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matrix 9x9 with duplicate values

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goran
goran on 19 Jan 2014
Answered: Harry Commin on 9 Feb 2014
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
goran on 19 Jan 2014
i changed matrix to 4x4 and i sould received like solution duplicate above main diagonal nuber 4, i receive solution 3. why?

  2 Comments

Mischa Kim
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).
goran
goran on 19 Jan 2014
sorry. code is ok. can you explain your code? what is FLAG?

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More Answers (5)

Mischa Kim
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

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goran
goran on 19 Jan 2014
what's code would be if i search in range above main diagonal?this code does not work with real numbers
goran
goran on 19 Jan 2014
when my matrix has elements like on pictures then no duplicates why?
i think that duplicate is 1.1
goran
goran on 23 Jan 2014
my solution for this problem is
if true
A=[1 3 2.2 130 5.4 60 8.9 8 45;2 58 4 8 6.5 69 78 23 1.1;7 8 4 69 65.3 6.7 89.3 102.3 45;3 4 3 1 8.9 1.1 59 4.5 65;
1 23 2.5 45 5.6 7.9 8.78 65 89; 12 897 67 0.5 102.3 7.8 9 11 567; 494 123 256 28.6 2.2 123 891 111 11.1;
45 12 1.2 2.6 58.4 36 45.7 12.1 78.6; 1.1 2.3 4.5 36 25 11.6 89.3 1.1 1]
%A=rand(9,9)
Agd=(triu(A, 1)); %selektuje elemente iznad glavne diagonale
Agdv=Agd(:); %kreira vektor
Agdv(Agdv==0)= []; %brise nule iz vektora
Aj=unique(Agd); %selektuje jedinstvene vrednosti iz Agd
Aj(Aj==0)=[]; %brise nule iz jedinstvenih vrednosti
if (length(Aj)~=length(Agdv)) %proverava da li broj jedinstvenih vrednosti razlicit u odnosu na Agdv, ako jesu onda postoje duplikati, ako ne stampa nema istih elemenata
[~,~, IAgdv] = unique(Agdv, 'stable'); %odredjuju se indeksi jedinstvenih vrednosti u nepromenjenom redosledu u Agdv
idx = find(diff(IAgdv)-1, 1)+1; %nalazi indeks koji se dva puta ponavlja
el=Agdv(idx); %dodeljuje vrednost tog indeksa promenljivoj el
fprintf('Prvi isti element koji se pojavio je: %.4f \n',el) % stampa vednost promenljive el
else
disp('Nema istih elemenata')
end
end

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goran
goran on 19 Jan 2014
the searching need to work just on elements above main diagonal. your code search entire matrix

  1 Comment

Mischa Kim
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.

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goran
goran on 19 Jan 2014
also, i need that duplicate values is printed, not how much duplicate values have?

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Mischa Kim
Mischa Kim on 19 Jan 2014
It is. See the display command. Simply run the code above, change the matrix and verify.

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Andrei Bobrov
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

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goran
goran on 20 Jan 2014
i need print message if no duplicates. what i do?
goran
goran on 20 Jan 2014
thank's sorry can you explain your code?

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Harry Commin
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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