why does parfor execute loops in a random order?
3 ビュー (過去 30 日間)
古いコメントを表示
I can't see any reason why, and if my PC crashes during a 20,000 loop iteration, it's a pain to figure out which iterations are still to be done.
Cheers Mike
0 件のコメント
採用された回答
Jason Ross
2013 年 2 月 28 日
Just to be clear -- parfor loops are of independent of iteration order, and do not guarantee deterministic results. It's not random.
The functional reason for this is if iteration 2 finishes before iteration 1 then iteration 3 can start work, and so on, since each iteration is independent of the other.
4 件のコメント
Jason Ross
2013 年 3 月 1 日
You might also want to investigate using the job/task interface, which gives you more control over the iterations and tracking down errors. A simple example:
c=parcluster();
alloutputs = [];
for ii=1:10
job = c.createJob;
for jj=1:10
createTask(job,@rand, 1, {3,3});
end
job.submit;
job.wait;
outputs = job.fetchOutputs;
alloutputs = [alloutputs ; outputs];
disp(ii);
job.destroy;
end
Note that things like preallocation, fancy formatting and error checking have been left out. Things that will likely be interesting to you:
- The job object contains the Task ID of errors. You could use this information to know what failed.
- You will likely need to figure out what work for you in terms of iteration display, number of tasks per job, dealing with the returns, etc.
その他の回答 (0 件)
参考
カテゴリ
Help Center および File Exchange で Startup and Shutdown についてさらに検索
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!