Parfor Execution time variation

2 Ansichten (letzte 30 Tage)
Anshika Goel
Anshika Goel am 6 Jun. 2024
Kommentiert: Anshika Goel am 21 Aug. 2024
Hi,
I am using parfor for reading 600 .raw files.
c=zeros(1536,1536,600,'uint16');
parpool('threads',4);
parfor i=1:600
fileName=[folder,'/',fileList(i).name];
a=fopen(fileName,'r');
Z=fread(a,[1536 1536],'uint16');
fclose(a);
c(:,:,i)=Z;
end
However, I am observing significant variability in the execution time, which ranges from 14 seconds to 110 seconds across different runs.
Why is this discrepancy occurring? Is there a way to achieve more consistent execution times?
  4 Kommentare
Christopher Mirfin
Christopher Mirfin am 12 Jun. 2024
Do you observe the same variability when running with a standard for-loop, or a process-based pool parpool("Processes",4) ?
Also, are you reading from your local hard drive or a network location?
Anshika Goel
Anshika Goel am 13 Jun. 2024
Bearbeitet: Anshika Goel am 13 Jun. 2024
In the standard for loop, I am not getting any variability it is taking 35-40 sec.
Whereas, in a process based pool, the variability is less (55-63 sec), but it is taking more time than the standard for loop.
And I am reading from a local hard drive,not from network location.
Is there any other solution, where I can reduce this execution time to less than 15 sec.

Melden Sie sich an, um zu kommentieren.

Akzeptierte Antwort

Swastik Sarkar
Swastik Sarkar am 21 Aug. 2024
I also have an Intel Xeon processor (4 cores) with 16GB RAM. I executed your code after generating 600 files as follows:
matrix = uint16(ones(1536));
folder = 'nums';
for i=1:600
fid = fopen([folder '/' num2str(i)], 'w');
mat = matrix .* i;
fwrite(fid, mat, 'uint16');
fclose(fid);
end
In my tests, the variability in execution time was not as significant as you mentioned; it ranged from 189 seconds to 200 seconds. the main bottleneck is likely due to file I/O operations.
To optimize execution time, consider performing file reads asynchronously. I developed the following code using parfeval to read 600 files asynchronously:
c = zeros(1536, 1536, 600, 'uint16');
folder = "nums";
pool = gcp('nocreate');
if isempty(pool)
pool = parpool('threads');
end
futures = parallel.FevalFuture.empty(600, 0);
for i = 1:600
fileName = fullfile(folder, num2str(i));
futures(i) = parfeval(@readFile, 1, fileName);
end
for i = 1:600
c(:, :, i) = fetchOutputs(futures(i));
end
function Z = readFile(fileName)
a = fopen(fileName, 'r');
Z = fread(a, [1536 1536], 'uint16');
fclose(a);
end
In my tests, this approach reduced the execution time to between 14-16 seconds.
You can learn more about “parfeval” from here:
I hope this helps.

Weitere Antworten (0)

Kategorien

Mehr zu Parallel for-Loops (parfor) finden Sie in Help Center und File Exchange

Produkte


Version

R2023a

Community Treasure Hunt

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

Start Hunting!

Translated by