Sum of Square Differences (SSD) in numpy/scipy

I am confused why you are taking i in range(3). Is that supposed to be the whole array, or just part?

Overall, you can replace most of this with operations defined in numpy:

def ssd(A,B):
    squares = (A[:,:,:3] - B[:,:,:3]) ** 2
    return numpy.sum(squares)

This way you can do one operation instead of three and using numpy.sum may be able to optimize the addition better than the builtin sum.

Just

s = numpy.sum((A[:,:,0:3]-B[:,:,0:3])**2)

(which I expect is likely just sum((A-B)**2) if the shape is always (,,3))

You can also use the sum method: ((A-B)**2).sum()

Right?

Just to mention that one can also use np.dot:

def ssd(A,B):
  dif = A.ravel() - B.ravel()
  return np.dot( dif, dif )

This might be a bit faster and possibly more accurate than alternatives using np.sum and **2, but doesn't work if you want to compute ssd along a specified axis. In that case, there might be a magical subscript formula using np.einsum.