[Numpy-discussion] Computing the norm of an array of vectors
Tue Feb 8 12:19:57 CST 2011
On Tue, 08 Feb 2011 18:06:48 +0000, Andrew Jaffe <firstname.lastname@example.org> wrote:
> For this shape=(N,3) vector, this is not what you mean: as Robert Kern
> also has it you want axis=1, which produces a shape=(N,) (or the
> [:,newaxis] version which produces shape=(N,1).
> But what is the point of the ones(3)? I think you intend to make a new
> (N,3) array where each row duplicates the norm, so that you can then
> divide out the norms. But through the magic of broadcasting, that's not
> v/np.sqrt(sum(v**2, axis=1)[:,newaxis])
> does what you want.
Thanks! I've since realized the error in my ways. I had completely
forgotten that newaxis existed. Robert Kern's proposal is perfect.
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