[SciPy-Dev] Akima Interpolation

Evgeni Burovski evgeny.burovskiy@gmail....
Tue Sep 10 12:13:22 CDT 2013


Christoph Gohlke has a python-friendly implementation:
http://www.lfd.uci.edu/~gohlke/code/akima.c.html

Quick googling also shows
http://code.google.com/p/miyoshi/source/browse/trunk/common/common.f90?r=88
Which seems to be MIT-licensed.
Haven't read the code though.




On Tue, Sep 10, 2013 at 10:54 AM, Andreas Hilboll <lists@hilboll.de> wrote:

> On 30.08.2013 18:15, Pauli Virtanen wrote:
> > 29.08.2013 19:01, Andreas Hilboll kirjoitti:
> >> I'd like to have Akima interpolation in scipy.interpolate, as I
> >> couldn't find any.
> >
> >> 1.) Did I miss something and there actually *is* an Akima
> >> interpolation in scipy? 2.) Is there interest in having Akima
> >> interpolation in Scipy? 3.) Does anyone have a good recommendation
> >> which implementation I should consider for pulling into Scipy?
> >
> > 1) I don't think so, unless pchip is related.
> >
> > 2) Perhaps. Is it useful?
>
> I would claim yes. Akima splines are more robust to outliers, and I
> think they would make a nice addition
>
> >
> > 3) No idea.
> >
> > One thing to look out: based on a quick look, Akima interpolation is a
> > form of spline interpolation, so it is probably possible to represent
> > the result as a B-spline. If so, FITPACK's spline representation
> > should be used, i.e., the same (t, c, k) format as in splrep.
> >
> > This allows reusing `splev` for evaluating the spline values, which
> > will make life easier later on when we clean up scipy.interpolate.
>
> I agree it would be nice to store the spline in the same tck format as
> used by FITPACK. Unfortunately, all references I could find directly
> compute the polynomial coefficients of the interpolants, and I couldn't
> find a refernce about how to derive the B-spline coefficients, i.e., the
> c, from the polynomial coefficients. I checked the bspline work by
> Chuck, but there isn't any B-spline coeffient calculation from the pp
> representation (yet?). Any ideas where to look?
>
> Andreas.
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