Matching values in different columns?

Hi everyone,

I have a matrix with two columns of data. The data in the first column vary from 1 to 120. The data in the second column vary from 1 to 12.

A= [10 13 10 1 111 102 10 13 7 10 112 100 14 116 102 10 14 120].

B= [2 4 5 7 12 1 2 6 7 9 12 2 4 7 9 11 12 12].

I would like to calculate the number of times each element of column A matches with different elements of column B. For example, the number 10 matches two times with 2, and one time with 5, 9 and 11.

How can I do that?

Thanks.

Answer 1

[Adam Danz]

If you have r2019a, use groupcounts().

%Original data
A= [10 13 10 1 111 102 10 13 7 10 112 100 14 116 102 10 14 120];
B= [2 4 5 7 12 1 2 6 7 9 12 2 4 7 9 11 12 12];
% Put data into a table
T = table(A.', B.','VariableNames',{'A','B'}); 
matchTable = groupcounts(T,{'A','B'}); % summary table 

Result

matchTable =
  17×3 table
     A     B     GroupCount
    ___    __    __________
      1     7        1     
      7     7        1     
     10     2        2     
     10     5        1     
     10     9        1     
     10    11        1     
     13     4        1     
     13     6        1     
     14     4        1     
     14    12        1     
    100     2        1     
    102     1        1     
    102     9        1     
    111    12        1     
    112    12        1     
    116     7        1     
    120    12        1  

If your version of matlab is prior to r2019a

You can use unique() and histcounts() to get the same result as above.

%Original data
A= [10 13 10 1 111 102 10 13 7 10 112 100 14 116 102 10 14 120];
B= [2 4 5 7 12 1 2 6 7 9 12 2 4 7 9 11 12 12];

[sorted, ~, groups] = unique(sortrows([A.',B.']),'rows','stable');
counts = histcounts(categorical(groups),categorical(unique(groups,'stable'))); 
matchTable = array2table([sorted,counts.'],'VariableNames',{'A','B','Count'});  % summary table 

[Aragorn23]

Thank you Adam Danz. The first one, fit perfectly.

[Adam Danz]

Nice! I just learned about that function myself today!

[Julian]

BTW, you can get exactly the same answer with groupsummary

groupsummary(T,{'A','B'}); % summary table

that was introduced 1 year before groupcounts (since R2018a).