How to search for a previous element that is equal in prior loop iteration

I have a Matrix called A as below

[1  1  3;
1  2  2;
1  3  3;
2  1  1;
2  2  3;
2  3  2;
2  4  4]

I want to produce a column vector that counts the number of times a value occurs in the 2nd column, I have tried the following function :

function count_stars(A)
total = 0;
for i = 1:size(A,1)
    total = total + (A(i,2)==A(:,2));
   end
total

This counts the number of matching values in the entire column e.g.

       2
       2
       2
       2
       2
       2
       1

What I am wanting is to match to values between 1:i only rather than the entire column so that the output looks like this

1
1
1
2
2
2
1

Any insights you are able to share are greatly appreciated!

[Image Analyst]

It's too late at night for me to work on this, but I'm thinking it might involve accumarray(), cumsum() or histcounts(). A question though, will the numbers always be integers? Or might they be floating point numbers with fractional parts?

[Tim Baker]

Thanks for your response, yes the numbers are always integers. Am interested in discovering other ways to solve the problem if it can be done more efficiently. Harjeet's solution works well.

Answer 1

[harjeet singh]

hello tim try this code

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

for i=1:size(a,1)
    [r,c]=find(a(1:i,2)==a(i,2));
    b(i,1)=length(r);
end
b

[Tim Baker]

Thanks Harjeet for your prompt answer, its works a treat.

Answer 2

[Guillaume]

For small arrays, it does not matter but this is likely to be more efficient than harjeet's answer for large arrays:

a = [1 1 3;
     1 2 2;
     1 3 3;       
     2 1 1;
     2 2 3;
     2 3 2;
     2 4 4];
[values] = unique(a(:, 3));
counts = zeros(size(a, 1), 1);
for val = values';
    occurences = cumsum(a(:, 3) == val);
    counts(a(:, 3) == val) = occurences(a(:, 3) == val);
end

Rather than looping over all the rows of a, it only loops over the unique values of a (4 iterations instead of 7 in this example).

[Tim Baker]

Wow! Your approach is significantly faster, thanks for your response.