Fast aggregates over overlapping subsets of a vector

2 views (last 30 days)
I have a long vector of values say x=rand(1e5,1), and a long but shorter cellarray with indices to the elements of x. I just want to compute means or sums over the vector subsets using these indices. There is a variable number of indices in cells and the indices are exclusive within each cell but may not be exclusive among the cells i.e. different cells may contain the same indices. I am aware of the accumarray but I assumed it can work for mutually exclusive sets of indices i.e. each element is used only once in aggregation. Is there any faster way to do that than in a loop:
n=1e5;
%Vector
x=rand(n,1);
%Random length indices
id=arrayfun(@(i) randperm(n, randi(n)), 1:round(n/10), 'UniformOutput',false)';
%And a loop test scenario
tic; y=zeros(size(id)); for i=1:numel(id) y(i)=mean(q(id{i})); end; t=toc
%t=2.788
Cheers
  2 Comments
Dyuman Joshi
Dyuman Joshi on 14 Feb 2023
Is there a need to store those indices? If not then, you can call them in each iteration and store mean accordingly -
n=1e5;
%Vector
y=zeros(1,round(n/10));
for k=1:round(n/10)
y(i)=mean(q(randperm(n,randi(n))));
end
However, You have not used x in the code and we don't know what q is, thus it's not clear what you exactly want to do.
dymitr ruta
dymitr ruta on 14 Feb 2023
Edited: dymitr ruta on 14 Feb 2023
Unfortunately yes, these indices come out from kdtree rangesearch() function and in a form of cellarray of variable length indices' vectors. Given the vector x and indices id as input I just need to get fastest computed means

Sign in to comment.

Answers (1)

Swaraj
Swaraj on 7 Mar 2023
You are correct in saying that “accumarray” only functions for sets of indices that are mutually exclusive. You can apply a function (like mean or sum) to each cell in your cell array that contains indices by using the “cellfun” function.
Please go through the below documentation for more details.

Products


Release

R2022b

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!