Re: Applying function to vector by index

[Andy Adler]

I've got most of a patch in place, but there's all
sorts of weird side cases.

For example, what should sparse(eye(2)).^(1+1i) give?

Matlab gives:
>> full(sparse(eye(2)).^(1+1i))

ans =

     1     0
     0     1

>> full(full(eye(2)).^(1+1i))

ans =

   1.0000             NaN +    NaNi
      NaN +    NaNi   1.0000


That's because a^(b+ci) = e^(ln(a))*(b+ci)
and ln(a) is NaN

Should we follow this behaviour?


Andy
--
Andy Adler,
Assistant Professor, School of Information Technology and Engineering,
University of Ottawa, Tel:1(613)562-5800 X 2345, Email:address@hidden


On Mon, 9 Dec 2002, Paul Kienzle wrote:

> 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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