Cross product for vectors with more than 10 dimensions in Python

Question

I currently have two vectors of shape (300L,) each. However, when I am trying to compute cross product between them through numpy cross function, it's throwing up the following error: z = np.cross(x,y)

C:\Users\SMG059\AppData\Local\Continuum\Anaconda\lib\site-packages\numpy\core\numeric.pyc in cross(a, b, axisa, axisb, axisc, axis) 1525 "(dimension must be 2 or 3)") 1526 if a.shape[-1] not in (2, 3) or b.shape[-1] not in (2, 3): -> 1527 raise ValueError(msg) 1528 1529 # Create the output array

ValueError: incompatible dimensions for cross product (dimension must be 2 or 3)

Can someone please tell me how to compute cross product for large vectors?

The vector cross product is only defined in three dimensions. Do you mean you have two arrays, each of 300, 3-dimensional vectors? — xnx
I think what you're looking for is the determinant of a matrix. The cross product of two 3-length vectors is calculated using a determinant. Is that correct? docs.scipy.org/doc/numpy/reference/generated/… — Charlie Haley
Also worth noting: Cross product only exists for vectors of 3 and 7 dimensions. math.stackexchange.com/questions/424482/… math.stackexchange.com/questions/720813/… — Charlie Haley
I actually want to determine the angle between the arrays. Is there a way to do that? That's why I needed cross product. — Salamander
Are you sure it isn't the dot product that you're after, then? np.dot will work with an arbitrary number of dimensions. — xnx

BlivetWidget BlivetWidget · Accepted Answer · 2015-08-04T15:19:29

Based on your comment about wanting to know the angle between the arrays, I think you do in fact want the dot product. For example, here is a function that will give you the angle between two vectors.

def angle(vector1, vector2):
    # cos(theta) = v1 dot v2 / ||v1|| * ||v2||
    import numpy
    numerator = numpy.dot(vector1, vector2)
    denominator = numpy.linalg.norm(vector1) * numpy.linalg.norm(vector2)
    x = numerator / denominator if denominator else 0
    return numpy.arccos(x)

Cross product for vectors with more than 10 dimensions in Python

1 Answers