Multiply several matrices in numpy

Question

Suppose you have n square matrices A1,...,An. Is there anyway to multiply these matrices in a neat way? As far as I know dot in numpy accepts only two arguments. One obvious way is to define a function to call itself and get the result. Is there any better way to get it done?

Answer 1

If you compute all the matrices a priori then you should use an optimization scheme for matrix chain multiplication. See this Wikipedia article.

Answer 2

Another way to achieve this would be using einsum, which implements the Einstein summation convention for NumPy.

To very briefly explain this convention with respect to this problem: When you write down your multiple matrix product as one big sum of products, you get something like:

P_im = sum_j sum_k sum_l A1_ij A2_jk A3_kl A4_lm

where P is the result of your product and A1, A2, A3, and A4 are the input matrices. Note that you sum over exactly those indices that appear twice in the summand, namely j, k, and l. As a sum with this property often appears in physics, vector calculus, and probably some other fields, there is a NumPy tool for it, namely einsum.

In the above example, you can use it to calculate your matrix product as follows:

P = np.einsum( "ij,jk,kl,lm", A1, A2, A3, A4 )

Here, the first argument tells the function which indices to apply to the argument matrices and then all doubly appearing indices are summed over, yielding the desired result.

Note that the computational efficiency depends on several factors (so you are probably best off with just testing it):

• Why is numpy's einsum slower than numpy's built-in functions?
• Why is numpy's einsum faster than numpy's built in functions?

Answer 3

This might be a relatively recent feature, but I like:

A.dot(B).dot(C)

or if you had a long chain you could do:

reduce(numpy.dot, [A1, A2, ..., An])

Update:

There is more info about reduce here. Here is an example that might help.

>>> A = [np.random.random((5, 5)) for i in xrange(4)]
>>> product1 = A[0].dot(A[1]).dot(A[2]).dot(A[3])
>>> product2 = reduce(numpy.dot, A)
>>> numpy.all(product1 == product2)
True

Update 2016: As of python 3.5, there is a new matrix_multiply symbol, @:

R = A @ B @ C

Answer 4

A_list = [np.random.randn(100, 100) for i in xrange(10)]
B = np.eye(A_list[0].shape[0])
for A in A_list:
    B = np.dot(B, A)

C = reduce(np.dot, A_list)

assert(B == C)

Answer 5

Resurrecting an old question with an update:

As of November 13, 2014 there is now a np.linalg.multi_dot function which does exactly what you want. It also has the benefit of optimizing call order, though that isn't necessary in your case.

Note that this available starting with numpy version 1.10.