fastest way to obtain cross product

Question

It looks like calculating the cross-product of an array of vectors explicitly is a lot faster than using np.cross. I've tried vector-first and vector-last, it doesn't seem to make a difference, though that was proposed in an answer to a similar question. Am I using it wrong, or is it just slower?

The explicit calculation seems to take about 60ns per cross-product on a laptop. Is that ~roughly~ as fast as it's going to get? In this case, there doesn't seem to be any reason to go to Cython or PyPy or writing a special ufunc yet.

I also see references to the use of einsum, but I don't really understand how to use that, and suspect it is not faster.

a = np.random.random(size=300000).reshape(100000,3) # vector last
b = np.random.random(size=300000).reshape(100000,3)
c, d = a.swapaxes(0, 1),  b.swapaxes(0, 1)          # vector first

def npcross_vlast():        return np.cross(a, b)
def npcross_vfirst():       return np.cross(c, d, axisa=0, axisb=0)
def npcross_vfirst_axisc(): return np.cross(c, d, axisa=0, axisb=0, axisc=0)
def explicitcross_vlast():
    e = np.zeros_like(a)
    e[:,0] = a[:,1]*b[:,2] - a[:,2]*b[:,1]
    e[:,1] = a[:,2]*b[:,0] - a[:,0]*b[:,2]
    e[:,2] = a[:,0]*b[:,1] - a[:,1]*b[:,0]
    return e
def explicitcross_vfirst():
    e = np.zeros_like(c)
    e[0,:] = c[1,:]*d[2,:] - c[2,:]*d[1,:]
    e[1,:] = c[2,:]*d[0,:] - c[0,:]*d[2,:]
    e[2,:] = c[0,:]*d[1,:] - c[1,:]*d[0,:]
    return e
print "explicit"
print timeit.timeit(explicitcross_vlast,  number=10)
print timeit.timeit(explicitcross_vfirst, number=10)
print "np.cross"
print timeit.timeit(npcross_vlast,        number=10)
print timeit.timeit(npcross_vfirst,       number=10)
print timeit.timeit(npcross_vfirst_axisc, number=10)
print all([npcross_vlast()[7,i] == npcross_vfirst()[7,i] ==
           npcross_vfirst_axisc()[i,7] == explicitcross_vlast()[7,i] ==
           explicitcross_vfirst()[i,7] for i in range(3)]) # check one

explicit
0.0582590103149
0.0560920238495
np.cross
0.399816989899
0.412983894348
0.411231040955
True

Answer 1

First off, if you're looking to speed up your code, you should probably try and get rid of cross-products altogether. That's possible in many cases, e.g., when used in connection with dot products <a x b, c x d> = <a, c><b, d> - <a, d><b, c>.

Anyways, in case you really need explicit cross products, check out

eijk = np.zeros((3, 3, 3))
eijk[0, 1, 2] = eijk[1, 2, 0] = eijk[2, 0, 1] = 1
eijk[0, 2, 1] = eijk[2, 1, 0] = eijk[1, 0, 2] = -1

np.einsum('ijk,aj,ak->ai', eijk, a, b)
np.einsum('iak,ak->ai', np.einsum('ijk,aj->iak', eijk, a), b)

These two are equivalent to np.cross, where the second uses two einsums with two arguments each, a technique suggested in a similar question.

The results are disappointing, though: Both of these variants are slower than np.cross (except for tiny n):

The plot was created with

import numpy as np
import perfplot

eijk = np.zeros((3, 3, 3))
eijk[0, 1, 2] = eijk[1, 2, 0] = eijk[2, 0, 1] = 1
eijk[0, 2, 1] = eijk[2, 1, 0] = eijk[1, 0, 2] = -1


b = perfplot.bench(
    setup=lambda n: np.random.rand(2, n, 3),
    n_range=[2 ** k for k in range(23)],
    kernels=[
        lambda X: np.cross(X[0], X[1]),
        lambda X: np.einsum("ijk,aj,ak->ai", eijk, X[0], X[1]),
        lambda X: np.einsum("iak,ak->ai", np.einsum("ijk,aj->iak", eijk, X[0]), X[1]),
    ],
    labels=["np.cross", "einsum", "double einsum"],
    xlabel="len(a)",
)

b.save("out.png")