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How to do element wise matrix multiply using numpy

Tags:

python

numpy

I have two numpy arrays, a in size (20*3*3) and b in size (3*3). Let a=(a1, a2, ..., a20). I want to calculate the matrix product element wise like this: c=(c1, c2, ..., c20), ci=b.Taib, i=1~20. How can I do it efficiently using numpy?

A slow version using for loop is like this:

a = np.random.sample((20, 3, 3))
b = np.random.sample((3, 3))
c = np.zeros_like(a)
for i0, ai in enumerate(a):
    c[i0] = np.dot(b.T, np.dot(ai, b))
like image 260
Ji Zhang Avatar asked Jun 19 '26 05:06

Ji Zhang


2 Answers

You can try np.matmul(b.T, np.dot(a,b)):

import numpy as np
import pandas as pd

a = np.random.sample((4, 3, 3))
b = np.random.sample((3, 3))
c = np.zeros_like(a)

# using for loop
for i0, ai in enumerate(a):
    c[i0] = np.dot(b.T, np.dot(ai, b))

# alternative method
e = np.zeros_like(a)
e = np.matmul(b.T, np.dot(a,b))

# checking for equal
print(np.array_equal(c, e))
like image 160
student Avatar answered Jun 24 '26 12:06

student


You can just put your operation in a vectorized form because your inputs are NumPy arrays. No need of explicit for loop and indexing.

P.S: Thanks to @yatu who found that the answer was not the same shape. Now I added the swapaxes to get the consistent answer as OP's approach

np.random.seed(1)

a = np.random.sample((4, 3, 3))
b = np.random.sample((3, 3))

c = np.dot(b.T, np.dot(a, b)).swapaxes(0,1)
print (c)

[[[0.96496962 1.30807122 0.55382266]
  [1.42300972 1.98975139 0.81871374]
  [0.32358338 0.45493059 0.1346777 ]]

 [[1.46772447 2.15650254 0.87555186]
  [2.26335921 3.33689922 1.28679305]
  [0.71561413 0.96507585 0.54309736]]

 [[1.50660527 2.36946435 0.59771395]
  [2.49705244 3.76328176 1.06274954]
  [0.96090846 1.43636151 0.31807679]]

 [[1.03706878 1.94107476 0.61884642]
  [1.74739926 3.07419808 1.03537019]
  [0.59565039 1.09721382 0.37283626]]]
like image 37
Sheldore Avatar answered Jun 24 '26 13:06

Sheldore