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

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.

Answer 1

[David Hill]

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'));

[Radu Mihail]

Thank you David,

I do have hundreds of such matrices and predicting a type of column vector(like you did for a) is not always possible, to use your answer. I would like some formula to simply count the frequency of every vector present without interrogating a special one

[David Hill]

You could multiply by different primes to get unique sums

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=sort(sum(R.*[13;17;19]));
k=histcounts(r,[unique(r),r(end)+1]);

[Radu Mihail]

I am sorry this is just to identify unique vectors. I do not follow and I do not understand how to get the table of all column vector frequencies distributions in the matrix. I need something like this

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

Frequency of column vector(mentioned once)

------------------------------------------------------------------------------------

3 1 2 3 3 2 - - 1 1 1 - - - - 1 1 1 1 - -

[David Hill]

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)