fastest way to obtain cross product - python-2.7

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

The performance of np.cross improved significantly in the 1.9.x release of numpy.
%timeit explicitcross_vlast()
%timeit explicitcross_vfirst()
%timeit npcross_vlast()
%timeit npcross_vfirst()
%timeit npcross_vfirst_axisc()
These are the timings I get for 1.8.0
100 loops, best of 3: 4.47 ms per loop
100 loops, best of 3: 4.41 ms per loop
10 loops, best of 3: 29.1 ms per loop
10 loops, best of 3: 29.3 ms per loop
10 loops, best of 3: 30.6 ms per loop
And these the timings for 1.9.0:
100 loops, best of 3: 4.62 ms per loop
100 loops, best of 3: 4.19 ms per loop
100 loops, best of 3: 4.05 ms per loop
100 loops, best of 3: 4.09 ms per loop
100 loops, best of 3: 4.24 ms per loop
I suspect that the speedup was introduced by merge request #4338.

Answer 2

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")

Answer 3

Simply changing your vlast to
def stacked_vlast(a,b):
x = a[:,1]*b[:,2] - a[:,2]*b[:,1]
y = a[:,2]*b[:,0] - a[:,0]*b[:,2]
z = a[:,0]*b[:,1] - a[:,1]*b[:,0]
return np.array([x,y,z]).T
i.e. replacing the column assignment with stacking, as the (old) cross does, slows the speed by 5x.
When I use a local copy of the development cross function, I get a minor speed improvement over your explicit_vlast. That cross uses the out parameter in an attempt to cut down on temporary arrays, but my crude tests suggest that it doesn't make much difference in speed.
https://github.com/numpy/numpy/blob/master/numpy/core/numeric.py
If you explicit version works, I wouldn't upgrade numpy just to get this new cross.