Is it possible to use use `parfor` for parallel computing in Matlab in these codes?
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Jamais avenir
el 4 de Mzo. de 2015
Comentada: Jamais avenir
el 5 de Mzo. de 2015
I am using parfor for parallel computing in Matlab. I am not familiar with this command. If that is possible, please look at my code below and tell me if I can write it with parfor. These errors and warnings are appear in Matlab Editor:
The parfor loop cannot be run due to the way variable 'Dat' is used.
The entire array or structure 'Bound' is broadcast variable. This might result in unnecessary communication overhead.
parfor pj = 1:size(normXpj,1)
Dat.normXpj = normXpj(pj,:);
if size(Dat.InitialGuess)==0
X = (Bound(:,1)+(Bound(:,2)-Bound(:,1)).*rand(Nvar,1))';
else
X = Dat.InitialGuess;
end
[Xsqp, ~, FLAG,Options] = mopOPT(X,Dat);
FEVALS = Options.funcCount;
FES = FES+FEVALS;
PSet(pj,:) = Xsqp;
PFront(pj,:) = mop(Xsqp,Dat,0);
%disp(['ExitFlag: ' num2str(FLAG)]);
if FLAG==-2
disp('.......... Algo paso...');
else
F = PFront(pj,:);
if Nobj==2
plot(F(1,1),F(1,2),'*r'); grid on; hold on;
elseif Nobj==3
end
end
end
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Edric Ellis
el 5 de Mzo. de 2015
The problem here is that it we can see that you're not using `Dat` in a way that is order-dependent, but the static analysis machinery of `parfor` cannot deduce that because of the way you're assigning into it. I think you can work around this by instead creating a whole new `Dat` for each iteration of the loop, like so:
Dat = struct('normXpj', rand(10,1), 'InitialGuess', 3);
normXpj = rand(10);
parfor idx = 1:10
tmpDat = struct('normXpj', normXpj(:,idx), 'InitialGuess', Dat.InitialGuess);
% use 'tmpDat'
disp(tmpDat);
end
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Matt J
el 4 de Mzo. de 2015
Editada: Matt J
el 4 de Mzo. de 2015
You should familiarize yourself with the material on classification of variables in parfor loops,
This line, in particular,
Dat.normXpj = normXpj(pj,:);
is not allowed. Since Dat is a broadcast variable, you cannot assign to it. It also doesn't make sense to try to do so. Assigning different parallel pj-dependent values to the same value would lead to an ill-defined value when the parfor loop ends.
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