# How to find the row numbers of similar elements

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Rashmi Mohan Kumar on 19 Nov 2019
Hi,
I needed a little help with finding the row numbers of similar elements in a matrix. It involves the use of unique,find and sort commnds but I could not clear the errors that I was facing. My code should do the following:
I have a matrix as follows:
1 3 2
2 4 1
4 3 6
5 1 8
2 8 9
1 7 5
6 6 2
7 9 1
1 3 2
I need my code to first check for similar values in column 1, for example in the above matrix let us consider for the number "1", it occurs three times in column1. So I want my code to check for all the rows in column 1 with a value 1 and then check for the same rows if the corresponding values in column 2 are also same, meaning in the 1st column above the value 1 occurs on rows 1, 6 and 9, then for the same rows the code should check what are the values in column 2, from the matrix it can be noticed that the corresponding values in column 2 for 1 is 3, 7 and 3. After this I need my code to check for the row numbers of similar elemnts that is 3. From the matrix above it is the row numbers 1 and 9. Then I need the code to check for the values in column 3 for rows 1 and 9, if the numbers in column 3 are same then set a variable to 1 otherwise 0. I need the code in a looping structure since I have to check for 50 values.
Any help or suggestions on how to do it is appreciated.
Thanks.

Rashmi Mohan Kumar on 20 Nov 2019
Thanks Rik, it was very helpful and it worked too. I had another question after generating the output rows instead of setting a variable to 1, is it possible to place a value 1 at the same rows that was generated but in the 4th column of A. For example, suppose the code generated [1,7] as the row numbers how do I a place a value 1 in both these rows but in the 4th column of A.
Rik on 20 Nov 2019
Happy to help. If my answer helped you, feel free to mark it as accepted answer.
As for your follow-up question: that would just be simple indexing.
A([1 7],end)=1;
Rashmi Mohan Kumar on 20 Nov 2019
Thanks again, but I did try implementing the same thing as you mentioned for the follow up question but it gave an error. I think is it because I am dealing with an 50X3 elements and here in the exmaple I have mentioned only for one repeated element but there can be row numbers generated too. Meaning to say the row numbers generated are[ 3, 7; 5, 1] then a 1 has to be placed at all these row numbers but in the 4th column of A.So should it be A([ans],4)=1; ??

Rik on 19 Nov 2019
Assuming you want a matrix with the duplicate row indices, the code below should work. It pads the columns with NaN. I slightly adapted your data to test the optimizations.
A=[1 3 2
2 4 1
2 4 1
2 4 1
4 3 6
5 1 8
2 8 9
1 7 5
6 6 2
7 9 1
1 3 2];
[a,b,c]=unique(A,'rows','stable');
z=NaN( numel(c)-numel(b) +1 );%pre-allocate worst-case
hasDuplicates=find( 1 < accumarray([c ones(size(c))],1) );
%slower option:
%for n=reshape(b,1,[])
for n=reshape(hasDuplicates,1,[])
tmp=find(c==n);
if numel(tmp)~=1
z(n,1:numel(tmp))=tmp;
end
end
%remove NaN rows and cols
z(all(isnan(z),2),:)=[];
z(:,all(isnan(z),1))=[];
clc
if ~isempty(z)
disp(z);
else
disp('nothing is found');
end