Re: Applying function to vector by index

[Paul Kienzle]

That's because S.^2 implicitly expands S to a full matrix 8-(
Anyone care to patch it?

Meanwhile, how about the following:

function [count, mean, diff] = spstats(v,j)

    i = [1:length(v)];
    S = sparse(i,j,v);
    count = spsum(sparse(i,j,1));
    mean = spsum(S) ./ count;
    diff = S - sparse(i,j,mean(j));
    var = spsum ( diff .* diff ) ./ (count - 1);

end

Paul Kienzle
address@hidden

On Mon, Dec 09, 2002 at 05:01:47PM -0000, Iago Mosqueira wrote:
> Thanks, it works fine. I was doing just this for estimating the mean, but I
> was somehow getting into memory problems (a 65000x1 vector, turning into a
> sparse matrix with 15 datapoints). It's fine now. Yes, some general way
> 'sapply' style would be nice...
> 
> Thanks,
> 
> 
> iago
> 
> 
> ----- Original Message -----
> From: "Paul Kienzle" <address@hidden>
> To: "Iago Mosqueira" <address@hidden>
> Cc: "octave-help" <address@hidden>
> Sent: 09 December 2002 16:56
> Subject: Re: Applying function to vector by index
> 
> 
> > Iago,
> >
> > Sparse matrices are a likely candidate here.  Split your data with
> > one column per bin:
> >
> > n = length(d);
> > S = sparse(1:n,idx,d);
> >
> > Then the stats are:
> >
> > count = spsum(sparse(1:n,idx,1));
> >         mean = spsum(S) ./ count;
> >         var = spsum( (S - sparse(1:n,idx,mean(idx))).^2 ) ./ (count-1);
> >
> > Not as good as a generic split/apply operations I agree, but even
> > with split/apply performance wouldn't be much better than a loop since
> > you would still be interpreting var(x) for each bin separately.
> >
> > Paul Kienzle
> > address@hidden
> >
> > On Mon, Dec 09, 2002 at 04:22:17PM -0000, Iago Mosqueira wrote:
> > > Hi,
> > >
> > > I have a long vector classified in ten bins of different length
> according to
> > > another vector. I want to estimate the variance, or any other function,
> of
> > > the values from the first vector for each of the categories described by
> the
> > > second, without using a loop. Is there any obvious way of doing this I
> am
> > > overlooking? In R one could do
> > >
> > >     sapply(split(d,idx),var)
> > >
> > > where d is my data vector and idx the index vector.
> > >
> > > Many thanks,
> > >
> > >
> > > iago
> > >
> > >
> > >
> > >
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> 
> 
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