matlabpool open n; doesn't work.

조회 수: 5 (최근 30일)
Martin
Martin 2014년 5월 7일
댓글: Bruno Silva 2020년 1월 17일
Hi,
I'd like to open matlabpool with a variable number of cores, to use the maximum number of cores available. So n would equal features('numCores') here.
But matlabpool open n, matlabpool open features('numCores'), matlabpool open maxNumCompThreads don't work here. matlabpool open seems to accept only numbers and not variables.
I could do something stupid like
if features('numCores') == 4
matlabpool open 4;
elseif features('numCores') == 2
matlabpool open 2;
end
but I'd like to know if something more elegant exists.
Thanks!

채택된 답변

Niklas Nylén
Niklas Nylén 2014년 5월 7일
편집: Niklas Nylén 2014년 5월 7일
This is because writing arguments like this:
matlabpool open 2
is the same as passing the strings 'open' and '2' to the matlabpool function, i.e. the equivalent to
matlabpool('open','2')
What you can do is to call matlabpool like this:
matlabpool('open', num2str(features('numCores')))
Possibly you can also pass the number of cores directly, but since I do not have parallell computing toolbox I can't test it, like so:
matlabpool('open', features('numCores'))

추가 답변 (2개)

Matt J
Matt J 2014년 5월 7일
편집: Matt J 2014년 5월 7일
If you have R2013b or later, you shouldn't be using matlabpool . You should be using parpool , which does take a variable as input directly,
parpool(n)
  댓글 수: 3
Haseeb Hassan
Haseeb Hassan 2018년 10월 7일
Thank you Matt for your useful hint.
Bruno Silva
Bruno Silva 2020년 1월 17일
Thanks!

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Malshikho
Malshikho 2018년 5월 25일
Unfortunately I have Matlab R2010b
  댓글 수: 1
Steven Lord
Steven Lord 2018년 5월 25일
Then use matlabpool using the syntax described in the accepted answer by Niklas Nylén.

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