error using parfor loop - Dimensions of arrays being concatenated are not consistent

1 vue (au cours des 30 derniers jours)
red salud
red salud le 20 Juin 2019
I got the parallel toolbox to help me with this part of my code (takes weeks to compute). I am not sure where and how best to use the parfor..
I get errors like "Dimensions of arrays being concatenated are not consistent" when i implement it as below
KD=WQ';
hh=[];
HH=[];
gf=size(KD,2);
FF=length(Ped);toc;
%%
for l=1:FF
for i =1:gf
YY{i}=KD{l,i};
r=size(YY{i},1);
% U=zeros(1,length(Arc));
parfor kk=1:r
u=length(YY{i}(kk,:));
uu=YY{i}(kk,:);
Find1=[uu(1:u-1);uu(2:u)];Find1=Find1';
[C,ia,ib] = intersect(Arc,Find1,'rows');
U(ia)=1;
HH=[HH;U];
hh=[hh;length(Find1)-1];
U=zeros(1,length(Arc));
end
end
end

Réponses (1)

Lokesh
Lokesh le 4 Nov 2024
Hello red,
In a 'parfor' loop, temporary variables are cleared at the beginning of each iteration.
In your code, the variable 'U' is cleared at the start of each iteration, leading to issues. To resolve this, ensure that 'U' is initialized before it is used.
Refer to the following code for guidance:
parfor kk = 1:r
u = length(YY{i}(kk, :));
uu = YY{i}(kk, :);
Find1 = [uu(1:u-1); uu(2:u)];
Find1 = Find1';
[C, ia, ib] = intersect(Arc, Find1, 'rows');
U = zeros(1, length(Arc)); % Initialize U here
U(ia) = 1;
HH = [HH; U];
hh = [hh; length(Find1) - 1];
end
For more information, please refer to the documentation on Temporary variables in 'parfor':
  1 commentaire
Walter Roberson
Walter Roberson le 4 Nov 2024
Also, I would recommend
uu = YY{i}(kk, :);
Find1 = [uu(1:end-1); uu(2:end)].';
[C, ia, ib] = intersect(Arc, Find1, 'rows');
U = zeros(1, length(Arc)); % Initialize U here
U(ia) = 1;
HH = [HH; U];
hh = [hh; size(Find1,2)];
I have to wonder whether Find1 is being constructed properly; it is currently being constructed as a single row.

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