Parfor Loop Error with Classification

조회 수: 2 (최근 30일)
Paul Safier
Paul Safier 2021년 2월 28일
댓글: Paul Safier 2021년 3월 2일
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

채택된 답변

Edric Ellis
Edric Ellis 2021년 3월 1일
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.
  댓글 수: 4
Paul Safier
Paul Safier 2021년 3월 2일
@Edric Ellis : Thanks so much!! That worked. What a benign hang-up! Thank you for your help, I appreciate it!
Paul Safier
Paul Safier 2021년 3월 2일
@Edric Ellis : on a side note, I converted the double loop to a single and that seems to work, and obviously MUCH faster:
ntot = nr*nc;
parfor k = 1:ntot
% Determine i and j
i = fix(k/nr);
if k/nr - fix(k/nr) ~= 0
i = i + 1;
end
j = nr - (i*nr - k);
stuff...
end

댓글을 달려면 로그인하십시오.

추가 답변 (0개)

카테고리

Help CenterFile Exchange에서 Loops and Conditional Statements에 대해 자세히 알아보기

제품


릴리스

R2020a

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by