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

Question

I recently moved to Python 3.5 and noticed the new matrix multiplication operator (@) sometimes behaves differently from the numpy dot operator. In example, for 3d arrays:

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)

The @ operator returns an array of shape:

c.shape
(8, 13, 13)

while the np.dot() function returns:

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

How can I reproduce the same result with numpy dot? Are there any other significant differences?

Answer 1

The @ operator calls the array's __matmul__ method, not dot. This method is also present in the API as the function np.matmul.

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

From the documentation:

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.

The last point makes it clear that dot and matmul methods behave differently when passed 3D (or higher dimensional) arrays. Quoting from the documentation some more:

For 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.

For 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