Find the value which is repeated in each row of a matrix (without loop for)
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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]
% 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'
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采纳的回答
Voss
2022-12-9
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]
[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))
2 个评论
Voss
2022-12-12
You're welcome! You can accept at most one answer, but you can vote for as many as you like.
更多回答(1 个)
Jon
2022-12-9
编辑:Jon
2022-12-9
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))
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