Re: Interpolacja Liniowa

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#27545

From: rav
Date: Tue, 29 Jun 2010 09:08

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On 28 Cze, 18:49, Bart Ogryczak <b.ogryc...@gmail.com> wrote:
> On Jun 26, 8:41 pm, rav <rafalgulin...@gmail.com> wrote:
>
> > Nie dziala kiedy x2-x1 == 0 (div by zero)
> > dlatego dodalbym sprawdzenie roznicy tych wartosci do funkcji
>
> > def getInterpolated(x0,y0, x1, y1):
> >     if x1 - x0 == 0:
> >         print 'x1 i x0 can\'t be equal'
> >         return None
> >     a = (y1 - y0) / (x1 - x0)
> >     b = y0 - a * x0
> >     def interpolate(x):
> >         return a * x + b
> >     return None
>
> Pythonistycznie, to raczej będzie tak:
>
> try:
>     a = (y2 - y1) / (x2 - x1)
>     b = ((-x1) * (y2 - y1) - (x2 - x1) * (-y1)) / (x2 - x1)
>     def interpolate(x):
>         return a * x + b
>     return interpolate
> except ZeroDivisionError:
>     print 'x1 i x0 can\'t be equal'
>     return None

Przy czestym x2-x1 == 0 "if" bedzie mniej kosztowny ale jezeli ten
warunek bedzie rzadko spelniony to faktycznie wyjatek jest lepszym
rozwiazaniem.

zrodelko:
http://stackoverflow.com/questions/2522005/cost-of-exception-handlers-in-python

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From: rav <rafalgulinski@gmail.com>
Newsgroups: pl.comp.lang.python
Subject: Re: Interpolacja Liniowa
Date: Tue, 29 Jun 2010 02:08:58 -0700 (PDT)
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