[Scilab-users] ?==?utf-8?q? {EXT} need a more efficient and faster code: suggestions welcome

[Antoine Monmayrant]

Hello Stéphane,

Sorry to hijack the discussion but I didn't know that there was such a difference between A.*A and A.^2.
Could you tell us more about it?
Why is is twice faster to use the A.*A form?
Is this documented somewhere?

Cheers,

Antoine 
 
 
Le Mercredi, Janvier 31, 2018 10:53 CET, Stéphane Mottelet <stephane.mottelet at utc.fr> a écrit: 
 
> Replacing
> 
>      MinDist=[MinDist sqrt(min(sum(DIFF.^2,2)))];
> 
> by
> 
>      MinDist=[MinDist sqrt(min(sum(DIFF.*DIFF,2)))];
> 
> will be at least twice faster. Crunching elapsed time could be done by 
> using parallel_run (with 5.5.2 version) if you have a multi-core processor.
> 
> S.
> 
> Le 31/01/2018 à 09:36, Dang Ngoc Chan, Christophe a écrit :
> > Hello,
> >
> > The following suggestions will probably not have a drastic influence
> > (I don't see how it could be more vectorised)
> > but his a little thing I see:
> >
> >> De : users [mailto:users-bounces at lists.scilab.org] De la part de Heinz Nabielek
> >> Envoyé : mercredi 31 janvier 2018 00:13
> >>
> >>     MinDist=[MinDist sqrt(min(sum(DIFF.^2,2)))];
> > Maybe you could concatenate the squares of the distance
> > and then compute the square root of the whole vector in the end:
> >
> > sqMinDist=[sqMinDist min(sum(DIFF.^2,2))];
> >
> > …
> >
> > end
> >
> > …
> >
> > MinDist = sqrt(sqMinDist)
> >
> > Hope this helps,
> >
> > Regards
> >
> > --
> > Christophe Dang Ngoc Chan
> > Mechanical calculation engineer
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> 
> -- 
> Stéphane Mottelet
> Ingénieur de recherche
> EA 4297 Transformations Intégrées de la Matière Renouvelable
> Département Génie des Procédés Industriels
> Sorbonne Universités - Université de Technologie de Compiègne
> CS 60319, 60203 Compiègne cedex
> Tel : +33(0)344234688
> http://www.utc.fr/~mottelet
> 
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