Re: Easy about Matrices

[Mike Miller]

On Tue, 19 Apr 2005, Geordie McBain wrote:

> I won't claim this is less awkward, but it does do the job without a 
> loop:
>
>      octave> X = [1, 3, 9, 2, 4, 2, 7, 5, 3, 2, 9, 7]';
>      octave> Y = [2, 3, 9]';
>      octave> mod (find (kron (X, ones (1, length (Y))) == kron (Y', ones 
> (length (X), 1))), length (X))
>      ans =
>
>         4
>         6
>        10
>         2
>         9
>         3
>        11
>
>      octave>
>
> which gives the same ans as your index, below.


Thank you, Geordie.  That is a step forward.  It would work nicely for 
certain matrix sizes but it won't scale well if the X and Y vectors get 
very long.  By the way, we noted some years ago on this list that 
multiplication is faster than kron, when multiplication can be used, so I 
think this might work faster than the above:

mod (find (X*ones(1,length(Y)) == ones(length(X),1)*Y'), length (X))

That was a nice tip, Geordie.  Thanks again.  Of course, if anyone else 
has more ideas, I'm all ears!

Mike


> > I can see how to do it with a loop...
> >
> > index=find(X==Y(1)); for i=2:length(Y), index=[index ; find(X==Y(i))]; end
> >
> > ...but can it be done without a loop?  (Also, my method is a little 
> > awkward and I wouldn't mind hearing about how I could do it better.)



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