Increase counter for each element in array

Hey,

given A = [1 1 3 4 5 6 6 7 7 7]. How could I return the sequence B =[1 2 1 1 1 1 2 1 2 3] such that each duplicate is counted and B is the same size as A?

I appreciate your help!

Answer 1

[madhan ravi]

ix = A(:) == unique(A);
B = nonzeros(cumsum(ix) .* ix)

[Dario Walter]

Really nice solution madhan. It works nicely for small arrays of A. Yet, A is a vector of 70000 entries in my case. Your approach results in an out of memory error :/

[madhan ravi]

u = unique(A);
B = cell(1,numel(u));
for k = 1:numel(u)
    ix = A == u(k);
    B{k} = nonzeros(cumsum(ix) .* ix).';
end
B = [B{:}];

[KSSV]

+1 I don't know this....(For the first answer)

[Dario Walter]

Hey, thanks for your help so far. This approach only works for sorted data (which I am not allowed to do in my approach). Bad example I chose, sorry for that. What about B = [4 4 3 3 3 2 2 1 1 7 7 9 9] (each unique number only appears in a sequence in my problem, so that, for instance, C = [4 4 3 4] does not exist.

[B,I] = sort(A,'ascend') does neither help.

[madhan ravi]

I knew you would come up with that question. That’s why you should experiment with 'stable' option in unique function.

[Dario Walter]

Well done Madhan :). Thanks a lot!