Is there a workaround besides using the vectorized version of rand? (Doing the analgous would be difficult in the application I have in mind.)
What's wrong with this parfor loop?
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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 Comments
Accepted Answer
Edric Ellis
on 11 Dec 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 Comments
Matt J
on 11 Dec 2015
Edited: Matt J
on 11 Dec 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
More Answers (1)
Walter Roberson
on 11 Dec 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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