如何分割每N个元素的Python列表 [英] How to Split Python list every Nth element

问题描述

我想做的事情很简单，但是我找不到如何做.

What I am trying to do is pretty simple, but I couldn't find how to do it.

• 从第一个元素开始，将每个第四个元素放入一个新列表.
• 重复第2、3和4个元素.

发件人:

list = ['1', '2', '3', '4', '5', '6', '7', '8', '9', 'a', 'b']

收件人:

list1 = ['1', '5', '9']
list2 = ['2', '6', 'a']
list3 = ['3', '7', 'b']
list4 = ['4', '9']

换句话说，我需要知道如何:

In other words, I need to know how to:

• 从列表中获取第N个元素(循环)
• 将其存储在新数组中
• 重复

推荐答案

具体解决方案是大步使用切片:

The specific solution is to use slicing with a stride:

source = ['1', '2', '3', '4', '5', '6', '7', '8', '9', 'a', 'b']
list1 = source[::4]
list2 = source[1::4]
list3 = source[2::4]
list4 = source[3::4]

source[::4]从索引0开始每隔第4个元素；其他切片只会更改起始索引.

source[::4] takes every 4th element, starting at index 0; the other slices only alter the starting index.

generic 解决方案是使用循环进行切片，并将结果存储在外部列表中；列表理解可以很好地做到这一点:

The generic solution is to use a loop to do the slicing, and store the result in an outer list; a list comprehension can do that nicely:

def slice_per(source, step):
    return [source[i::step] for i in range(step)]

演示:

>>> source = ['1', '2', '3', '4', '5', '6', '7', '8', '9', 'a', 'b']
>>> source[::4]
['1', '5', '9']
>>> source[1::4]
['2', '6', 'a']
>>> def slice_per(source, step):
...     return [source[i::step] for i in range(step)]
... 
>>> slice_per(source, 4)
[['1', '5', '9'], ['2', '6', 'a'], ['3', '7', 'b'], ['4', '8']]
>>> slice_per(source, 3)
[['1', '4', '7', 'a'], ['2', '5', '8', 'b'], ['3', '6', '9']]

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