Counting occurrences in order

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Matt Fig
Matt Fig on 13 Sep 2012
I am wondering if anyone knows of a toolbox function that can perform this task quickly. I am not familiar with the Stats toolbox, but I have a feeling something could be there. Or maybe even the Image Processing Toolbox?
Here is the basic problem, and a naive solution. I want to count the occurrences of each unique item in a vector in the order in which they appear.
N = 5e5; % Typical _minimum_ size, max N = 5e7.
C = 50; % Best case shown, worst case: C = 1.5;
A = randi(floor(N/C),1,N); % Data looks like this.
% Now to produce our results. Need to find B.
Au = unique(A);
H = histc(A,Au);
B = zeros(size(A));
for ii = 1:length(Au)
B(A==Au(ii)) = 1:H(ii);
end
As a simple example for clarity, for N=10, C=3:
[A(:) B(:)]
ans =
3 1
2 1
3 2
3 3
3 4
2 2
3 5
3 6
3 7
1 1
I have written code that reduces the run-time by a large factor with the worst case C and minimum N, but it is a bit convoluted. After doing so I thought there might already be something like this out there, perhaps in a toolbox, but I couldn't find it. Perhaps there is a way to make further improvements for larger N.
Thanks!
EDIT I have not found any other solution to the this, so I may delete the question....
  2 Comments
Sean de Wolski
Sean de Wolski on 4 Oct 2012
I would hazard a guess that removing unique() would be the biggest improvement. Though the one time cost might be worth it..
Ashish Uthama
Ashish Uthama on 5 Oct 2012
This is interesting. Thinking along the lines of using some kind of hash table. Please dont delete!

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Accepted Answer

Image Analyst
Image Analyst on 4 Oct 2012
I'm not sure why you want that but anyway it seems kind of esoteric/specialized. I can't think of anything in the Image Processing Toolbox that would get you the same thing as your code.
  1 Comment
Matt Fig
Matt Fig on 4 Oct 2012
Thanks, IA. I thought there might be something either in the IPT or Stats. Perhaps not. Yes, it is very specialized ;-).

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More Answers (1)

Matt J
Matt J on 5 Oct 2012
Edited: Matt J on 5 Oct 2012
  1 Comment
Matt Fig
Matt Fig on 5 Oct 2012
Haha, Matt J! That is very similar. Perhaps my question is not as specialized as I thought. Thanks for the link, I don't know how I missed that question.

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