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 From: Mike Miller Subject: Re: Easy about Matrices Date: Mon, 18 Apr 2005 22:15:54 -0500 (CDT)

```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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```