matlabpool open n; doesn't work.

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!

 Respuesta aceptada

Niklas Nylén
Niklas Nylén el 7 de Mayo de 2014
Editada: Niklas Nylén el 7 de Mayo de 2014
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'))

Más respuestas (2)

Matt J
Matt J el 7 de Mayo de 2014
Editada: Matt J el 7 de Mayo de 2014
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)

3 comentarios

Martin
Martin el 7 de Mayo de 2014
Ok, thank you for your answers. I use Matlab 2012, does parpool work too?
Haseeb Hassan
Haseeb Hassan el 7 de Oct. de 2018
Thank you Matt for your useful hint.
Bruno Silva
Bruno Silva el 17 de En. de 2020
Thanks!

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Malshikho
Malshikho el 25 de Mayo de 2018

0 votos

Unfortunately I have Matlab R2010b

1 comentario

Steven Lord
Steven Lord el 25 de Mayo de 2018
Then use matlabpool using the syntax described in the accepted answer by Niklas Nylén.

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