- Required (static): If you use a nested for-loop to index into a sliced array, you cannot use that array elsewhere in the parfor-loop.

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Hey everyone I've been looking at parfor and going through some previously given answers and documentation but I'm having a hard time to figure out what my mistake is.

Error: The variable CFxL_Norm in a parfor cannot be classified.

I do assume that CFxLconst_Norm will give the same error. In my script I first assign 2 empty cell-matrix of size (XX,10) and each cell will contain a matrix (501x401x401), except in the last column the cells will contain 1 simple number.

normalisePerc=[0.2, 0.5, 0.8, 1, 0.01, 0.05, 0.1, 0.3, 0.4, 0.6, 0.7, 0.9]; %values can be changed later on

Lnlp=length(normalisePerc); %Length of the normalise percentage array

CFxL_Norm=cell(Lnlp,8+2); % CFxLight400nm + normaliseperc

CFxLconst_Norm=cell(Lnlp,8+2); % CFxLight400nm + normaliseperc

some other code and then this

A=Ef_Norm{9}; %%% assign variable as temporary for parfor, throwaway afterwards

%Lnlp=length(normalisePerc); %Same, to not use length but a fixed constant

parfor kk=1:Lnlp

fprintf('number %i of %i cycles for normalising \n', kk, Lnlp);

NLP=normalisePerc(kk); %get value from array

NLPinv=1/NLP; %allows to use multiplication in a few places for an increase in speed

D=A.*NLPinv;

D(D > 1) = 2./(1+1./D(D > 1));

for tt=1:8

fprintf('number %i of %i steps for normalising \n', tt+(kk-1)*8, Lnlp*8);

tic

B=Ef_Norm{tt}.*NLPinv; %Gives a warning about overhead but not causing problems

B(B > 1) = 2./(1+1./B(B > 1));

for qq=1:lyq2 %Would have prefered to use parfor for this loop, lyq2=401. Matlab recommends outer loops only

CFs=Cafluo_Norm{tt}(:,:,qq)'; %Gives a warning about overhead but not causing problems

CFxL_Norm{kk,tt}(:,:,qq)=CFs.*B(:,:,qq);

CFxLconst_Norm{kk,tt}(:,:,qq)=CFs.*D(:,:,qq);

end

fprintf('Time needed for this step was: \n');

toc

end

tt=9;

for qq=1:lyq2 %Also would have prefered to use parfor for this loop but Matlab recommends outer loops only.

tmp1=Cafluo_Norm{tt}(:,:,qq)'.*D(:,:,qq);

CFxL_Norm{kk,tt}(:,:,qq)=tmp1;

CFxLconst_Norm{kk,tt}(:,:,qq)=tmp1;

end

end

tt=10;

tmpnlp=num2cell(normalisePerc');

CFxL_Norm(:,10)=tmpnlp;

CFxLconst_Norm(:,tt)=tmpnlp;

clear tt kk NLP A B D NLPinv qq CFs tmp1 tmpnlp

fprintf('\n \n \n');

I don't see any of the problems regarding things, like addressing the same cell at the same time or even multiple times. I am also pretty sure the variables are independent of each other. Thus I don't really understand why I'm still getting this error and what I'm doing wrong.

Edric Ellis
on 29 Jul 2020

The code you've shown there isn't complete enough for us to attempt to run and see the error you're encountering. It would be helpful if you could simplify the code to the point where we can try and run it as a for loop before trying a parfor loop.

That said, I think you're hitting one of the restrictions of assigning into sliced variables with nested for loops inside parfor. This doc page has the details. I think the relevant restriction is:

- Required (static): If you use a nested for-loop to index into a sliced array, you cannot use that array elsewhere in the parfor-loop.

In your code, you've got multiple cases where you're assigning into the sliced outputs. I would recommend trying to change things so that you have only a single assignment statement into your sliced outputs.

Walter Roberson
on 29 Jul 2020

In the below, I have made the relevant changes only for CFxL_Norm, without looking at the other variables. You probably need to make similar changes for CFxLconst_Norm .

The key here is to use temporary variables to accumulate all partial results, and then have a single assignment statement at the end that changes everything having to do with the variable indexed by the parfor index.

A=Ef_Norm{9}; %%% assign variable as temporary for parfor, throwaway afterwards

parfor kk=1:Lnlp

fprintf('number %i of %i cycles for normalising \n', kk, Lnlp);

NLP=normalisePerc(kk); %get value from array

NLPinv=1/NLP; %allows to use multiplication in a few places for an increase in speed

D=A.*NLPinv;

D(D > 1) = 2./(1+1./D(D > 1));

AAAt = cell(1,9);

for tt=1:8

fprintf('number %i of %i steps for normalising \n', tt+(kk-1)*8, Lnlp*8);

tic

B=Ef_Norm{tt}.*NLPinv; %Gives a warning about overhead

B(B > 1) = 2./(1+1./B(B > 1));

AAA=zeros(lzq2,lxq2,lyq2); %CHANGE TO ADD THIS, temporary variable

AAA2=zeros(lzq2,lxq2,lyq2); %CHANGE TO ADD THIS, temporary variable

for qq=1:lyq2

CFs=Cafluo_Norm{tt}(:,:,qq)'; %Gives a warning about overhead

AAA(:,:,qq)=CFs.*B(:,:,qq); %Introduced to remove slicing and indexing issues

AAA2(:,:,qq)=CFs.*D(:,:,qq); %Introduced to remove slicing and indexing issues

end

AAAt{tt}=AAA; %Assign here to actual variable

CFxLconst_Norm{kk,tt}=AAA2; %Assign here to actual variable

fprintf('Time needed for this step was: \n');

toc

end

tt=9;

AAA3=zeros(lzq2,lxq2,lyq2); %Same changes repeated as before

AAA4=zeros(lzq2,lxq2,lyq2); %same change

for qq=1:lyq2

tmp1=Cafluo_Norm{tt}(:,:,qq)'.*D(:,:,qq);

AAA3(:,:,qq)=tmp1; %same change as before

AAA4(:,:,qq)=tmp1; %same change

end

AAAt{tt}=AAA3; %same change

CFxLconst_Norm{kk,tt}=AA4; %Same change

CFxL_Norm(kk,:) = AAAt;

end

tt=10;

tmpnlp=num2cell(normalisePerc');

CFxl_Norm(:,tt)=tmpnlp;

CFxLconst_Norm(:,tt)=tmpnlp;

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