Can numpy einsum() perform a cross-product between segments of a trajectory

Question

I perform the cross product of contiguous segments of a trajectory (xy coordinates) using the following script:

In [129]:
def func1(xy, s):
    size = xy.shape[0]-2*s
    out = np.zeros(size)
    for i in range(size):
        p1, p2 = xy[i], xy[i+s]     #segment 1
        p3, p4 = xy[i+s], xy[i+2*s] #segment 2
       out[i] = np.cross(p1-p2, p4-p3)
    return out

def func2(xy, s):
    size = xy.shape[0]-2*s
    p1 = xy[0:size]
    p2 = xy[s:size+s]
    p3 = p2
    p4 = xy[2*s:size+2*s]

    tmp1 = p1-p2
    tmp2 = p4-p3
    return tmp1[:, 0] * tmp2[:, 1] - tmp2[:, 0] * tmp1[:, 1]


In [136]:
xy = np.array([[1,2],[2,3],[3,4],[5,6],[7,8],[2,4],[5,2],[9,9],[1,1]])
func2(xy, 2)

Out[136]:
array([ 0, -3, 16,  1, 22])

func1 is particularly slow because of the inner loop so I rewrote the cross-product myself (func2) which is orders of magnitude faster.

Is it possible to use the numpy einsum function to make the same calculation?

Answer 1

einsum computes sums of products only, but you could shoehorn the cross-product into a sum of products by reversing the columns of tmp2 and changing the sign of the first column:

def func3(xy, s):
    size = xy.shape[0]-2*s
    tmp1 = xy[0:size] - xy[s:size+s]
    tmp2 = xy[2*s:size+2*s] - xy[s:size+s]
    tmp2 = tmp2[:, ::-1]
    tmp2[:, 0] *= -1
    return np.einsum('ij,ij->i', tmp1, tmp2)

---

But func3 is slower than func2.

In [80]: xy = np.tile(xy, (1000, 1))

In [104]: %timeit func1(xy, 2)
10 loops, best of 3: 67.5 ms per loop

In [105]: %timeit func2(xy, 2)
10000 loops, best of 3: 73.2 µs per loop

In [106]: %timeit func3(xy, 2)
10000 loops, best of 3: 108 µs per loop

Sanity check:

In [86]: np.allclose(func1(xy, 2), func3(xy, 2))
Out[86]: True

I think the reason why func2 is beating einsum here is because the cost of setting of the loop in einsum for just 2 iterations is too expensive compared to just manually writing out the sum, and the reversing and multiplying eat up some time as well.