numpy dot()和Python 3.5+矩陣乘法@的區別

我最近遷移到了Python 3.5，注意到new matrix multiplication operator（新的矩陣乘法雲算符） (@)的行為與numpy dot操作符有所不同。例如，對於3D數組：

import numpy as np

a = np.random.rand(8,13,13)
b = np.random.rand(8,13,13)
c = a @ b  # Python 3.5+
d = np.dot(a, b)

@運算符返回一個形狀數組：

c.shape
(8, 13, 13)

而np.dot()函數返回：

d.shape
(8, 13, 8, 13)

我怎樣才能重現與numpy點乘相同的結果？還有其他的重大差異嗎？

最佳解決方法

@運算符調用數組的__matmul__方法，而不是dot。該方法也作為函數np.matmul在API中提供。

>>> a = np.random.rand(8,13,13)
>>> b = np.random.rand(8,13,13)
>>> np.matmul(a, b).shape
(8, 13, 13)

從文檔：

matmul differs from dot in two important ways.

• Multiplication by scalars is not allowed.
• Stacks of matrices are broadcast together as if the matrices were elements.

最後一點清楚地表明dot和matmul方法在傳遞3D(或更高維)數組時行為不同。更多的引用文檔：

對於matmul：

If either argument is N-D, N > 2, it is treated as a stack of matrices residing in the last two indexes and broadcast accordingly.

對於np.dot：

For 2-D arrays it is equivalent to matrix multiplication, and for 1-D arrays to inner product of vectors (without complex conjugation). For N dimensions it is a sum product over the last axis of a and the second-to-last of b

參考資料

• Difference between numpy dot() and Python 3.5+ matrix multiplication @