python - 使用 Python 对矩阵进行切片

Question

我需要将 12*12 矩阵切成 24 个 2*3 block 。输入矩阵是:

arr = [
[1,0,1,1,1,0,1,1,1,0,1,0],
[0,0,0,1,0,1,1,1,1,0,1,0],
[0,1,1,0,0,0,1,0,1,0,0,0],
[1,0,0,1,0,0,1,1,1,0,1,1],
[0,0,0,1,0,0,0,1,1,1,1,0],
[0,0,0,0,0,1,1,0,0,0,0,1],
[1,0,1,0,0,0,1,1,0,0,1,1],
[0,0,1,1,0,1,0,1,1,0,1,0],
[0,1,0,0,0,0,1,0,1,0,0,1],
[1,1,0,1,0,1,0,1,0,1,0,0],
[0,0,1,1,1,1,0,1,0,1,1,1],
[0,0,0,0,1,0,0,0,1,1,0,0]]

我尝试用 numpy Matrix 完成任务:

from sympy import Matrix
Matrix(arr)[:3,:2]

但它只会给出原始矩阵的一片。

Matrix([
[1, 0],
[0, 0],
[0, 1]])

将 12*12 矩阵切成 2*3 片的便捷方法是什么？我还需要原始尺寸的 3*2，但假设在准备好第一个尺寸后就很容易了。

Answer 1

您可以使用 numpy.reshape() 函数或直接将 numpy 矩阵的形状从 (12,12) 更改为 (24,3, 2) ，这应该会给你你想要的结果。

例子-

In [25]: arr
Out[25]: 
[[1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 0],
 [0, 0, 0, 1, 0, 1, 1, 1, 1, 0, 1, 0],
 [0, 1, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0],
 [1, 0, 0, 1, 0, 0, 1, 1, 1, 0, 1, 1],
 [0, 0, 0, 1, 0, 0, 0, 1, 1, 1, 1, 0],
 [0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 1],
 [1, 0, 1, 0, 0, 0, 1, 1, 0, 0, 1, 1],
 [0, 0, 1, 1, 0, 1, 0, 1, 1, 0, 1, 0],
 [0, 1, 0, 0, 0, 0, 1, 0, 1, 0, 0, 1],
 [1, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 0],
 [0, 0, 1, 1, 1, 1, 0, 1, 0, 1, 1, 1],
 [0, 0, 0, 0, 1, 0, 0, 0, 1, 1, 0, 0]]

In [26]: n = np.array(arr)

In [28]: n.shape
Out[28]: (12, 12)

In [29]: n.shape = (24,3,2)

In [30]: n
Out[30]: 
array([[[1, 0],
        [1, 1],
        [1, 0]],

       [[1, 1],
        [1, 0],
        [1, 0]],

       [[0, 0],
        [0, 1],
        [0, 1]],

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