Extract lines of a three dimensional matrix using an array of indices and NO for-loop

I have a three dimensional 10x5x2 array. Example:
r(:,:,1)  = [1 0 2 1 1; 2 0 3 1 1; 3 0 4 1 1; 4 0 1 1 -1; 5 0 -1 1 1; 1 1 3 1 -1; 2 1 2 1 1; 3 1 5 0 -1; 4 1 4 1 -1; 5 1 1 0 -1]; 
r(:,:,2) = [1 0 2 1 -1;2 0 3 1 1;3 0 1 1 -1;4 0 1 1 -1;5 0 -1 1 1;1 1 1 1 -1;2 1 2 1 1;3 1 4 1 1;4 1 5 1 1;5 1 3 0 1]

and logical vector m like for instance: m =

10×2 logical array

1 0

0 0

0 1

0 0

0 0

0 0

0 0

0 0

0 0

0 0

how can I access those lines of r, when m(:,k) is used as index vector of r(:,:,k). In this example the result should be line 1 of r(:,:,1) and line 3 of r(:,:,2) or

1 0 2 1 1

3 0 1 1 -1

important: for performance reasons I want to do it without a for-loop.

Thanks for anyone's help!

Answer 1

[BobH]

Use column 1 of m to select rows of r in the first

>> r( m(:,1), :,1)
ans =
     1     0     2     1     1

Use column 2 of m to select rows of r in the other

>> r( m(:,2), :,2)
ans =
     3     0     1     1    -1

[Thomas]

But what would I do if my matrix is r(10,5,100000) instead of r(10,5,2) ? The expression as you describe would become too big...

Or what oucld I do, if the size of the third dimension is unknown or changes?

Doing it with a foor loop would be simple, but is very slow...

[BobH]

oops. commented in wrong place.

You can use arrayfun and pass in numbers which you'll use as indexes

a = arrayfun(@(X)  r( m(:,X), :, X ), 1:size(r,3), 'un', 0 );

The result comes out as an array of cells, but that's easy to make usable again

b = vertcat( a{:} )
b =
     1     0     2     1     1
     3     0     1     1    -1

[Thomas]

Thank you very much! Works!