find duplicated rows in matlab without for loop

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Mina Mino
Mina Mino el 21 de Jul. de 2023
Movida: Matt J el 21 de Jul. de 2023
Hello Friedns,
I have a very large matrix with 2 columns. I need to find the location of duplicated rows (the position of them) . However, I don't want to solve this problem with a for loop because I've tried it before (see the attached code) and it takes a long time. I'm looking for an alternative way to do this. I would be grateful if you could suggest me an idea.
Best,
Mina
x = [File(:,1) File(:,2)];
Grid=unique(x,'rows');
for j=1:length(DD)
idx=find(day_of_year==DD(j));
File2=File(idx,:);
for g=1:length(Grid)
[index1] = (ismember(File2(:,[1 2]),Grid(g,:),'rows'));
idx2=find(index1==1);
Total=[Total;Grid(g,1) Grid(g,2) DD(j) mean(File2(idx2,3)) mean(File2(idx2,4)) mean(File2(idx2,5)) mean(File2(idx2,6))];
end
end

Respuesta aceptada

Matt J
Matt J el 21 de Jul. de 2023
Editada: Matt J el 21 de Jul. de 2023
[~,I]=unique(x,'rows');
locations=setdiff(1:height(x),I) %locations of duplicate rows
  2 comentarios
Mina Mino
Mina Mino el 21 de Jul. de 2023
Hi Matt J,
Many thanks for your help! It seems that for each line your code is only finding the position of one of the duplicated lines. However, there are more than one duplicates for each row. How can I find all duplicate rows of each row?
Thanks in advance for your answer and time.
Matt J
Matt J el 21 de Jul. de 2023
Editada: Matt J el 21 de Jul. de 2023
It seems that for each line your code is only finding the position of one of the duplicated lines.
I don't think so. It should return the indices of all rows that have been seen before. As you can see below, the final locations list includes all rows except for 1 and 3, which is where a new row is encountered.
x=[ 1 2;
1 2;
0 4;
1 2;
0 4
0 4];
[~,I]=unique(x,'rows');
locations=setdiff(1:height(x),I) %locations of duplicate rows
locations = 1×4
2 4 5 6

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Más respuestas (1)

Walter Roberson
Walter Roberson el 21 de Jul. de 2023
Movida: Matt J el 21 de Jul. de 2023
The third output of unique gives the "group number" for each entry. There are different ways of handling that. one of ways is
[unique_rows, ~, ic] = unique(x,'rows');
appears_in_rows = accumarray(ic, (1:size(x,1)).', [], @(v) {v});
T = table(unique_rows, appears_in_rows);
This would create a table in which the first variable is each unique row, and the second variable is a cell array of row indices that are that unique row. The cell array will always have at least one entry, but might have more.
  1 comentario
Mina Mino
Mina Mino el 21 de Jul. de 2023
Movida: Matt J el 21 de Jul. de 2023
@Walter Roberson Thansk for your help! I exactly need it:)
Best,

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