配对列表列表中的元素以形成新列表 [英] Pair elements from lists of list to form a new list

问题描述

我有一个类似以下的列表

I have a list of lists like the following,

a = [[ [1,2], [10, 3]], [[4,5], [6, 7]]]

我需要以这种方式将最里面的列表元素配对

I need to pair the inner most list elements this way,

a = [[[1, 10], [2, 3]], [[4, 6], [5, 7]]]. 

简单的方法如下:

pairings_ = []
for ind in a:
    pairings_.append([[x, y] for x in ind[0] for y in ind[1])

如果ind中的列表大于2，则将导致内存错误.
例如， 如果ind [0]，[1、2]，[10、3]，[7、8]中有三个内部列表，则配对将为[1、10、7]和[2、3、8] . 假设[[1，10]，[2，3]]和[[4,6]，[5，7]]的内部列表的长度将始终相等.

This would cause memoryerror if the lists inside ind are more than 2.
For example, If there were three inner lists in ind[0], [1, 2], [10, 3], [7, 8], then the pairing would be [1, 10, 7] and [2, 3, 8]. The assumption is the length of inner list of [[1, 10], [2, 3]] and [[4,6], [5, 7]] will always be equal.

我将如何以尽可能多的pythonic/numpy/最有效的方式进行此操作?

How would I go about doing this in the most pythonic / numpy / efficient way possible?

推荐答案

我将如何以尽可能多的pythonic/numpy/最有效的方式进行此操作?

How would I go about doing this in the most pythonic / numpy / efficient way possible?

您可以使用 np.transpose 和将其转换回列表.

You can use np.transpose and convert it back to list.

In [1]: import numpy as np


In [2]: a = [[ [1,2], [10, 3]], [[4,5], [6, 7]]]


In [3]: np.transpose(a, (0,2,1)).tolist()

Out[3]: [[[1, 10], [2, 3]], [[4, 6], [5, 7]]]

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