将列表拆分为单独但重叠的块 [英] Split list into separate but overlapping chunks

问题描述

假设我有一个列表A

A = [1,2,3,4,5,6,7,8,9,10]

我想按以下顺序使用上面的列表创建一个新列表(例如B).

I would like to create a new list (say B) using the above list in the following order.

B = [[1,2,3], [3,4,5], [5,6,7], [7,8,9], [9,10,]]

即前三个数字为A[0,1,2]，后三个数字为A[2,3,4]，依此类推.

i.e. the first 3 numbers as A[0,1,2] and the second 3 numbers as A[2,3,4] and so on.

我相信numpy中具有用于此类操作的功能.

I believe there is a function in numpy for such a kind of operation.

推荐答案

只需将Python的内置列表推导与列表切片结合使用即可:

Simply use Python's built-in list comprehension with list-slicing to do this:

>>> A = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
>>> size = 3
>>> step = 2
>>> A = [A[i : i + size] for i in range(0, len(A), step)]

这为您提供了您想要的东西:

This gives you what you're looking for:

>>> A
[[1, 2, 3], [3, 4, 5], [5, 6, 7], [7, 8, 9], [9, 10]]

但是您必须写几行代码，以确保代码不会因大小/步长的空前值而中断.

But you'll have to write a couple of lines to make sure that your code doesn't break for unprecedented values of size/step.

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