On Sun, Jul 21, 2019 at 5:51 PM shall689 <address@hidden> wrote:
shivax wrote
> hi, see this code:
>
> input for example:
>
> ris=[1 0 0;0 1 0;1 0 0; 0 1 1]
> tit=[0 1 0 0]
>
> tic
> bb=0;
> for hh=1:columns(ris)
> bb=0;
> for gg=1:rows(ris)
> if tit(gg)
> bb=0;
> if ris(gg,hh)
> bb=1;
> endif
> endif
> ris(gg,hh)=bb;
> endfor
> endfor
> toc
>
>
>
> it's possible to avoid loop?
>
>
>
> --
> Sent from:
> http://octave.1599824.n4.nabble.com/Octave-General-f1599825.html
If you explain what you are trying to do, there might be a better way.
Here is what I understand of your very interesting problem. Each row in the
tit vector is associated with a row in the ris matrix. All ris rows up
until the first tit row that equals 1 should be set to zero. When a tit row
= 1, the associated row in the ris vector should be copied and over write
all ris rows until tit row equals 1 again.
I haven't been able to solve the problem, but I think repmat and find might
be helpful.
find(tit==1) will find the index of all the rows in tit that are equal to 1
and repmat(ris(row,:),row,1) can be used to repeat a specific row in ris.
If you can find the difference between the row index values of tit==1, you
can determine how many times to repeat a specific row.
If you find out how to vectorize this problem, could you please post the
solution.
Thanks,
Stephen
--
Sent from: http://octave.1599824.n4.nabble.com/Octave-General-f1599825.html