matrix 9x9 with duplicate values

2014 1 19
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更新时间：2014 2 9
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0 个投票

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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goran
goran 2014-1-19

0 个投票

i changed matrix to 4x4 and i sould received like solution duplicate above main diagonal nuber 4, i receive solution 3. why?

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Mischa Kim
Mischa Kim 2014-1-19
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 2014-1-19
sorry. code is ok. can you explain your code? what is FLAG?

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Mischa Kim
Mischa Kim 2014-1-19
编辑：Mischa Kim 2014-1-19

0 个投票

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 2014-1-19
can you explain me your code, i must print message 'no duplicate' if there're not duplicates?
Mischa Kim
Mischa Kim 2014-1-19
First, find the unique values of A and write them into a vector. Next enter the nested for loops where you are searching for duplicates: add the current matrix entry to the vector of unique values and test if that changes the vector. If it does not, this means that the entry must be a duplicate. In this case set the FLAG variable to true, which is necessary to end ( break ) the nested loop.
Add
if ~FLAG
display('No duplicates')
end
goran
goran 2014-1-19
again i puted image
first duplicate is 2. number 2 is found under main diagonal. if i need searching above main diagonal what i do?
Mischa Kim
Mischa Kim 2014-1-19
The algorithm only searches for duplicates above the diagonal. The 2 found is the one in the first row, second column.
goran
goran 2014-1-19
what's code would be if i search in range above main diagonal?this code does not work with real numbers
goran
goran 2014-1-19
when my matrix has elements like on pictures then no duplicates why?
i think that duplicate is 1.1
goran
goran 2014-1-23
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 2014-1-19

0 个投票

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

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Mischa Kim
Mischa Kim 2014-1-19
编辑：Mischa Kim 2014-1-19
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 2014-1-19

0 个投票

also, i need that duplicate values is printed, not how much duplicate values have?

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Mischa Kim
Mischa Kim 2014-1-19
It is. See the display command. Simply run the code above, change the matrix and verify.

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Andrei Bobrov
Andrei Bobrov 2014-1-19
编辑：Andrei Bobrov 2014-1-20

0 个投票

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 2014-1-19
can you explain me this code, if in matrix there're not duplicates, i need print message 'no duplicates'
goran
goran 2014-1-20
i need print message if no duplicates. what i do?
Andrei Bobrov
Andrei Bobrov 2014-1-20
编辑：Andrei Bobrov 2014-1-20
use file test1.m:
goran
goran 2014-1-20
thank's sorry can you explain your code?

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Harry Commin
Harry Commin 2014-2-9

0 个投票

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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