python矩阵转置和zip [英] python matrix transpose and zip

问题描述

如何得到这个矩阵的转置......任何更简单的算法方法来做到这一点......

第一个问题:

 输入 a=[[1,2,3],[4,5,6],[7,8,9]]预期输出 a=[[1, 4, 7], [2, 5, 8], [3, 6, 9]]

第二个问题:

Zip 给了我下面的输出，当我不知道数组中有多少元素时，我如何压缩，在这种情况下我知道 3 个元素 a[0],a[1],a[2]但是我如何压缩 [n] 个元素

 >>>zip(a[0],a[1],a[2])[(1, 4, 7), (2, 5, 8), (3, 6, 9)]

解决方案

问题解答:

<预><代码>>>>将 numpy 导入为 np>>>first_answer = np.transpose(a)>>>second_answer = [list(i) for i in zip(*a)]

感谢 afg 的帮助

How to get the transpose of this matrix..Any easier ,algorithmic way to do this...

1st question:

 Input  a=[[1,2,3],[4,5,6],[7,8,9]]
 Expected output a=[[1, 4, 7], [2, 5, 8], [3, 6, 9]] 

2nd Question:

Zip gives me the following output said below,how can i zip when i dont know how many elements are there in the array,in this case i know 3 elements a[0],a[1],a[2] but how can i zip a[n] elements

 >>> zip(a[0],a[1],a[2])
 [(1, 4, 7), (2, 5, 8), (3, 6, 9)]

解决方案

question answers:

>>> import numpy as np
>>> first_answer = np.transpose(a)
>>> second_answer = [list(i) for i in zip(*a)]

thanks to afg for helping out

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