如何分割每N个元素的Python列表 [英] How to Split Python list every Nth element
本文介绍了如何分割每N个元素的Python列表的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
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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