Identifying the repeated rows in a matrix and comparing them to another matrix

I need to check which rows of a matrix A are inside another matrix B and if there are repeated rows from the matrix A, then identify the repeated ones and count how many repeated rows are found but for each different set of values. I have tried with ismember and unique functions but i don't know how to keep track of which are the repeated rows and count how many times thoserows are repeated.

A= [4 4; 2 3; 4 2; 3 3; 2 3; 1 3; 3 3]
B=[1 1; 2 1; 3 1; 4 1; 1 2; 2 2; 3 2; 4 2; 1 3; 2 3; 3 3; 4 3; 1 4; 2 4; 3 4; 4 4]

Answer 1

[infinity]

Hello,

you can refer my answer as follows

clear

A= [4 4; 2 3; 4 2; 3 3; 2 3; 1 3; 3 3];
B=[1 1; 2 1; 3 1; 4 1; 1 2; 2 2; 3 2; 4 2; 1 3; 2 3; 3 3; 4 3; 1 4; 2 4; 3 4; 4 4];

n = size(A,1);

rowAinB = [];

for i = 1:n
    [LIA,LOCB] = ismember(A(i,:),B,'rows','legacy');
    if ~isempty(LOCB)
        rowAinB = [rowAinB i];
    end
end

repeatA = [];
C = unique(A,'rows','stable');

for i = 1:n
    [LIA,LOCB] = ismember(A(i,:),C,'rows','legacy');
    repeatA = [repeatA LOCB];
end

m = size(C,1);
coutrepeatA = zeros(m,1);
for i = 1:m
    idx = find(repeatA == repeatA(i));
    coutrepeatA(i) = length(idx)-1;
end

the unique of matrix A is C and number of repeated row of C is stored in vector "coutrepeatA". As you can see the results below

A =

     4     4
     2     3
     4     2
     3     3
     2     3
     1     3
     3     3
     
C =

     4     4
     2     3
     4     2
     3     3
     1     3
     
coutrepeatA =

     0
     1
     0
     1
     1     
     

Best regards,

Trung

[Osvaldo Bouche]

Hello Trung, this really solved my problem!

I have another question: i tried sorting the rows to have an specific order with the repeated rows. I'm using

A= [4 4; 2 3; 4 2; 3 3; 2 3; 1 3; 3 3; 4 4; 3 3; 4 2; 1 2];

A=sortrows(A)

but once i run the script , this results appear and they number of repeated rows is not correct; but i don't find where do i need to modify the code or if i should be trying something else.

C =

     1     2
     1     3
     2     3
     3     3
     4     2
     4     4


coutrepeatA =

     1
     1
     2
     2
     3
     3

[infinity]

Hello,

For the case of sorting matrix A. We should modify a little bit in the code since I have written it in very specific way. Here you can refer the modify

clear

% A= [4 4; 2 3; 4 2; 3 3; 2 3; 1 3; 3 3];
A= [4 4; 2 3; 4 2; 3 3; 2 3; 1 3; 3 3; 4 4; 3 3; 4 2; 1 2];
B=[1 1; 2 1; 3 1; 4 1; 1 2; 2 2; 3 2; 4 2; 1 3; 2 3; 3 3; 4 3; 1 4; 2 4; 3 4; 4 4];

A = sortrows(A);

n = size(A,1);

rowAinB = [];

for i = 1:n
    [LIA,LOCB] = ismember(A(i,:),B,'rows','legacy');
    if ~isempty(LOCB)
        rowAinB = [rowAinB i];
    end
end

repeatA = [];
C = unique(A,'rows','stable');

for i = 1:n
    [LIA,LOCB] = ismember(A(i,:),C,'rows','legacy');
    repeatA = [repeatA LOCB];
end

m = size(C,1);
coutrepeatA = zeros(m,1);

for i = 1:m
    idx = find(repeatA == i);
    coutrepeatA(i) = length(idx);
end

I have tested for the new matrix A that you provided in the above comment.

Hope it could help you.