[Numpy-discussion] finding close together points.

[Christopher Barker]

James Bergstra wrote:
> In some cases a brute-force approach is also good.

true.

> If r is a matrix of shape Nx2:
> 
> (r*r).sum(axis=1) -2 * numpy.dot(r, r.T) +
> (r*r).sum(axis=1).reshape((r.shape[0], 1)) < thresh**2
> 
> It's brute force, but it takes advantage of fast matrix multiplication.

I'm more concerned about memory -- doesn't this use N^2 memory? Which 
could be an issue here.

-Chris


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Christopher Barker, Ph.D.
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