methods that return cross product of two 3xN arrays (e.g. time-dependent vectors)

Question

I have two vectors evaluated at say 10,000 points as two [3, 10000] arrays. Reading the definitions for np.cross and np.einsum I don't understand how to use them without a loop to get the cross product here.

This is what I do now, just hard-wired arrays. Can either of those, or anything else within the numpy or scipy universe do this as fast? I'm thinking that it would be nicer to use something standard like np.cross or np.einsum even if it's the same speed as my method, but I don't see how without looping.

def cross_these(a, b):

    c0 = a[1]*b[2] - a[2]*b[1]
    c1 = a[2]*b[0] - a[0]*b[2]
    c2 = a[0]*b[1] - a[1]*b[0]

    return np.vstack((c0, c1, c2))

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| python   | arrays   | scipy   | numpy   2016-12-20 13:12 2 Answers

Answers to methods that return cross product of two 3xN arrays (e.g. time-dependent vectors) ( 2 )

  1. 2016-12-20 14:12

    You need the optional keyword arguments to np.cross that specify which axes of the input (and possibly output) arrays correspond to Cartesian coordinates, and you're done:

    import numpy as np
    a = np.random.rand(3,5)
    b = np.random.rand(3,5)
    c = np.cross(a, b, axisa=0, axisb=0, axisc=0)
    
    # check if each row is equal to the simple cross-product
    all(np.allclose(np.cross(a,b,axisa=0,axisb=0)[k],
                    np.cross(a[:,k],b[:,k])) for k in range(a.shape[1]))
    
  2. 2016-12-20 14:12

    Here's a vectorized approach -

    d0 = np.array([[1,2,0],[2,0,1]]) # Scaling values
    d1 = np.array([[2,0,1],[1,2,0]])
    out = a[d0[0]]*b[d1[0]] - a[d0[1]]*b[d1[1]]
    

    As one-liner -

    a[[1,2,0]]*b[[2,0,1]] - a[[2,0,1]]*b[[1,2,0]]
    

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