CODEWITHSUNDEEP
pythonnumpy
Eric Laufer
Eric Laufer

Reputation: 1

Python: Using Tensordot for tensor x matrix multiplication

Hi im tying to multiply a tensor with a matrix in the following fashion:

dimensions

W: a x b x c

V: a x c

I want Z such that

Z[i]=dot(W[i],V[i])

Z is then of dimension a x ( (b x c) . (c x 1)), so (a x b)

Ive tried numpy.tensordot to do this but havent been able to. Can it do what I want? If not how can I do this WITHOUT loops.

Basically the equivalent of

def f(W,V):    
    Z=[]    
    for i in range(len(W)):    
        Z.append(dot(W[i],V[i]))    
    return Z

Thanks

edit: Specifically is this achievable with tensordot?

Upvotes: 0

Views: 2296

Answers (2)

Elmar Zander
Elmar Zander

Reputation: 1516

What about

import numpy as np

a,b,c=3,5,6

r=np.random.random
W = r((a,b,c))
V = r((a,c))

Z = np.sum(W*V[:,np.newaxis,:],axis=2)

Doesn't use loops or newer features and should be reasonably fast. Comparing with "z_loop" from J.F. Sebastian's post:

print np.sum(np.abs(Z-z_loop(W,V)))

gives 4.99600361081e-16.

Upvotes: 2

jfs
jfs

Reputation: 414865

np.einsum("abc,ac -> ab", w, v):

import numpy as np

def z_loop(w,v): # define it to check that `einsum()` gives necessary result
    z = np.empty(w.shape[:-1], dtype=w.dtype)
    for i in range(z.shape[0]):
        z[i,:] = np.dot(w[i,:], v[i,:])
    return z

w = np.random.uniform(size=(3,4,5))
v = np.random.uniform(size=w.shape[::2])
assert np.allclose(z_loop(w, v), np.einsum('abc,ac -> ab', w, v))

There might be simpler variants (via dot(), .reshape()) but einsum() is the most obvious for the task description.

def z_dot(w, v):
    z = np.dot(w, v[:,...,np.newaxis])
    z = z.reshape(z.shape[:-1])
    return np.diagonal(z, axis2=-1).T

assert np.allclose(z_dot(w, v), np.einsum('abc,ac -> ab', w, v))

Upvotes: 5

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