Fastest Way to Keep the last N Outputs of a Loop in a Matrix
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Hey everyone, I have some code with a for loop where I would like to store the last N outcomes and do calculations intermittently with the last N evaluations within the loop. I am storing each evaluation result in a column of the matrix mydata. When the number of steps passes N, I remove the first column of data (as it is the oldest) and add my result at N+1th step to the final column as following:
mydata(:,1)=[];
mydata(:,N)=newdata;
While this works fine, it is very time-consuming. Is there a faster way to remove the first column of data and add the new data to the last column of my matrix? Note that I cannot keep all the data as it would end up being a very large matrix.
4 commentaires
dpb
le 28 Fév 2023
Not surprising but thought just outside chance the optimizer might be able to make it not too bad...not much lost to try and I had to go at the time...
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Jan
le 28 Fév 2023
Modifié(e) : Jan
le 28 Fév 2023
If you do not know the number of iterations in advance, a circulare buffer is more efficient than shrinking an expanding an array repeatedly:
N = 10;
Buffer = nan(N, 5);
iBuffer = 0;
while rand > 0.05
data = rand(1, 5);
iBuffer = mod(iBuffer, N) + 1; % Wrap around cursor
Buffer(iBuffer, :) = data;
end
LastData = Buffer([iBuffer+1:N, 1:iBuffer], :);
4 commentaires
Jan
le 1 Mar 2023
The circular buffer is filled in each iteration, but only the last N iterations matter. If the number of iterations is known in advance, it is cheaper to fill the buffer only in the last N iterations. The offset is adjusted dynamically to match the number of iterations and the wanted length of the buffer.
The circular buffer is working also, but it wastes time for overwriting memory.
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