Parfor Loop Error with Classification
显示 更早的评论
When trying to convert the outer loop of the code below to a parfor loop, the error is:
Error: The variable z_new in a parfor cannot be classified.
See Parallel for Loops in MATLAB, "Overview"
Following the instruction on another post, I used a temporary variable to represent z_new, but that did not work either. Can anyone see what needs to be changed to permit this to run with parfor?
x = linspace(0,6,100); y = x; [xx,yy] = meshgrid(x,y);
z = sin(xx + 0.25*yy);
dx = 1; dy = 1; Dep = 100*ones(size(z));
z_new = z + Dep;
[nr,nc] = size(z);
nnx = round(Dep/dx,0); nny = round(Dep/dy,0);
parfor i = 1:nc
for j = 1:nr
for ii = i-nnx(j,i):i+nnx(j,i)
for jj = j-nny(j,i):j+nny(j,i)
iinew = ii;
jjnew = jj;
Thk = Dep.*Dep - dx*dx*(i-ii)*(i-ii) - dy*dy*(j-jj)*(j-jj);
if (ii > 0 && ii <= nc && jj > 0 && jj <= nr && Thk(j,i) >= 0)
Th_tmp = z(j,i) + sqrt(Thk(j,i));
if (iinew > 0 && iinew <= nc && jjnew > 0 && jjnew <= nr)
if z_new(jjnew,iinew) < Th_tmp
z_new(jjnew,iinew) = Th_tmp;
end
end
end
end
end
end
end
采纳的回答
Edric Ellis
2021-3-1
I think you can adapt this to work with parfor, but it's going to be a little bit of a stretch. Your variable z_new can be considered to be a parfor reduction variable. Normally, simple "reduction variables" are fine for computing a maximum value. Consider the simplest case I can think of:
largest = -Inf;
parfor i = 1:10
newValue = rand();
largest = max(newValue, largest);
end
This looks slightly different to your example in a couple of ways: firstly, there's no indexing into largest - we'll deal with that later; secondly - I've used max rather than if to convince parfor that largest is indeed a reduction variable.
So, next we need to deal with the indexing. This is a bit more complicated. We can do this like so:
largest = -Inf(1, 3); % Starting point
parfor i = 1:10
% Allocate an update for this iteration of the parfor loop
update = -Inf(size(largest));
for j = 1:3
% Fill out an element of 'update'
update(j) = rand();
end
% Finally, apply the update all at once
largest = max(largest, update);
end
The trick here is to create a single update array that we can use with a single "reduction" operation max. In your case, you'll need to allocate your equivalent of update just ahead of the loops over ii and jj and then apply it just after those loops.
4 个评论
Paul Safier
2021-3-1
Edric Ellis
2021-3-2
That's really close to working, but unfortunately, the rules for reduction variables are very stringent - you cannot access the reduction variable in any way at all other than the reduction expression. It took me a while to spot it, but it's the innocuous size(largest) that's tripping you over. Fortunately, in this case, it's simple to "hoist" that expression outside the loop like this:
szLargest = size(largest)
parfor ...
update = -Inf(szLargest);
end
Paul Safier
2021-3-2
Paul Safier
2021-3-2
更多回答（0 个）
类别
在 帮助中心 和 File Exchange 中查找有关 Loops and Conditional Statements 的更多信息
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!
Translated by 
