matrix 9x9 with duplicate values

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

Answer 1

[goran]

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

[Mischa Kim]

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]

sorry. code is ok. can you explain your code? what is FLAG?

Answer 2

[Mischa Kim]

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

[goran]

can you explain me your code, i must print message 'no duplicate' if there're not duplicates?

[Mischa Kim]

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]

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]

The algorithm only searches for duplicates above the diagonal. The 2 found is the one in the first row, second column.

[goran]

what's code would be if i search in range above main diagonal?this code does not work with real numbers

[goran]

when my matrix has elements like on pictures then no duplicates why?

i think that duplicate is 1.1

[goran]

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

Answer 3

[goran]

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

[Mischa Kim]

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.

Answer 4

[goran]

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

[Mischa Kim]

It is. See the display command. Simply run the code above, change the matrix and verify.

Answer 5

[Andrei Bobrov]

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

[goran]

can you explain me this code, if in matrix there're not duplicates, i need print message 'no duplicates'

[goran]

i need print message if no duplicates. what i do?

[Andrei Bobrov]

use file test1.m:

[goran]

thank's sorry can you explain your code?

Answer 6

[Harry Commin]

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.