# Why the weird nested arrayfun is most time effective?

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Maxim on 21 Aug 2014
Edited: Maxim on 21 Aug 2014
I was writing a simple Autocorrelation function for the matrix array. Each row is a separate time series and we autocorrelate it with itself with certain set of lags. I have found out that one of the most contrintuitive things works best.
% 17 second benchmark
A = cell2mat(arrayfun(@(i) AutoCorrelation(array(i,:),lags)',1:size(array,1),'UniformOutput',false))';
where the function itself is the following
function ans = AutoCorrelation(array,lags)
arrayfun(@(x) dot(array(lags(x)+1:end),array(1:end-lags(x)))/(length(array)-lags(x)),1:length(lags));
end
Other things I have tried:
A = zeros(size(array,1),length(lags));
T = size(array,2);
% 97 seconds benchmark
A = cell2mat(arrayfun(@(i) arrayfun(@(x) dot(array(i,lags(x)+1:end),array(i,1:end-lags(x)))/(size(array,2)-lags(x)),1:length(lags))',1:size(array,1),'UniformOutput',false))';
% 100 seconds benchmark
A = arrayfun(@(i,x) dot(array(i,lags(x)+1:end),array(i,1:end-lags(x)))/(size(array,2)-lags(x)),repmat((1:size(array,1))',1,length(lags)),repmat(1:length(lags),size(array,1),1));
% 27 second benchmark
for i = 1:length(lags)
A(:,i) = dot(array(:,lags(i)+1:end),array(:,1:end-lags(i)),2)/(T-lags(i));
end
% 95 second benchmark
for i = 1:length(lags)
for j = 1:size(array,1);
A(j,i) = dot(array(j,lags(i)+1:end),array(j,1:end-lags(i)),2)/(T-lags(i));
end
end
It is more a curiosity question. If you ask me I would bet a direct dot product method would work the best. Also if arrayfun works that well, why then the double argument arrayfun doesn't perform well?
My array was 512*100000 array of doubles.

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