获取 N 维数组的所有索引作为列表

Get all indices of an N-dimensional array as a list

有没有办法在Python中快速高效地获取N维数组中所有索引的列表或数组?

例如,图像我们有以下数组:

import numpy as np

test = np.zeros((4,4))

array([[0., 0., 0., 0.],
       [0., 0., 0., 0.],
       [0., 0., 0., 0.],
       [0., 0., 0., 0.]])

我想获取所有元素索引如下:

indices = [ [0,0],[0,1],[0,2] ... [3,2],[3,3] ]

你可以试试itertools.product:

>>> from itertools import product
>>> 
>>> [list(i) for i in product(range(4), range(4))]
[[0, 0], [0, 1], [0, 2], [0, 3], [1, 0], [1, 1], [1, 2], [1, 3], [2, 0], [2, 1], [2, 2], [2, 3], [3, 0], [3, 1], [3, 2], [3, 3]]

如果您可以使用列表理解

test = np.zeros((4,4))
indices = [[i, j] for i in range(test.shape[0]) for j in range(test.shape[1])]
print (indices)

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

我建议使用 np.ones_like 制作一个与 test 数组形状相同的 1 数组,然后使用 np.where:

>>> np.stack(np.where(np.ones_like(test))).T
# Or np.dstack(np.where(np.ones_like(test)))
array([[0, 0],
       [0, 1],
       [0, 2],
       [0, 3],
       [1, 0],
       [1, 1],
       [1, 2],
       [1, 3],
       [2, 0],
       [2, 1],
       [2, 2],
       [2, 3],
       [3, 0],
       [3, 1],
       [3, 2],
       [3, 3]])

只是列举应该做的:

test = [[0., 0., 0., 0.],
       [0., 0., 0., 0.],
       [0., 0., 0., 0.],
       [0., 0., 0., 0.],
       [0., 0., 0., 0.]]

indices = [[i, j] for i, row in enumerate(test) for j, col in enumerate(row)]
print(indices)

>>> [[0, 0], [0, 1], [0, 2], [0, 3], [1, 0], [1, 1], [1, 2], [1, 3], [2, 0], [2, 1], [2, 2], [2, 3], [3, 0], [3, 1], [3, 2], [3, 3], [4, 0], [4, 1], [4, 2], [4, 3]]

使用 np.indices 并进行一些整形:

np.indices(test.shape).reshape(2, -1).T

array([[0, 0],  
       [0, 1],  
       [0, 2],  
       [0, 3],  
       [1, 0],  
       [1, 1],  
       [1, 2],  
       [1, 3],  
       [2, 0],  
       [2, 1],  
       [2, 2],  
       [2, 3],  
       [3, 0],  
       [3, 1],  
       [3, 2],  
       [3, 3]])