Indexing duplicates in a matrix: Matlab

Consider the matrix

 X = [ 1 2 0 1; 
       1 0 1 2;                                          
       1 2 3 4;                                     
       2 4 6 8;
          .           
          .                          
       1 2 0 1                  
          .                 
          .    ]

      

I want to create a new column so that I can type the ith

occurrence of each row.

Ans:

   X = [ 1 2 0 1;   y =  [1
         1 0 1 2;         1                                 
         1 2 3 4;         1                            
         2 4 6 8;         1
           .             .
           .             .             
         1 2 0 1          2        
           .             .    
           .    ]        .]

      

Any ideas?

+3


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3 answers


How about this?

y = sum(triu(squareform(pdist(X))==0)).';

      

It works by counting how many previous lines are equal to each line. Two lines are equal if their distance (calculated with squareform

and pdist

) is 0. triu

ensures that only the previous lines are counted.



To shorten the computation time and avoid depending on the statistics toolbar, you can use @ user1735003's suggestion:

y = sum(triu((bsxfun(@plus, sum(X.^2,2), sum(X.^2,2)') - 2*X*X.')==0));

      

+3


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Approach # 1

%// unique rows
unqrows = unique(X,'rows'); 

%// matches for each row against the unique rows and their cumsum values
matches_perunqrow = squeeze(all(bsxfun(@eq,X,permute(unqrows,[3 2 1])),2));
cumsum_unqrows = cumsum(matches_perunqrow,1);

%// Go through a row-order and get the cumsum values for the final output
[row,col] = find(matches_perunqrow);
[sorted_row,ind] = sort(row);
y=cumsum_unqrows(sub2ind(size(cumsum_unqrows),[1:size(cumsum_unqrows,1)]',col(ind)));

      

Example run -

X =
     1     2     0     1
     1     0     1     2
     1     2     3     4
     2     4     6     8
     1     2     0     1
     1     2     3     4
     1     2     3     4
     1     2     3     4
     1     2     3     4
     1     2     0     1
out =
     1
     1
     1
     1
     2
     2
     3
     4
     5
     3

      


Approach # 2

%// unique rows
unqrows = unique(X,'rows');

%// matches for each row against the unique rows
matches_perunqrow = all(bsxfun(@eq,X,permute(unqrows,[3 2 1])),2)

%// Get the cumsum of matches and select only the matches for each row.
%// Since we need to go through a row-order, transpose the result
cumsum_perrow = squeeze(cumsum(matches_perunqrow,1).*matches_perunqrow)' %//'

%// Select the non zero values for the final output
y = cumsum_perrow(cumsum_perrow~=0)

      




Approach # 3

%// label each row based on their uniqueness
[~,~,v3] = unique(X,'rows')
matches_perunqrow = bsxfun(@eq,v3,1:size(X,1))

cumsum_unqrows = cumsum(matches_perunqrow,1);

%// Go through a row-order and get the cumsum values for the final output
[row,col] = find(matches_perunqrow);
[sorted_row,ind] = sort(row);
y=cumsum_unqrows(sub2ind(size(cumsum_unqrows),[1:size(cumsum_unqrows,1)]',col(ind)));

      


Approach # 4

%// label each row based on their uniqueness
[~,~,match_row_id] = unique(X,'rows');

%// matches for each row against the unique rows and their cumsum values
matches_perunqrow = bsxfun(@eq,match_row_id',[1:size(X,1)]');
cumsum_unqrows = cumsum(matches_perunqrow,2);

%// Select the cumsum values for the ouput based on the unique matches for each row
y = cumsum_unqrows(matches_perunqrow);

      

+2


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A solution involving a for loop can be done quite easily, it might be fast enough already. I'm sure there is a faster solution that you can use cumsum

, but you might not even need to. Basic idea: Find the indices of unique strings first, so that you can handle scalar indices instead of full strings (vectors). Then collapse by indices and find the number of previous occurrences:

X = [ 1 2 0 1; 
   1 0 1 2;                                          
   1 2 3 4;                                     
   2 4 6 8;                        
   1 2 0 1;                 
   1 3 3 7;                 
   1 2 0 1];

[~,~,idx] = unique(X, 'rows'); %// find unique rows

%// loop over indices and accumulate number of previous occurences
y = zeros(size(idx));
for i = 1:length(idx)
   y(i) = sum(idx(1:i) == idx(i)); %// this line probably scales horrible with length of idx.
end

      

Result for example:

y =

 1
 1
 1
 1
 2
 1
 3

      

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