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RE: vectorization challenge
From: |
Tim Rueth |
Subject: |
RE: vectorization challenge |
Date: |
Sat, 7 Aug 2010 03:55:35 -0700 |
> -----Original Message-----
> From: Jaroslav Hajek [mailto:address@hidden
> Sent: Thursday, August 05, 2010 11:44 PM
> To: address@hidden
> Cc: Octave-ML
> Subject: Re: vectorization challenge
>
> On Fri, Aug 6, 2010 at 6:00 AM, Tim Rueth <address@hidden> wrote:
> > I'm desperately trying to vectorize a calculation that I'm
> currently
> > doing in a for-loop. I haven't been able to figure this
> one out, and
> > my brain now hurts.
> > I have two column vectors A and B as inputs:
> >
> > A B C
> > -----------------------------------------
> > 43.94 0 0.0000
> > 44.25 0 0.7055
> > 44.34 0 0.9103
> > 44.81 0 1.9799
> > 45.00 1 2.4123
> > 44.97 0 -0.0666
> > 44.97 0 -0.0666
> > 44.66 1 -0.7555
> > 44.72 0 0.1343
> > 44.94 1 0.6269
> > 44.59 0 -0.7788
> > 43.47 0 -3.2710
> > 43.44 0 -3.3377
> > 43.41 1 -3.4045
> > 43.56 0 0.3455
> > 43.72 1 0.7141
> > 43.69 1 -0.0686
> > ...
> >
> > What I'm trying to do is calculate C without using a
> for-loop. C is
> > simply the running percent change in A since the last time
> there was a "1" in B.
> > As you might have noticed, whenever there's a "1" in B,
> there's either
> > a peak or a trough in A. I can easily break B into two
> vectors, one
> > for peaks and one for troughs, but I don't think that will help in
> > figuring out how to create C.
> >
> > If there really were only a few elements in these vectors,
> a for-loop
> > would be no big deal for me, but these vectors are very
> long and the
> > for-loop itself gets run often, hence the need to
> vectorize. Any help
> > is greatly appreciated as always.
> >
>
> Here's a solution that destroys B:
>
> b = find (B);
> B(b) = diff ([0; b-1]); B(1) = 1;
> C = 100*(A ./ A(cumsum (B)) - 1);
>
> hth
>
> --
> RNDr. Jaroslav Hajek, PhD
> computing expert & GNU Octave developer
> Aeronautical Research and Test Institute (VZLU) Prague, Czech Republic
> url: www.highegg.matfyz.cz
Thanks Jordi and Jaroslav. You guys are brilliant.
I used Jaroslav's solution shown above. I only had to fix one minor thing:
I replaced A(cumsum(B)) with A(cumsum([1; B(1:end-1)])) to get everything to
line up properly.
Thanks again,
--Tim