A loopless 3D matrix multiplication in python

I am looking to do the following operation in python (numpy).

Matrix A is M x N x R
Matrix B is N x 1 x R

Matrix multiply AB = C, where C is a M x 1 x R matrix. Essentially each M x N layer of A (R of them) is matrix multiplied independently by each N x 1 vector in B. I am sure this is a one-liner. I h开发者_开发百科ave been trying to use tensordot(), but I that seems to be giving me answers that I don't expect.

I have been programming in Igor Pro for nearly 10 years, and I am now trying to convert pages of it over to python.

Sorry for the necromancy, but this answer can be substantially improved upon, using the invaluable np.einsum.

import numpy as np

D,M,N,R = 1,2,3,4
A = np.random.rand(M,N,R)
B = np.random.rand(N,D,R)

print np.einsum('mnr,ndr->mdr', A, B).shape

Note that it has several advantages: first of all, its fast. np.einsum is well-optimized generally, but moreover, np.einsum is smart enough to avoid the creation of an MxNxR temporary array, but performs the contraction over N directly.

But perhaps more importantly, its very readable. There is no doubt that this code is correct; and you could make it a lot more complicated without any trouble.

Note that the dummy 'D' axis can simply be dropped from B and the einsum statement if you wish.

numpy.tensordot() is the right way to do it:

a = numpy.arange(24).reshape(2, 3, 4)
b = numpy.arange(12).reshape(3, 1, 4)
c = numpy.tensordot(a, b, axes=[1, 0]).diagonal(axis1=1, axis2=3)

Edit: The first version of this was faulty, and this version computes more han it should and throws away most of it. Maybe a Python loop over the last axis is the better way to do it.

Another Edit: I've come to the conclusion that numpy.tensordot() is not the best solution here.

c = (a[:,:,None] * b).sum(axis=1)

will be more efficient (though even harder to grasp).

Another way to do it (easier for those not familiar with Einstein notation, like me) is np.matmul(). The important thing is just to have the matching dimensions ((M, N) x (N, 1)) in the last two indices. For this use np.transpose() Example:

M, N, R = 4, 3, 10
A = np.ones((M, N, R))
B = np.ones((N, 1, R))

# have the matching dimensions at the very end
C = np.matmul(np.transpose(A, (2, 0, 1)), np.transpose(B, (2, 0, 1))) 
C = np.transpose(C, (1, 2, 0))

print(A.shape)
# out: #(4, 3, 10)
print(B.shape)
# out: #(3, 1, 10)
print(C.shape)
# out: #(4, 1, 10)