# More Ways to Find Matching Data

Today on the newsgroup, a user wanted help finding when values in a matrix matched some other values (see the post). There was already a solution posted when I was reading, but something about this problem kept nagging at me. So I've invested a little bit of time thinking more about the problem.

### Contents

### Sample Data

Here's sample data, and the user wants to find all the places in `A` which have values that match values in `B`. Simple enough statement.

A = [11 22 34 56 89 23 44 11 20 66 79 54 32 17 89 11 66 21 45 90] B = [11 66 44 40 90]

A = 11 22 34 56 89 23 44 11 20 66 79 54 32 17 89 11 66 21 45 90 B = 11 66 44 40 90

### Bruno's Solution - Column-wise Solution

As I said, there was already a solution when I was reading. Here it is.

RESULTS = zeros(size(A)); for i = 1: size(B,2) RESULTS = RESULTS + ( A == B(1,i) ); end RESULTS

RESULTS = 1 0 0 0 0 0 1 1 0 1 0 0 0 0 0 1 1 0 0 1

The idea here is to create the right size output, and cycle through the values in `B` (the smaller array for the user's example). Check to see where a given value in `B` matches one in `A`, and add a 1 to the `RESULTS` when those hits are found.

### Find Exact Location Matches

To be honest, I misread the question at first, and came up with the following code. However, it does **not** solve the problem as stated!

C = ~(A-repmat(B,size(A,1),1))

C = 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 1

The reason I tried the solution with `repmat` was so I could first get the right answer, and then find a solution instead with `bsxfun`. Instead, this solution (which could be quite costly due to the `repmat`) looks to match values in specific column locations. In other words, the wrong problem solved.

### Find Presence - Row-wise Solution

Wising up a very tiny amount, that I was solving the wrong problem, I next tried finding matches by row in `A`.

Z = zeros(size(A)); for k = 1:size(A,1) Z(k,:) = ismember(A(k,:),B); end Z

Z = 1 0 0 0 0 0 1 1 0 1 0 0 0 0 0 1 1 0 0 1

At least this time I get the right answer and am solving the right problem! But this normally would take longer than Bruno's
approach because the user premise was that `A` was **huge** and `B` wasn't nearly so large.

isequal(Z,RESULTS)

ans = 1

The reason I tried this approach was to then see if could convert it to something using `arrayfun`, or perhaps `cellfun`.

### Another Solution

Finally I had some coffee however! And thank goodness. The answer was in front of me all along. And I was already using
it before: `ismember`.

FinalAnswer = ismember(A,B)

FinalAnswer = 1 0 0 0 0 0 1 1 0 1 0 0 0 0 0 1 1 0 0 1

isequal(FinalAnswer, RESULTS)

ans = 1

In one fell swoop I can compute the entire result because I don't care about matching those locations. It's enough to say
that a value in `A` matches some value in `B`. Voila!

### Other Approaches

I just showed you a few approaches for solving this problem. Do you have similar problems, perhaps ones that don't yield as simple a solution? Or do you have other approaches to solving this problem that might be useful, especially in more complicated situations? Please share them here.