I need to use information from two columns in a matrix, to build a matrix using information from a third column.

I wish to build a 4x6 matrix from information given as a matrix. The given matrix is 24x3, the first column has numbers, only four numbers, repeated over and over. The second column has six distinct IDs repeated four times. The final column contains data, which is used to build my final matrix.

MatrixA=[ 1  96310  1.1
          2  96311  1.2
          3  96312  1.3
          4  96313  1.4
          1  96314  1.5
          2  96315  1.6
          3  96310  1.7
          4  96311  1.8
          1  96312  1.9
          2  96313  2.0
          3  96314  2.1
          4  96315  2.2
          1  96311  2.3
          2  96312  2.4
          3  96313  2.5
          4  96314  2.6
          1  96315  2.7
          2  96310  2.8
          3  96311  2.9
          4  96312  3.0
          1  96313  3.1
          2  96314  3.2
          3  96315  3.3
          4  96310  3.4]

My desired end result is:

[1.1  2.3  1.9  3.1  1.5  2.7
 2.8  1.2  2.4  2  3.2  1.6
 1.7  2.9  1.3  2.5  2.1  3.3
 3.4  1.8  3  1.4  2.6  2.2]

The numbers have all been made up, but the concept is still the same. I need to somehow build a code which allows me to create that final matrix using information presented in the style of MatrixA.

I have been advised that apparently I could do this by making use of the functions unique, ismember and sub2ind . I have been playing around with this since Tuesday, still haven't been able to complete this.

[Jan]

You forgot to reveal the wanted pattern. Why is the element (1,2) set to 2.3? Why is the element (2,1) set to 2.8? It is not reliable or efficient, when we guess what you want.

Answer 1

[Walter Roberson]

[uid, ~, ididx] = unique(MatrixA(:,2));
result = accumarray([MatrixA(:,1), ididx(:)], MatrixA(:,3));

[Star Strider]

@Walter —

The first line should be:

[uid, ~, ididx] = unique(MatrixA(:,2));

Then it works.

[Richard Lea]

Omg! It works perfectly! Can you please explain why the first line works, and where I can get more information about that. Thanks!

[Stephen23]

@Walter Roberson: I added the missing unique.

Answer 2

[Jan]

A bold guess:

A = [ 1  96310  1.1; ...
   2  96311  1.2; ...
   3  96312  1.3; ...
   4  96313  1.4; ...
   1  96314  1.5; ...
   2  96315  1.6; ...
   3  96310  1.7; ...
   4  96311  1.8; ...
   1  96312  1.9; ...
   2  96313  2.0; ...
   3  96314  2.1; ...
   4  96315  2.2; ...
   1  96311  2.3; ...
   2  96312  2.4; ...
   3  96313  2.5; ...
   4  96314  2.6; ...
   1  96315  2.7; ...
   2  96310  2.8; ...
   3  96311  2.9; ...
   4  96312  3.0; ...
   1  96313  3.1; ...
   2  96314  3.2; ...
   3  96315  3.3; ...
   4  96310  3.4];

ID     = unique(A(:, 2));
Result = zeros(4, numel(ID));
for kID = 1:numel(ID)
   match = (A(:, 2) == ID(kID));
   Result(A(match, 1), kID) = A(match, 3);
end