How to transpose a matrix

Matrix A as follows:

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

I want to transpose matrix A based on the unique ID in the first column. Add 0 at the end wherever its needed in order to keep matrix dimension consistent.

out = [1  8  5  4  0  0
2  6  7  2  5  0
7  6  4  8  0  0
9  9  1  2  6  2
];

[Jan]

The procedure is not explained uniquely. Surely this is not a transposing. With some guessing a method can be invented, but it would be safer, if you explain it clearly.

Answer 1

[JESUS DAVID ARIZA ROYETH]

you can use this:

A = [1  8
1  5
1  4
2  6
2  7
2  2
2  5
7  6
7  4
7  8
9  9
9  1
9  2
9  6
9  2
];
s=unique(A(:,1));
[~,v]=mode(A(:,1));
out=zeros(length(s),v+1);
for k=1:length(s)
    value=[s(k) A(A(:,1)==s(k),2)'];
    out(k,:)=[value zeros(1,v+1-length(value))];
end

Answer 2

[Jan]

With some guessing:

A = [1 8; ...
     1 5; ...
     1 4; ...
     2 6; ...
     2 7; ...
     2 2; ...
     2 5; ...
     7 6; ...
     7 4; ...
     7 8; ...
     9 9; ...
     9 1; ...
     9 2; ...
     9 6; ...
     9 2];
[Key, iKey, iA] = unique(A(:, 1));
R = zeros(numel(Key), 1 + mode(iA));   % Pre-allocate
for k = 1:numel(Key)
  index = (iA == k);
  R(k, 1:sum(index) + 1) = [Key(k), A(index, 2).'];
end

Is the 1st column of A sorted? Then an alternative with FEX: RunLength:

[B, N, Index] = RunLength(A(:, 1));
R = zeros(numel(B), 1 + max(N));   % Pre-allocate
for k = 1:numel(Key)
  R(k, 1)          = B(k);
  R(k, 2:N(k) + 1) = A(Index(k):Index(k)+N(k)-1, 2).';
end