Hello, all.
I got a question on using parfor in matlab. When I tried to run the following code, I got an error showing:"Error: The variable Resp in a parfor cannot be classified.". Is this because Resp will be generated simultaneously when jjjj will be executed parallelly? If so, does this mean I can not perform this since different Phi use the same Resp as the first structure name? Will there be any idea on solving this problem? Thank you very much.
The following are the codes:
sizePhiValuep = 7;
sizeMassp = 20;
parfor (jjjj=1:sizePhiValuep,7)
Resp.(sprintf('Phi%d',jjjj)).MaxX2RespBegin_beta= zeros(sizeMassp,1);
Resp.(sprintf('Phi%d',jjjj)).MaxX2RespEnd_beta= zeros(sizeMassp,1);
end

 Akzeptierte Antwort

Alexandra Harkai
Alexandra Harkai am 22 Nov. 2016

1 Stimme

The error message refers to the variable classification when using parfor: classification-of-variables-in-parfor-loops. It doesn't know if you'll operate on segments in each loop or accumulate values regardless of the iteration order.
You could make Resp-something an array (or two arrays, if you need Begin and End) and index into that from the corresponding loop using jjjj.

Weitere Antworten (1)

Baozai
Baozai am 23 Nov. 2016

0 Stimmen

Thank you, Alexandra, When I change the code to the following, it works.
sizePhiValuep = 7;
sizeMassp = 20;
parfor (jjjj=1:sizePhiValuep,7)
Resp(jjjj).Phi.MaxX2RespBegin_beta= zeros(sizeMassp,1);
Resp(jjjj).Phi.MaxX2RespEnd_beta= zeros(sizeMassp,1);
end

1 Kommentar

Alexandra Harkai
Alexandra Harkai am 24 Nov. 2016
Cool.
Dynamic variable name generation for indexing purposes ( sprintf('Phi%d',jjjj) ) should be avoided anyway, so it's always good practice to use 'actual' indexing instead.

Melden Sie sich an, um zu kommentieren.

Kategorien

Mehr zu Loops and Conditional Statements finden Sie in Hilfe-Center und File Exchange

Tags

Community Treasure Hunt

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

Start Hunting!

Translated by