使用索引矩阵检索值的 Pythonic 方法？

Question

我有一个 h-by-w-by-3 numpy 数组A和另一个 h-by-w-by-2 numpy 数组B，它告诉我A在每个 (h, w) 位置从哪里获取值。用C(h-by-w-by-3)表示结果矩阵。天真，我可以C = np.zeros(A.shape)for i in range(B.shape[0]):    for j in range(B.shape[1]):        C[i, j, :] = A[B[i, j, 0], B[i, j, 1], :]但我相信有更高效、更 Pythonic 和更快的方式！

Answer 1

确实有：

>>> import numpy as np

>>> 

>>> h, w = 4, 6

>>> 

>>> A = np.random.randint(0, 100, (h, w, 3))

>>> B = np.random.randint(0, h*w, (h, w, 2)) % [h, w]

>>> 

>>> C = np.zeros(A.shape)

>>> for i in range(B.shape[0]):

...     for j in range(B.shape[1]):

...         C[i, j, :] = A[B[i, j, 0], B[i, j, 1], :]

... 

>>> 

>>> C2 = A[(*np.moveaxis(B, 2, 0),)]

>>> np.all(C == C2)

True