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

1 vue (au cours des 30 derniers jours)
Sim le 9 Déc 2022
Commenté : Sim le 14 Déc 2022
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
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### Réponse acceptée

Voss le 9 Déc 2022
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
##### 2 commentairesAfficher AucuneMasquer Aucune
Sim le 12 Déc 2022
Modifié(e) : Sim le 12 Déc 2022
Thank you very much @Voss! :-)
Both solutions are great and I do not know which one to accept! Possible to accept both ?
Voss le 12 Déc 2022
You're welcome! You can accept at most one answer, but you can vote for as many as you like.

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### Plus de réponses (1)

Jon le 9 Déc 2022
Modifié(e) : Jon le 9 Déc 2022
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
##### 4 commentairesAfficher 2 commentaires plus anciensMasquer 2 commentaires plus anciens
Jon le 12 Déc 2022
Sim le 14 Déc 2022
Thanks a lot!
I would say it is a pity that there is not the possibility to accept multiple answers :-) :-)

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