find duplicated rows in matlab without for loop

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Mina Mino
Mina Mino am 21 Jul. 2023
Verschoben: Matt J am 21 Jul. 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

Akzeptierte Antwort

Matt J
Matt J am 21 Jul. 2023
Bearbeitet: Matt J am 21 Jul. 2023
[~,I]=unique(x,'rows');
locations=setdiff(1:height(x),I) %locations of duplicate rows
  2 Kommentare
Mina Mino
Mina Mino am 21 Jul. 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 am 21 Jul. 2023
Bearbeitet: Matt J am 21 Jul. 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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Weitere Antworten (1)

Walter Roberson
Walter Roberson am 21 Jul. 2023
Verschoben: Matt J am 21 Jul. 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 Kommentar
Mina Mino
Mina Mino am 21 Jul. 2023
Verschoben: Matt J am 21 Jul. 2023
@Walter Roberson Thansk for your help! I exactly need it:)
Best,

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