[Numpy-discussion] efficient norm of a vector

[Bill Baxter]

There is numpy.linalg.norm.

Here's what it does:

def norm(x, ord=None):
    x = asarray(x)
    nd = len(x.shape)
    if ord is None: # check the default case first and handle it immediately
        return sqrt(add.reduce((x.conj() * x).ravel().real))
    if nd == 1:
        if ord == Inf:
            return abs(x).max()
        elif ord == -Inf:
            return abs(x).min()
        elif ord == 1:
            return abs(x).sum() # special case for speedup
        elif ord == 2:
            return sqrt(((x.conj()*x).real).sum()) # special case for speedup
        else:
            return ((abs(x)**ord).sum())**(1.0/ord)
    elif nd == 2:
        if ord == 2:
            return svd(x, compute_uv=0).max()
        elif ord == -2:
            return svd(x, compute_uv=0).min()
        elif ord == 1:
            return abs(x).sum(axis=0).max()
        elif ord == Inf:
            return abs(x).sum(axis=1).max()
        elif ord == -1:
            return abs(x).sum(axis=0).min()
        elif ord == -Inf:
            return abs(x).sum(axis=1).min()
        elif ord in ['fro','f']:
            return sqrt(add.reduce((x.conj() * x).real.ravel()))
        else:
            raise ValueError, "Invalid norm order for matrices."
    else:
        raise ValueError, "Improper number of dimensions to norm."



--bb



On 3/14/07, lorenzo bolla <lbolla at gmail.com> wrote:
> Hi all,
> just a quick (and easy?) question.
> what is the best (fastest) way to implement the euclidean norm of a vector,
> i.e. the function:
>
> import scipy as S
> def norm(x):
>    """normalize a vector."""
>    return S.sqrt(S.sum(S.absolute(x)**2))
>
> ?
>
> thanks in advance,
> Lorenzo.
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