python:独立移动张量内的每个矩阵

Question

这个问题对于矩阵非常相似Roll rows of a matrix independently

但我没能将其调整为 3D 张量

我有一个张量

0 0 0
1 1 1
0 0 0

0 0 0
1 1 1
0 0 0 

和一个向量，指定我想将矩阵按列移动多少

1 2

我想要一个新的张量，其中每个矩阵都像这样按列移动

0 0 0
0 0 0
1 1 1

1 1 1
0 0 0
0 0 0

到目前为止我已经能够得到一个潜在的映射索引

import numpy as np

# Input
A = np.array([[0, 0, 0],
              [1, 1, 1],
              [0, 0, 0],
              [0, 0, 0]])
B = np.array([[0, 0, 0],
              [1, 1, 1],
              [0, 0, 0],
              [0, 0, 0]])
AB = np.array([A, B])

# Shifting amount
r = np.array([-1, 1])

d1, d2, d3 = np.ogrid[:AB.shape[0], :AB.shape[1], :AB.shape[2]]

r[r < 0] += AB.shape[1]
r = np.array([r, ]*AB.shape[1]).transpose()
r = r[:, np.newaxis]

# New column indices?
d2 = d2 - r
d2[d2 < 0] += AB.shape[1]

result = AB[d2]
print(result)

但是我得到这个错误:

~/Work/ethz/iml/task2 $ python test.py
Traceback (most recent call last):
  File "test.py", line 27, in <module>
    result = AB[d2]
IndexError: index 2 is out of bounds for axis 0 with size 2

这就是 d2 的样子:

[[[1 1 1 1]
  [2 2 2 2]
  [3 3 3 3]
  [0 0 0 0]]

 [[3 3 3 3]
  [0 0 0 0]
  [1 1 1 1]
  [2 2 2 2]]]

Answer 1

方法 #1

适应strided-based solution来自同一个链接的性能问答 -

from skimage.util.shape import view_as_windows

def roll_along_second_axis_3dar(a, r):
    r = np.asarray(r)
    a_ext = np.concatenate((a,a[:,:-1,:]),axis=1)
    n = a.shape[1]
    idx = (n-r)%n
    w = view_as_windows(a_ext,(1,n,1))[...,0,:,0]
    return w[np.arange(len(idx)),idx].swapaxes(1,2)

sample 运行-

In [11]: a
Out[11]: 
array([[[44, 47, 64],
        [67, 67,  9],
        [83, 21, 36],
        [87, 70, 88]],

       [[88, 12, 58],
        [65, 39, 87],
        [46, 88, 81],
        [37, 25, 77]]])

In [12]: roll_along_second_axis_3dar(a, r=[-1,1])
Out[12]: 
array([[[67, 67,  9],
        [83, 21, 36],
        [87, 70, 88],
        [44, 47, 64]],

       [[37, 25, 77],
        [88, 12, 58],
        [65, 39, 87],
        [46, 88, 81]]])

方法 #2

继续尝试，看来您已经足够接近了。我们可以通过少量修改/更正获得最终输出 -

d1, d2, d3 = np.ogrid[:AB.shape[0], :AB.shape[1], :AB.shape[2]]
r[r < 0] += AB.shape[1]
D2 = ((d2 - r).transpose(2,1,0))%AB.shape[1]
out = AB[d1,D2,d3]