Indexing into a n-by-m matrix using values from a n-by-1 matrix.

Given an n-by-m matrix x, and a n-by-1 matrix y, how can I turn this into the n-by-1 matrix z where each row i of z is the value x(i, y(i))? That is, I want to assign each row in z the column of x given by y at the same row.

My current horrible loop code is:

   x = magic(6);
   y = [1; 3; 2; 4; 6; 1];

   z = zeros(length(y), 1);
   for i = 1 : length(y)
      z(i) = x(i, y(i));
   end

   % z is now [35, 7, 9, 17, 16, 4]'.

I hope this is clear enough, please let me know if it is not.

[Jan]

y has 3 elements only, therefore the loop cannot run until length(x), but until length(y) only. Then z does not need 100 elements, but "z = zeros(length(y), 1)" is better.

[Stephen]

Sorry, that was a very lazy updating of the question from me to attempt to make it clearer. Evidently it did much the latter!

Answer 1

[Sean de Wolski]

Okay, I'm not totally clear and your for-loop does not do what your problem describes.

We have an nxm matrix, x, for simplicity:

x = magic(3);

We have an nx1 vector y:

y = [1;3;2];

We want the columns of x to be indexed by y in to the rows of z:

z = x(:,y)';

This is what I understood from your description. Please clarify what is right or wrong with it.

More per comments

x = magic(3);
y = [1;3;2];
z = x(sub2ind(size(x),(1:numel(y))',y))

[Stephen]

My apologies, I must have been unclear. In your example, I would expect z to be [8, 7, 9], which is what my code fragment gives. I will attempt to update my question to make it clearer.

Answer 2

[Jan]

   x = magic(6);
   y = [1; 3; 2];
   z = x(sub2ind(size(x), 1:length(y), y);