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From: dave levine on 9 Feb 2010 13:47
"Oleg Komarov" <oleg.komarovRemove.this(a)hotmail.it> wrote in message <hks9qv$muq$1(a)fred.mathworks.com>...
> "dave levine"
> > Oleg,
> > Thank you. Do I just place this inside a single for loop? I need a whole vector of values of n1.
> >
> > Also, how does this histc function work? I need to be able to translate this code into C++ after I get this matlab algorithm working.
> >
> > Thanks
>
> The power of matlab is that you don't need loops here.
> n2 is already a vector of values, same as n1 (see te equal test above).
>
> histc is a built-in matlab fcn, so code isn't available.
>
> Matlab is definitely not the starting point to write in C++ style since its language tries to avoid as much as possible loops.
>
> Oleg


This histc works great when the number of values is small, but i need something that is zippy with 200,000+ samples to process. Right now it is doing less than 5k.

It ran with 48000 data points: Elapsed time is 389.979217 seconds.

Is there another way?
From: Oleg Komarov on 9 Feb 2010 13:58
> This histc works great when the number of values is small, but i need something that is zippy with 200,000+ samples to process. Right now it is doing less than 5k.
>
> It ran with 48000 data points: Elapsed time is 389.979217 seconds.
>
> Is there another way?


YOUR LOOP == > 34.923162 seconds.
Using time1 and time2 as you supplied them:
> edges = [time1 inf];
> [trash,n2] = histc(time2,edges);
HISTC == > Elapsed time is 0.005901 seconds.

Oleg


From: dave levine on 9 Feb 2010 14:05
"Oleg Komarov" <oleg.komarovRemove.this(a)hotmail.it> wrote in message <hksb7r$m7r$1(a)fred.mathworks.com>...
> > This histc works great when the number of values is small, but i need something that is zippy with 200,000+ samples to process. Right now it is doing less than 5k.
> >
> > It ran with 48000 data points: Elapsed time is 389.979217 seconds.
> >
> > Is there another way?
>
>
> YOUR LOOP == > 34.923162 seconds.
> Using time1 and time2 as you supplied them:
> > edges = [time1 inf];
> > [trash,n2] = histc(time2,edges);
> HISTC == > Elapsed time is 0.005901 seconds.
>
> Oleg
>

Oops, I apologize for being dense. This works great, even with huge sets! Even with 10x the points: Elapsed time is 0.087216 seconds. It does not ever increase in computational cost like nested for loops do.

Ill try to recreate the histc function in C++, or create an equivalent

Thanks Oleg
From: Oleg Komarov on 9 Feb 2010 14:12
"Oleg Komarov"
> > This histc works great when the number of values is small, but i need something that is zippy with 200,000+ samples to process. Right now it is doing less than 5k.
> >
> > It ran with 48000 data points: Elapsed time is 389.979217 seconds.
> >
> > Is there another way?
>
>
> YOUR LOOP == > 34.923162 seconds.
> Using time1 and time2 as you supplied them:
> > edges = [time1 inf];
> > [trash,n2] = histc(time2,edges);
> HISTC == > Elapsed time is 0.005901 seconds.
>
> Oleg

Now I understand, samples... 200 000 * 50 000 datapoints = 10e10...*0.006 secs (per loop) = 1666 hrs...

or 400 secs * 40 = 4,5 hrs...well it's not so bad after all... :)

If you give us more details about what you're trying to accomplish...maybe we can help.

Oleg
From: Oleg Komarov on 9 Feb 2010 14:13
"dave levine" <drp2_delete.this(a)unh.edu> wrote in message <hksblh$k59$1(a)fred.mathworks.com>...
> "Oleg Komarov" <oleg.komarovRemove.this(a)hotmail.it> wrote in message <hksb7r$m7r$1(a)fred.mathworks.com>...
> > > This histc works great when the number of values is small, but i need something that is zippy with 200,000+ samples to process. Right now it is doing less than 5k.
> > >
> > > It ran with 48000 data points: Elapsed time is 389.979217 seconds.
> > >
> > > Is there another way?
> >
> >
> > YOUR LOOP == > 34.923162 seconds.
> > Using time1 and time2 as you supplied them:
> > > edges = [time1 inf];
> > > [trash,n2] = histc(time2,edges);
> > HISTC == > Elapsed time is 0.005901 seconds.
> >
> > Oleg
> >
>
> Oops, I apologize for being dense. This works great, even with huge sets! Even with 10x the points: Elapsed time is 0.087216 seconds. It does not ever increase in computational cost like nested for loops do.
>
> Ill try to recreate the histc function in C++, or create an equivalent
>
> Thanks Oleg

Ah ok I got it right the first time.

Oleg
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