Early answers are reworking my for loop into a listcomp, which is not the idea; you're basically giving me my exact answer back in a different form. I'm seeing if there's an alternate means to accomplish this, like a hypothetical .split on lists or something. I also do use this as a generator in some code that I wrote last night:

def split_list(L, n):
    assert type(L) is list, "L is not a list"
    for i in range(0, len(L), n):
        yield L[i:i+n]

解决方案

A Python recipe (In Python 2.6, use 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)

Example usage:

>>> 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)]

If you want the last group to be shorter than the others instead of padded with fillvalue, then you could e.g. change the code like this:

>>> 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]]

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