# How to count the frequency of each type of column vector present in a matrix

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Radu Mihail on 18 Mar 2021
Commented: David Hill on 19 Mar 2021
I have a matrix R(21,3)
R=[1,1,1,2,2,2,1,2,2,1,4,2,2,2,2,1,2,1,3,1,1;1,1,2,2,2,1,2,2,1,4,2,2,2,2,1,2,1,3,1,1,1;1,2,2,2,1,2,2,1,4,2,2,2,2,1,2,1,3,1,1,1,1];
and I would like to count the frequency of each column vector present from the whole matrix. For instance column vector a=[2,2,2]' is present 3 times, vector b=[2,2,1]' is present 3 times, vector c=[1,4,2]' is present only once. A table with all column vectors and their frequency.

David Hill on 18 Mar 2021
R=[1,1,1,2,2,2,1,2,2,1,4,2,2,2,2,1,2,1,3,1,1;1,1,2,2,2,1,2,2,1,4,2,2,2,2,1,2,1,3,1,1,1;1,2,2,2,1,2,2,1,4,2,2,2,2,1,2,1,3,1,1,1,1];
a=[2 2 2];
nnz(ismember(R',a,'rows'));
David Hill on 19 Mar 2021
R=[1,1,1,2,2,2,1,2,2,1,4,2,2,2,2,1,2,1,3,1,1;1,1,2,2,2,1,2,2,1,4,2,2,2,2,1,2,1,3,1,1,1;1,2,2,2,1,2,2,1,4,2,2,2,2,1,2,1,3,1,1,1,1];
[r,idx]=sort(sum(R.*[13;17;19]));%choose primes such that the number of repetitions of each will not exceed
[rr,idx2]=unique(r);
numberOfeachUniqueR=histcounts(r,[rr,r(end)+1]);%generates the number of occurrances of each unique triplet
UniqueR=R(:,idx(idx2));%generates cooresponding unique tripets (just like your output above)

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