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


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:

(8, 13, 13)

while the np.dot() function returns:

(8, 13, 8, 13)

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

Asked By: blaz



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

Answered By: Alex Riley

The answer by @ajcr explains how the dot and matmul (invoked by the @ symbol) differ. By looking at a simple example, one clearly sees how the two behave differently when operating on ‘stacks of matricies’ or tensors.

To clarify the differences take a 4×4 array and return the dot product and matmul product with a 3x4x2 ‘stack of matricies’ or tensor.

import numpy as np
fourbyfour = np.array([

threebyfourbytwo = np.array([

print('4x4*3x4x2 dot:n {}n'.format(np.dot(fourbyfour,threebyfourbytwo)))
print('4x4*3x4x2 matmul:n {}n'.format(np.matmul(fourbyfour,threebyfourbytwo)))

The products of each operation appear below. Notice how the dot product is,

…a sum product over the last axis of a and the second-to-last of b

and how the matrix product is formed by broadcasting the matrix together.

4x4*3x4x2 dot:
 [[[232 152]
  [125 112]
  [125 112]]

 [[172 116]
  [123  76]
  [123  76]]

 [[442 296]
  [228 226]
  [228 226]]

 [[962 652]
  [465 512]
  [465 512]]]

4x4*3x4x2 matmul:
 [[[232 152]
  [172 116]
  [442 296]
  [962 652]]

 [[125 112]
  [123  76]
  [228 226]
  [465 512]]

 [[125 112]
  [123  76]
  [228 226]
  [465 512]]]
Answered By: Nathan

In mathematics, I think the dot in numpy makes more sense

dot(a,b)_{i,j,k,a,b,c} = formula

since it gives the dot product when a and b are vectors, or the matrix multiplication when a and b are matrices

As for matmul operation in numpy, it consists of parts of dot result, and it can be defined as

matmul(a,b)_{i,j,k,c} = formula

So, you can see that matmul(a,b) returns an array with a small shape,
which has smaller memory consumption and make more sense in applications.
In particular, combining with broadcasting, you can get

matmul(a,b)_{i,j,k,l} = formula

for example.

From the above two definitions, you can see the requirements to use those two operations. Assume a.shape=(s1,s2,s3,s4) and b.shape=(t1,t2,t3,t4)

  • To use dot(a,b) you need
  1. t3=s4;
  • To use matmul(a,b) you need
  1. t3=s4
  2. t2=s2, or one of t2 and s2 is 1
  3. t1=s1, or one of t1 and s1 is 1

Use the following piece of code to convince yourself.

import numpy as np
for it in range(10000):
    a = np.random.rand(5,6,2,4)
    b = np.random.rand(6,4,3)
    c = np.matmul(a,b)
    d = np.dot(a,b)
    #print ('c shape: ', c.shape,'d shape:', d.shape)
    for i in range(5):
        for j in range(6):
            for k in range(2):
                for l in range(3):
                    if c[i,j,k,l] != d[i,j,k,j,l]:
                        print (it,i,j,k,l,c[i,j,k,l]==d[i,j,k,j,l])  # you will not see them              
Answered By: Yong Yang

Just FYI, @ and its numpy equivalents dot and matmul are all equally fast. (Plot created with perfplot, a project of mine.)

enter image description here

Code to reproduce the plot:

import perfplot
import numpy

def setup(n):
    A = numpy.random.rand(n, n)
    x = numpy.random.rand(n)
    return A, x

def at(A, x):
    return A @ x

def numpy_dot(A, x):
    return numpy.dot(A, x)

def numpy_matmul(A, x):
    return numpy.matmul(A, x)

    kernels=[at, numpy_dot, numpy_matmul],
    n_range=[2 ** k for k in range(15)],
Answered By: Nico Schlömer

My experience with MATMUL and DOT

I was constantly getting “ValueError: Shape of passed values is (200, 1), indices imply (200, 3)” when trying to use MATMUL. I wanted a quick workaround and found DOT to deliver the same functionality. I don’t get any error using DOT. I get the correct answer


>>>(200, 3)




>>>array([0.37454012, 0.95071431, 0.73199394])

YY = np.matmul(X,w)

>>>  ValueError: Shape of passed values is (200, 1), indices imply (200, 3)"

with DOT

YY = np.dot(X,w)
# no error message
>>>array([ 2.59206877,  1.06842193,  2.18533396,  2.11366346,  0.28505879, …


>>> (200, )

Here is a comparison with np.einsum to show how the indices are projected

np.allclose(np.einsum('ijk,ijk->ijk', a,b), a*b)        # True 
np.allclose(np.einsum('ijk,ikl->ijl', a,b), a@b)        # True
np.allclose(np.einsum('ijk,lkm->ijlm',a,b), a.dot(b))   # True