When I try to run the following parfor loop, Matlab returns the error claiming that A cannot be classified.
N = 10;
T = 3;
A = zeros(N,T);
parfor i = 1:1:N
for t = 1:1:T
A(i,t) = rand(1,1);
end
end
Why isn't A considered to be a sliced output variable? The following modification works just fine.
N = 10;
T = 3;
A = zeros(N,T);
parfor i = 1:1:N
A(i,1) = rand(1,1);
end
So it seems to be something to do with the inner loop over t. From reading the help document on sliced variables, I don't understand why this inner loop would make a difference.

2 commentaires

even coil
even coil le 11 Déc 2015
Is there a workaround besides using the vectorized version of rand? (Doing the analgous would be difficult in the application I have in mind.)
Matt J
Matt J le 11 Déc 2015
Modifié(e) : Matt J le 11 Déc 2015
If you interchange the order of the loops you've shown, it will run. Whether that's the best solution is hard to tell, though, without seeing an example closer to the application you have in mind.

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 Réponse acceptée

Edric Ellis
Edric Ellis le 11 Déc 2015
In this case, the fix is simply to remove the increment from the inner for loop, like so:
N = 10;
T = 3;
A = zeros(N,T);
parfor i = 1:1:N
for t = 1:T
A(i,t) = rand(1,1);
end
end
(I'm not sure why this limitation exists, it doesn't appear to be explicitly mentioned in the documentation)

3 commentaires

Matt J
Matt J le 11 Déc 2015
Modifié(e) : Matt J le 11 Déc 2015
(I'm not sure why this limitation exists, it doesn't appear to be explicitly mentioned in the documentation)
Maybe for the same reasons that this isn't supported:
parfor i = 1:1:N
for t = 1:i:T
A(i,t) = rand(1,1);
end
end
If A isn't pre-allocated, I can see parfor having a hard time figuring out what size(A) should ultimately be.
even coil
even coil le 11 Déc 2015
Why isn't this considered a bug? Isn't "1:T" just shorthand for "1:1:T"?
Matt J
Matt J le 11 Déc 2015
Modifié(e) : Matt J le 11 Déc 2015
Hard to say if it can be called a bug because it's not yet clear what level of support is intended for nested loops that interact with sliced variables. Note, for example, that the following also produces a classification error,
parfor i = 1:1:N
for t = 1:2:T
A(i,t) = rand(1,1);
end
end

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Plus de réponses (1)

Walter Roberson
Walter Roberson le 11 Déc 2015
N = 10;
T = 3;
A = zeros(N,T);
parfor i = 1:1:N
v = zeros(1,T);
for t = 1:1:T
v(1,t) = rand(1,1);
end
A(i, :) = v;
end

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