如何按n个元素对python中的元素进行分组 [英] How to group elements in python by n elements

问题描述

<块引用>

可能的重复:
如何将列表平均拆分Python 中的大小块?

我想从列表 l 中获取大小为 n 个元素的组:

即:

[1,2,3,4,5,6,7,8,9] ->[[1,2,3], [4,5,6],[7,8,9]] 其中 n 为 3

解决方案

好吧，蛮力的答案是:

subList = [theList[n:n+N] for n in range(0, len(theList), N)]

其中 N 是组大小(在您的情况下为 3):

<预><代码>>>>theList = 列表(范围(10))>>>N = 3>>>subList = [theList[n:n+N] for n in range(0, len(theList), N)]>>>子列表[[0, 1, 2], [3, 4, 5], [6, 7, 8], [9]]

如果你想要一个填充值，你可以在列表理解之前这样做:

tempList = theList + [fill] * NsubList = [tempList[n:n+N] for n in range(0, len(theList), N)]

示例:

<预><代码>>>>填充 = 99>>>tempList = theList + [fill] * N>>>subList = [tempList[n:n+N] for n in range(0, len(theList), N)]>>>子列表[[0, 1, 2], [3, 4, 5], [6, 7, 8], [9, 99, 99]]

Possible Duplicate:
How do you split a list into evenly sized chunks in Python?

I'd like to get groups of size n elements from a list l:

ie:

[1,2,3,4,5,6,7,8,9] -> [[1,2,3], [4,5,6],[7,8,9]] where n is 3

解决方案

Well, the brute force answer is:

subList = [theList[n:n+N] for n in range(0, len(theList), N)]

where N is the group size (3 in your case):

>>> theList = list(range(10))
>>> N = 3
>>> subList = [theList[n:n+N] for n in range(0, len(theList), N)]
>>> subList
[[0, 1, 2], [3, 4, 5], [6, 7, 8], [9]]

If you want a fill value, you can do this right before the list comprehension:

tempList = theList + [fill] * N
subList = [tempList[n:n+N] for n in range(0, len(theList), N)]

Example:

>>> fill = 99
>>> tempList = theList + [fill] * N
>>> subList = [tempList[n:n+N] for n in range(0, len(theList), N)]
>>> subList
[[0, 1, 2], [3, 4, 5], [6, 7, 8], [9, 99, 99]]

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