Find the value which is repeated in each row of a matrix (without loop for)

In each row of the following matrix, one of the values is repeated. For example, in the first row, the value 523 is repeated twice. I would like to find all those values that, in each row, are repeated, but without using a loop for as in the following example. Is it possible ?

% Input
a = [    523        2920         523        1227
        8003        8343        5611        8343
        2066        5333        5333        5783
        1447        2331        2331        8810
         375        8083        8083        8343
        5611        6866        5611        8343
        2935        7026        5446        7026
        1409        6842        6614        6842
        2118        7208        4446        7208
        4055        4439        4439        4921
          42        8656        6691        8656
         725        8478         822        8478
        1003        1227        1227        6349
         921        6614        6614        6842]

a = 14×4

523 2920 523 1227 8003 8343 5611 8343 2066 5333 5333 5783 1447 2331 2331 8810 375 8083 8083 8343 5611 6866 5611 8343 2935 7026 5446 7026 1409 6842 6614 6842 2118 7208 4446 7208 4055 4439 4439 4921

% Would it be possible to perform the same calculation without the loop for ?
for i = 1 : size(a,1)
    [v, w] = unique( a(i,:), 'stable' );
    duplicate_indices = setdiff( 1:numel(a(i,:)), w );
    b(i) = a(i,duplicate_indices);
end

% Output
b'

ans = 14×1

523 8343 5333 2331 8083 5611 7026 6842 7208 4439

Answer 1

[Voss]

Assuming that there's exactly one repeat (i.e., one number appears two times) in each row, which the for-loop solution also assumes, here's one way to do it without a loop:

a = [    523        2920         523        1227
        8003        8343        5611        8343
        2066        5333        5333        5783
        1447        2331        2331        8810
         375        8083        8083        8343
        5611        6866        5611        8343
        2935        7026        5446        7026
        1409        6842        6614        6842
        2118        7208        4446        7208
        4055        4439        4439        4921
          42        8656        6691        8656
         725        8478         822        8478
        1003        1227        1227        6349
         921        6614        6614        6842]

a = 14×4

523 2920 523 1227 8003 8343 5611 8343 2066 5333 5333 5783 1447 2331 2331 8810 375 8083 8083 8343 5611 6866 5611 8343 2935 7026 5446 7026 1409 6842 6614 6842 2118 7208 4446 7208 4055 4439 4439 4921

[m,n] = size(a);
[a_sorted,idx] = sort(a,2);
is_repeat = diff(a_sorted,1,2) == 0;
idx = idx.';
c = idx([is_repeat.'; false(1,m)]);
b = a(sub2ind([m n],(1:m).',c))

b = 14×1

523 8343 5333 2331 8083 5611 7026 6842 7208 4439

[Sim]

Thank you very much @Voss! :-)

Both solutions are great and I do not know which one to accept! Possible to accept both ?

[Voss]

You're welcome! You can accept at most one answer, but you can vote for as many as you like.

Answer 2

[Jon]

Here's another way

% Input
A = [    523        2920         523        1227
        8003        8343        5611        8343
        2066        5333        5333        5783
        1447        2331        2331        8810
         375        8083        8083        8343
        5611        6866        5611        8343
        2935        7026        5446        7026
        1409        6842        6614        6842
        2118        7208        4446        7208
        4055        4439        4439        4921
          42        8656        6691        8656
         725        8478         822        8478
        1003        1227        1227        6349
         921        6614        6614        6842];

%[C,ia,ic] = unique(A)

Asrt = sort(A,2);


delta = [diff(Asrt,1,2) ones(size(A,1),1)];

[r,c] = find((delta==0)');

B = Asrt';

dup = B(sub2ind(size(B),r,c))

dup = 14×1

523 8343 5333 2331 8083 5611 7026 6842 7208 4439

[Jon]

Could probably be cleaned up even a little more regarding transposes, but this gives the idea

[Sim]

Thank you very much @Jon! :-)

Both solutions are great and I do not know which one to accept! Possible to accept both ?

[Jon]

Your welcome. Unfortunately you can only accept one answer.

[Sim]

Thanks a lot!

I would say it is a pity that there is not the possibility to accept multiple answers :-) :-)