Is z an object here? How to interpret the code?
Mostrar comentarios más antiguos
data = rand(4,4);
means = zeros(1,4);
parfor I = 1:4
% The below usage of structures
% is flagged by Code Analyzer
z.mean = mean(data(:,I));
means(I) = z.mean;
end
disp(means)
Hi all, I found the above code was used to demonstrate incorrect use of parfor construct with MATLAB which can flag compiler errors.
On executing as expected with the code I encountered the an error stating The variable z in a parfor cannot be classified. See Parallel for Loops in MATLAB, "Overview".
what I fail to understand is 1) what is the datatype of z? is it a class an object ??? 2) but the code can be easily modified to handle the problem by replacing z.mean by z and the program runs perfectly. That makes me wonder why using z.mean is required??
Respuestas (1)
James Tursa
el 13 de Sept. de 2017
Editada: James Tursa
el 13 de Sept. de 2017
0 votos
z is a struct. It has a field named "mean".
2 comentarios
Aketh Thimmasandra-Maregowda
el 13 de Sept. de 2017
Editada: Aketh Thimmasandra-Maregowda
el 13 de Sept. de 2017
James Tursa
el 13 de Sept. de 2017
Editada: James Tursa
el 13 de Sept. de 2017
I don't know why a struct would be needed for that either. Just use z.
Where did the z.mean code come from in the first place? Did you inherit this code from someone or download it from somewhere? Does it complain if you do a direct assignment:
means(I) = mean(data(:,I));
Is this just an exercise in finding out why parfor doesn't like the struct? (Seems counterproductive to use it for mean calculations)
Categorías
Más información sobre Loops and Conditional Statements en Centro de ayuda y File Exchange.
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!