[Numpy-discussion] efficient norm of a vector

[lorenzo bolla]

thanks. I hadn't seen it.

anyway, from very rough benchmarks I did, the quickest and easiest way of
computing the euclidean norm of a 1D array is:
n = sqrt(dot(x,x.conj()))
much faster than:
n = sqrt(sum(abs(x)**2))
and much much faster than:
n = scipy.linalg.norm(x)

regards,
lorenzo.


On 3/14/07, Bill Baxter <wbaxter at gmail.com> wrote:
>
> 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.
> > _______________________________________________
> > Numpy-discussion mailing list
> > Numpy-discussion at scipy.org
> > http://projects.scipy.org/mailman/listinfo/numpy-discussion
> >
> >
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