I am triying to rearranging the matrix and form one as explained below

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Hems
Hems 2016-10-13
编辑： James Tursa 2016-10-13
I would like to rearrange and form a new matrix, I have a matrix for example A=[1 2;2 16;3 5;1 12;2 6;3 7;1 5;2 7;3 6;1 7;2 10;3 18] this has repetitive 1st column "1 2 3 " I need to form a new martix with values in the 2nd row arranged corrosponding values resulting B=[1 2 12 5 7;2 16 6 7 10;3 5 7 6 18] Please help ,how this can be done? TIA!
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Alexandra Harkai
Alexandra Harkai 2016-10-13
编辑：Alexandra Harkai 2016-10-13
Using splitapply could be a good option, since your matrix seems to be defining the group indices (1,2,3) and the corresponding values:
splitapply(@(x) {x'}, A(:,2), A(:,1))
This does not give you a matrix yet, but is able to deal with the situation even when not all groups have the same number of elements. Plus you can then append the group indices in front of the result matrix, but this is a good direction to start in. Explanation: @(x) {x'} is basically an identity function, but transposing the result. You just want to apply the identity function to each group of elements, which is what splitapply does.

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Ganesh Hegade
Ganesh Hegade 2016-10-13
编辑：James Tursa 2016-10-13
Hi, Hope this below piece of code will help you.
aSize = size(A);
initialValue = A(1);
count = 1;
temp=[];
for i = 1: aSize(1)
if i >1 && A(i) == initialValue
[B] = A(count:i -1, :) ;
count = i;
temp = [temp B];
end;
end;
bSize = size(temp);
temp = [temp A(i- bSize(1) + 1:i, :)];
B = unique(temp.','rows', 'stable').';
Thanks.

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Guillaume
Guillaume 2016-10-13
Following on from Alexandra Harkai's comment to your question:
A = [1 2;2 16;3 5;1 12;2 6;3 7;1 5;2 7;3 6;1 7;2 10;3 18]
rows = splitapply(@(x) {x'}, A(:,2), A(:,1))
assert(all(diff(cellfun(@numel, rows)) == 0), 'Cannot make a matrix since each unique index does not have the same number of elements');
B = [unique(A(:, 1)), vertcat(rows{:})]

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