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Effacer les filtres

How can I decrease the runtime for a function (using cellfun or parfor)?

3 vues (au cours des 30 derniers jours)
AES
AES le 11 Mar 2024
Commenté : AES le 12 Mar 2024
I am trying to speed some code up. Right now each iteration takes about 6 seconds to run and over the course of many iterations the runtime drastically increases. I tried two different running strategies: using cellfun and parallelization. Will paralization ever be slower than using cellfun? I get hugely varying times for the parfor on different computers due to different number of cores. Will the cellfun be more consistent (but slower) across computers? Below is my code. Thank you in advance.
Here is the function call:
%First way
testOut = cellfun(@testDFF_trial, num2cell(fReflect, 2), num2cell(fReflect2, 2), ...
num2cell(fSom, 2), num2cell(ones(size(fReflect, 1), 1) .* numFrames), 'UniformOutput', false);
%Second way
parfor numCell = 1:size(fReflect, 1)
dffAll{numCell} = testDFF(fReflect, fReflect2, fSom, numFrames, numCell); %slightly altered function for parfor compatibility
end
Here is the function that I am trying to speed up:
function dffTest = testDFF(fReflect, fReflect2, fSom, numFrames)
stepSize = 1;
windowSize = 900 / 2; %12
startLoc = 451; %20
endLoc = numFrames + 450; %80
%Gets the starting numbers
usedRange = startLoc:stepSize:endLoc;
%Gets the index that matter for each window
aOut = arrayfun(@(a, window) a-window:a+window, usedRange, ones(size(fReflect(usedRange))) .* windowSize, 'UniformOutput', false);
aOut2 = arrayfun(@(a, window) a-window:a+window, usedRange, ones(size(fReflect2(usedRange))) .* windowSize, 'UniformOutput', false);
%Makes ones array for later input
onesAll = (ones(1, length(fReflect)) .* fReflect);
onesAll2 = (ones(1, length(fReflect2)) .* fReflect2);
%Gets the values within each index
myfun = @(input, index) input(index);
datOut = cellfun(@(index) myfun(onesAll, index), aOut, 'UniformOutput', false);
datOut2 = cellfun(@(index) myfun(onesAll2, index), aOut2, 'UniformOutput', false);
%Find the 8th percentile
prcDat = cellfun(@prctile, datOut, num2cell(ones(1, length(datOut)) .* 8));
prcDat2 = cellfun(@prctile, datOut2, num2cell(ones(1, length(datOut2)) .* 8));
%Find the dff
dffFunct = @(rawF, prcF, baseline) (rawF - prcF) ./ baseline;
dffTest = arrayfun(dffFunct, fSom, prcDat, prcDat2);
tInter = toc;
display(['Cell processed - time: ', num2str(tInter / 60, '%0.2f')]);
end
  4 commentaires
Steven Lord
Steven Lord le 12 Mar 2024
FYI, another tool to add to your programming arsenal is the Profiler. This can help you understand what the most expensive location or segment of your code is. Using that you can also experiment with different approaches to determine if a change is better, worse, or the same in terms of performance. [That reminds me of my last eye exam ;)]
AES
AES le 12 Mar 2024
That sounds familiar, but I have never used it before. Definitely will take a look!

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Voss
Voss le 11 Mar 2024
Modifié(e) : Voss le 12 Mar 2024
Before worrying about cellfun vs parfor to call the function, try to make the function itself more efficient.
Compare the function testDFF_modified below with the original testDFF.
I don't know what typical inputs to the function are, but with these made-up inputs:
fReflect = rand(1,1000);
fReflect2 = rand(1,1100);
fSom = rand(1,50);
numFrames = 50;
The result is the same:
out1 = testDFF(fReflect, fReflect2, fSom, numFrames);
out2 = testDFF_modified(fReflect, fReflect2, fSom, numFrames);
isequal(out1,out2)
ans = logical
1
And the modified function is about 3-4 times faster:
t1 = timeit(@()testDFF(fReflect, fReflect2, fSom, numFrames));
t2 = timeit(@()testDFF_modified(fReflect, fReflect2, fSom, numFrames));
fprintf('old: %5.3f ms\nnew: %5.3f ms\n',t1*1000,t2*1000);
old: 7.947 ms new: 2.114 ms
function dffTest = testDFF_modified(fReflect, fReflect2, fSom, numFrames)
stepSize = 1;
windowSize = 900 / 2; %12
startLoc = 451; %20
endLoc = numFrames + 450; %80
%Gets the starting numbers
usedRange = startLoc:stepSize:endLoc;
% matrix of indices:
idx = usedRange+(-windowSize:windowSize).';
% simple indexing, and taking advantage of the fact that prctile
% given a matrix operates by column:
dffTest = (fSom-prctile(fReflect(idx),8))./prctile(fReflect2(idx),8);
end
function dffTest = testDFF(fReflect, fReflect2, fSom, numFrames)
stepSize = 1;
windowSize = 900 / 2; %12
startLoc = 451; %20
endLoc = numFrames + 450; %80
%Gets the starting numbers
usedRange = startLoc:stepSize:endLoc;
%Gets the index that matter for each window
aOut = arrayfun(@(a, window) a-window:a+window, usedRange, ones(size(fReflect(usedRange))) .* windowSize, 'UniformOutput', false);
aOut2 = arrayfun(@(a, window) a-window:a+window, usedRange, ones(size(fReflect2(usedRange))) .* windowSize, 'UniformOutput', false);
%Makes ones array for later input
onesAll = (ones(1, length(fReflect)) .* fReflect);
onesAll2 = (ones(1, length(fReflect2)) .* fReflect2);
%Gets the values within each index
myfun = @(input, index) input(index);
datOut = cellfun(@(index) myfun(onesAll, index), aOut, 'UniformOutput', false);
datOut2 = cellfun(@(index) myfun(onesAll2, index), aOut2, 'UniformOutput', false);
%Find the 8th percentile
prcDat = cellfun(@prctile, datOut, num2cell(ones(1, length(datOut)) .* 8));
prcDat2 = cellfun(@prctile, datOut2, num2cell(ones(1, length(datOut2)) .* 8));
%Find the dff
dffFunct = @(rawF, prcF, baseline) (rawF - prcF) ./ baseline;
dffTest = arrayfun(dffFunct, fSom, prcDat, prcDat2);
% tInter = toc;
% display(['Cell processed - time: ', num2str(tInter / 60, '%0.2f')]);
end
  6 commentaires
AES
AES le 12 Mar 2024
Thank you. This is very helpul and learned a lot,.
Voss
Voss le 12 Mar 2024
You're welcome!

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