List of indexes for each row of a matrix without a for loop

Matrix DATA contains some data, and matrix INDSETS contains the indexes to be extracted from each row of DATA.

Example:

DATA = [8 9 3 1 2; 4 5 8 7 6; 1 5 9 7 3]
DATA =

     8   9   3   1   2
     4   5   8   7   6
     1   5   9   7   3

INDSETS = [2 4 3; 1 3 5; 5 2 1; 5 4 2]
INDSETS =

     2   4   3
     1   3   5
     5   2   1
     5   4   2

For EACH row of DATA, I would like to extract its N subsets as given by each row of INDSETS (where N = number of rows in INDSETS). For instance, for the first row of DATA, its subsets are:

DATA(1, :)(INDSETS)
ans =

     9   1   3
     8   3   2
     2   9   8
     2   1   9

For the 2nd row of DATA, its subsets are:

DATA(2, :)(INDSETS)
ans =

     5   7   8
     4   8   6
     6   5   4
     6   7   5

and so on. A possible slow solution would be to preallocate a 3D array C to store the results, and run a for loop row by row, saving the results of each iteration on the 3rd dimension of C:

Slow solution:

C = zeros(size(INDSETS, 1), size(INDSETS, 2), size(DATA, 1));
for i = 1:size(DATA, 1)
 C(:, :, i) = DATA(i, :)(INDSETS); 
end

Is there a more efficient way to do this without a for loop, perhaps a vectorized solution?

Answer 1

[Stephen23]

>> S = size(INDSETS);
>> reshape(DATA(:,INDSETS).',S(1),S(2),[])

or on one line:

>> reshape(DATA(:,INDSETS).',[size(INDSETS),size(DATA,1)])

both produce this output:

ans(:,:,1) =
     9     1     3
     8     3     2
     2     9     8
     2     1     9
ans(:,:,2) =
     5     7     8
     4     8     6
     6     5     4
     6     7     5
ans(:,:,3) =
     5     7     9
     1     9     3
     3     5     1
     3     7     5

[George Aipal]

Thanks Stephen, that works! I have a lot to work from you