Faster alternative to numpy.einsum for taking the "element-wise" dot product of two lists of vectors?

Question

Let's say you're given two arrays of vectors:

v1 = np.array([ [1, 2], [3, 4] ]) v2 = np.array([ [10, 20], [30, 40]])

We would like to generate an array that is equivalent to:

v3 = np.array([ np.dot(v1[0], v2[0]), np.dot(v1[1], v2[1]) ])

Currently I use:

v3 = np.einsum('ij,ij->i', v1, v2)

However, I do this a lot in my code, so speed ups here would be very helpful for me.

How could we speed it up? np.einsum is already quite efficient, but I wonder if for this particular use-case, there is a faster way?

If your arrays are really big, check stackoverflow.com/questions/17527340/… To see some comparisons of numpy.einsum to other ways — usethedeathstar
– usethedeathstar, Commented Jun 17, 2014 at 12:10

hpaulj · Accepted Answer · 2014-06-16 06:50:35Z

einsum does the best of 3 options that I can think of:

In [73]: timeit v3=np.einsum('ij,ij->i',v1,v2) 100000 loops, best of 3: 5.14 us per loop In [74]: timeit np.diag(np.dot(v1,v2.T)) 100000 loops, best of 3: 7.43 us per loop In [75]: timeit np.sum(v1*v2,axis=1) 100000 loops, best of 3: 16.8 us per loop

Several questions to ask:

is this calculation really that expensive ?
if it is relatively expensive do you have to doing so often ?
can you consolidate the einsum calls - concatenate arrays ?

Thanks for this helpful answer. Actually, suggestion 3 might be possible. Here is a post about the use-case: codereview.stackexchange.com/questions/54347/…

Fnord · Accepted Answer · 2016-07-26 04:03:31Z

Try inner1d

import numpy as np import cProfile from numpy.core.umath_tests import inner1d v1 = np.random.random((10**7,2,)) # 10 million random vectors v2 = np.random.random((10**7,2,)) # 10 million random vectors v3 = np.einsum('ij,ij->i', v1, v2) # einsum v4 = inner1d(v1,v2) # inner1d (~2x faster than einsum) cProfile.run("np.einsum('ij,ij->i', v1, v2)") # cProfile: 3 function calls in 0.065 seconds cProfile.run("inner1d(v1,v2)") # cProfile: 2 function calls in 0.033 seconds print np.allclose(v3,v4) # Returns True

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