Test of various cross products with NumPy

test_cross.py

from numpy import *

N = 10
a = zeros((3, N, N, N))+1
b = zeros((3, N, N, N))+2
c = zeros((3, N, N, N))

a1 = zeros((N, N, N, 3))+1
b1 = zeros((N, N, N, 3))+2
c1 = zeros((N, N, N, 3))

def cross1(c, a, b):
    c[0] = a[1]*b[2] - a[2]*b[1]
    c[1] = a[2]*b[0] - a[0]*b[2]
    c[2] = a[0]*b[1] - a[1]*b[0]
    return c

def cross2(c, a, b):
    c[:] = cross(a, b, axisa=0, axisb=0, axisc=0)
    return c

def cross3(c, a, b):
    c[..., 0] = a[..., 1]*b[..., 2] - a[..., 2]*b[..., 1]
    c[..., 1] = a[..., 2]*b[..., 0] - a[..., 0]*b[..., 2]
    c[..., 2] = a[..., 0]*b[..., 1] - a[..., 1]*b[..., 0]
    return c

def cross4(c, a, b):
    c[:] = cross(a, b, axisa=3, axisb=3, axisc=3)
    return c

if __name__ == "__main__":
    import timeit
    print timeit.timeit("c = cross1(c, a, b)", setup="from __main__ import cross1, c, a, b", number=100)
    print timeit.timeit("c = cross2(c, a, b)", setup="from __main__ import cross2, c, a, b", number=100)
    print timeit.timeit("c1 = cross3(c1, a1, b1)", setup="from __main__ import cross3, c1, a1, b1", number=100)
    print timeit.timeit("c1 = cross4(c1, a1, b1)", setup="from __main__ import cross4, c1, a1, b1", number=100)
    

Running it on my computer shows that first option is fastest
In [1]: run test_cross
0.00189399719238
0.0251979827881
0.00320816040039
0.0264158248901