Python 3.x在矩阵上获取奇数列 [英] Python 3.x get odd columns on matrix

问题描述

我正在使用python 3.7，我想获取矩阵的所有奇数列.

I am working with python 3.7 and I would like to get all the odd columns of a matrix.

举个例子，我现在有一个这种样式的4x4矩阵.

To give a example, I have a 4x4 matrix of this style right now.

[[0, 9, 1, 6], [0, 3, 1, 5], [0, 2, 1, 7], [0, 6, 1, 2]]

那是...

0 9 1 6
0 3 1 5
0 2 1 7
0 6 1 2

我想得到:

9 6
3 5
2 7
6 2

矩阵的数字和大小会改变，但结构始终为

The numbers and the size of the matrix will change but the structure will always be

[[0, (int), 1, (int), 2...], [0, (int), 1, (int), 2 ...], [0, (int), 1, (int), 2...], [0, (int), 1, (int), 2...], ...]

要获取行，我可以执行 [:: 2] ，但是这种出色的解决方案目前对我不起作用.我尝试使用以下方法访问矩阵:

To get the rows I can do [:: 2], but that wonderful solution does not work for me right now. I try to access the matrix with:

for i in matrix:
    for j in matrix:

但是，这一切都不起作用.我该怎么解决?

But none of this doesn't work either. How can I solve it?

谢谢.

推荐答案

如果不使用 numpy ，则可以使用类似于索引方案( [1 :: 2] )中的列表理解:

Without using numpy, you can use something similar to your indexing scheme ([1::2]) in a list comprehension:

>>> [i[1::2] for i in mat]
[[9, 6], [3, 5], [2, 7], [6, 2]]

使用 numpy ，您可以执行类似的操作:

Using numpy, you can do something similar:

>>> import numpy as np
>>> np.array(mat)[:,1::2]
array([[9, 6],
       [3, 5],
       [2, 7],
       [6, 2]])

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