i changed matrix to 4x4 and i sould received like solution duplicate above main diagonal nuber 4, i receive solution 3. why?
matrix 9x9 with duplicate values
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i have matrix C9x9 with duplicates. i must find duplicate above main diagonale. when i find first duplicate the searching stop and print this duplicate? thanks
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Accepted Answer
goran
on 19 Jan 2014
2 Comments
Mischa Kim
on 19 Jan 2014
Hello goran, could you please post follow-up questions as comments, not as answers? To your question: the first duplicate above the diagonal (searching from left to right, top to bottom) is the 3 at position (1,2).
More Answers (5)
Mischa Kim
on 19 Jan 2014
Edited: Mischa Kim
on 19 Jan 2014
This should do. For a 3x3 matrix, as an example:
A = [1 3 2; 4 5 1; 3 1 1]
A_unique = unique(tril(A, -1));
FLAG = false;
for ii = 1:length(A(1,:))
for jj = ii+1:length(A(1,:))
if (length(A_unique) == length(unique([A_unique; A(ii,jj)])))
display(A(ii,jj))
FLAG = true;
break;
end
end
if FLAG
break;
end
end
7 Comments
goran
on 19 Jan 2014
1 Comment
Mischa Kim
on 19 Jan 2014
Edited: Mischa Kim
on 19 Jan 2014
Hello goran, the algorithm only searches above the diagonal. That's because of the indexing of the for loop ( jj = ii+1! ):
for ii = 1:length(A(1,:))
for jj = ii+1:length(A(1,:))
You can verify by simply printing the indices of the matrix elements.
goran
on 19 Jan 2014
1 Comment
Mischa Kim
on 19 Jan 2014
It is. See the display command. Simply run the code above, change the matrix and verify.
Andrei Bobrov
on 19 Jan 2014
Edited: Andrei Bobrov
on 20 Jan 2014
function test1
A = randi(15,3)
B=A;
B(tril(B)>0)=nan;
C=B(~isnan(B));
[a,b] = unique(C,'first');
[~,ii] = sort(b);
c = histc(C,a);
out0 = [a(ii),c(ii)];
out = out0(find(out0(:,2)>1,1,'first'),1);
if isempty(out), disp('no duplicates'); end
end
4 Comments
Harry Commin
on 9 Feb 2014
To extract only the upper triangular numbers into a column vector, you could use:
Aupper = A(triu(ones(size(A)))==1);
I think it is easier to find all duplicates than just the 'first' one. (How do we even define "first"?). However, assuming we want to progress through A column-wise, we could use:
B = unique(Aupper,'stable');
first_duplicate = Aupper(find(Aupper(1:length(B)) ~= B, 1))
The first line finds unique values in the order they appear. The second line finds the first place where the input vector and the 'uniques' are different (i.e. the first duplicate) and prints out that value.
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