Question

是否有任何 inline 命令可以在numpy的python中生成移位的身份矩阵？

A=[ ...
    0, 1, 0, 0, 0, 0, 0, 0, 0, 0
    0, 0, 1, 0, 0, 0, 0, 0, 0, 0
    0, 0, 0, 1, 0, 0, 0, 0, 0, 0
    0, 0, 0, 0, 1, 0, 0, 0, 0, 0
    0, 0, 0, 0, 0, 1, 0, 0, 0, 0
    0, 0, 0, 0, 0, 0, 1, 0, 0, 0
    0, 0, 0, 0, 0, 0, 0, 1, 0, 0
    0, 0, 0, 0, 0, 0, 0, 0, 1, 0
    0, 0, 0, 0, 0, 0, 0, 0, 0, 1
    0, 0, 0, 0, 0, 0, 0, 0, 0, 0]

Answer 1

如果我理解正确，则可以使用np.eye的k参数：

import numpy as np

result = np.eye(10, 10, 1)
print(result)

输出

[[0. 1. 0. 0. 0. 0. 0. 0. 0. 0.]
 [0. 0. 1. 0. 0. 0. 0. 0. 0. 0.]
 [0. 0. 0. 1. 0. 0. 0. 0. 0. 0.]
 [0. 0. 0. 0. 1. 0. 0. 0. 0. 0.]
 [0. 0. 0. 0. 0. 1. 0. 0. 0. 0.]
 [0. 0. 0. 0. 0. 0. 1. 0. 0. 0.]
 [0. 0. 0. 0. 0. 0. 0. 1. 0. 0.]
 [0. 0. 0. 0. 0. 0. 0. 0. 1. 0.]
 [0. 0. 0. 0. 0. 0. 0. 0. 0. 1.]
 [0. 0. 0. 0. 0. 0. 0. 0. 0. 0.]]

来自链接的文档：

k：int，对角线的可选索引：0（默认值）是指主对角线，正值表示上对角线，a 负值到较低的对角线。

Python身份转换矩阵

1 个答案: