How can I map to each vector entry a number telling its occurence so far?

I have a vector

vec = [1,2,3,4,2,5,2,3];

and I want to have the output

[1,1,1,1,2,1,3,2]

so that each number tells me how often the value so far has occurred in the vector. Is there any possibility to do this without a for/while loop?

Answer 1

[Stephen23]

No third-party functions required, four lines of code:

>> vec = 5+[1,2,3,4,2,5,2,3];
>> [uni,~,idx] = unique(vec);
>> tmp = cell2mat(arrayfun(@(n)1:n,histc(vec,uni),'uni',0));
>> [~,idy] = sort(idx);
>> tmp(idy) % code output
ans =
   1   1   1   1   2   1   3   2
>> [1,1,1,1,2,1,3,2] % requested output
ans =
   1   1   1   1   2   1   3   2

[Guillaume]

I would argue that arrayfun is a loop.

While it is indeed shorter, it is also about 4 times as slow on my machine. On the other hand both take less than half a second for a 1e6 element vector, so it doesn't really matter.

Answer 2

[Guillaume]

It can be done without a loop, however for part of the code I recommend you use rcumsum from the FileExchange instead of what I've done here:

vec = [1,2,3,4,2,5,2,3];
[ordervec, indices ] = sort(vec); %group indentical values together and store their original position
issameasprevious = [0, ~diff(ordervec)];

%the following lines can be replaced by
%runlength = rcumsum(issameasprevious) + 1;
%otherwise:
zeroloc = ~issameasprevious;
runlength = cumsum(issameasprevious);   %compute temporary cumsum with no reset 
issameasprevious(zeroloc) = -diff([0 runlength(zeroloc)]);  % how much to subtract at 0 values?
runlength = cumsum(issameasprevious) + 1;  %recompute cumsum which no reset at 0

result = runlength(indices)  %reorder according to original order

edit: typo in the code

[Guillaume]

Note: my recommendation to use rcumsum apply if you're going to use the compiled mex version of it. It's optional and if you don't the code is exactly equivalent to what I've written (except it's going to be slower because of the input checks).

If you use the mex version of rcumsum then the above is going to be much faster than alternative solutions (and is also just 4 lines of code)

Answer 3

[Udit Dhand]

I wasnt able to do it without using while loop

vec = [1,2,3,4,2,5,2,3];

j=length(vec);k=0;
while k~=j
o = find(vec==vec(j-k));
i=1:length(o);
out(o)=i;
k=k+1;
end