Parfor variable cannot be classified
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I'm completely stumped by my inability to use parfor for a simple problem. I've boiled it down to a ridiculous case. Can someone please explain?
function b=wtf(a)
sz = size(a);
b=a;
parfor x=1:sz(1)
for y=1:sz(2)
b(x,y) = 5;
end
end
Calling this results in an error:
>> wtf(zeros(3))
Error: File: wtf.m Line: 8 Column: 9
The variable b in a parfor cannot be classified.
See Parallel for Loops in MATLAB, "Overview".
Is it actually not possible to use parfor when the size of the input is not absolutely fixed ahead of time?
采纳的回答
Edric Ellis
2019-6-28
编辑：Edric Ellis
2019-6-28
You've tripped over one of the limitations of for loops nested within parfor. This is described in the doc. The requirement is:
You must define the range of a for-loop nested in a parfor-loop by constant numbers or broadcast variables.
Unfortunately, even indexing a broadcast variable for the range of the for loop is not permitted, so what you need to do is pull out the elements of the size vector outside the loop like this:
function b = fixed(a)
[m,n] = size(a);
b=a;
parfor x=1:m
for y=1:n
b(x,y) = 5;
end
end
更多回答（1 个）
I don't know what rule your version violates, but this will work,
parfor x=1:sz(1)
b(x,:) = 5;
end
So will this,
parfor x=1:sz(1)
tmp=nan(1,sz(2));
for y=1:sz(2)
tmp(y)=5;
end
b(x,:) = tmp;
end
I suspect the problem has to do with the fact that b is a Sliced Variable. Sliced variables have some strict rules - some well documented, others not so well documented - about how they can be indexed.
1 个评论
Another version, where tmp's memory is pre-allocated:
buffer=nan(1,sz(2));
parfor x=1:sz(1)
tmp=buffer;
for y=1:sz(2)
tmp(y)=5;
end
b(x,:) = tmp;
end
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