How to vectorize a dot product of one matrix for every row of another?

Question

If I have a matrix A and I want to get the dot product of A with every row of B.

import numpy as np

a = np.array([[1.0, 2.0],
              [3.0, 4.0]])

b = np.array([[1.0, 1.0],
              [2.0, 2.0],
              [3.0, 3.0]])

If the goal was to do it manually (or in a loop):

c = np.array([np.dot(a, b[0])])
c = np.append(c, [np.dot(a, b[1])], axis=0)
c = np.append(c, [np.dot(a, b[2])], axis=0)

print(c)

c = [[  3.   7.]
     [  6.  14.]
     [  9.  21.]]

What is the second column? – Willem Van Onsem Mar 03 '17 at 21:32 — Willem Van Onsem, Mar 03 '17 at 21:32
I'm not following the question, second column of what? – dranobob Mar 03 '17 at 21:33 — dranobob, Mar 03 '17 at 21:33

Divakar · Accepted Answer · 2017-03-03T21:52:14.443

2

With some transposing and matrix-multiplication using np.dot -

a.dot(b.T).T
b.dot(a.T)

With np.einsum -

np.einsum('ij,kj->ki',a,b)

With np.tensordot -

np.tensordot(b,a,axes=((1,1)))

Runtime test -

In [123]: a = np.random.rand(2000, 2000)
     ...: b = np.random.rand(3000, 2000)
     ...: 

In [124]: %timeit a.dot(b.T).T
     ...: %timeit b.dot(a.T)
     ...: %timeit np.einsum('ij,kj->ki',a,b)
     ...: %timeit np.tensordot(b,a,axes=((1,1)))
     ...: 
1 loops, best of 3: 234 ms per loop
10 loops, best of 3: 169 ms per loop
1 loops, best of 3: 7.59 s per loop
10 loops, best of 3: 170 ms per loop

edited Mar 03 '17 at 21:52

answered Mar 03 '17 at 21:33

Divakar

218,885
19
262
358

so elegant and simple. now I'm going to go do this by hand to understand why this works. – dranobob Mar 03 '17 at 21:35
What's the matter with `einsum` on this one? – Paul Panzer Mar 04 '17 at 03:28
@PaulPanzer Just dot products being good at sum-reduction. `einsum` comes into play only when we need to keep one or more axes aligned, not the case here. – Divakar Mar 04 '17 at 08:39

How to vectorize a dot product of one matrix for every row of another?

1 Answers1