Just a Beginner's 'parfor' confusion
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Amnah
am 8 Feb. 2014
Beantwortet: Arslan Ahmad
am 9 Feb. 2014
I'm trying to execute this code and constantly getting the message that 'variable jVals is indexed but not sliced'. Can anyone kindly help
iVal=5:0.2:5.4;
jVal=2:0.5:2.5;
iLen=length(iVal)
jLen=length(jVal);
matrix=zeros(iLen,jLen);
parfor i=1:iLen
dummy=zeros(1,jLen);
for j=1:jLen
dummy(j)=jVal(j); %/// This line is in ERROR
end
matrix(i,:)=dummy;
end
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Matt J
am 8 Feb. 2014
Bearbeitet: Matt J
am 8 Feb. 2014
In your case, it will go away if you modify the loop as
parfor i=1:iLen
matrix(i,:)=jVal;
end
It's not an error, but rather a warning that you might not have efficient code. MATLAB sees that you are indexing a matrix variable inside the loop using another variable j. This usually means the matrix variable is a large array. If it were a small array, you would usually just index it with known, fixed constants, like jVal(1). If it is a large array, you usually want to find a way to have parfor send to the worker only a slice of the array that the worker will use, instead of broadcasting the whole array to every worker.
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Arslan Ahmad
am 9 Feb. 2014
It is better not to use another index inside the "parfor" use only the index of parfor in your case it's better to use i instead of j for more information I would like to refer you to MATLAB parfor documentation for more information and so for your can try this code for solving your problem. I modified it for you, it tooks only 0.194221 seconds on two matlab workers. If you find it useful than please accept the answer.
tic
iVal=5:0.2:5.4;
jVal=2:0.5:2.5;
iLen=length(iVal)
jLen=length(jVal);
matrix=zeros(iLen,jLen);
parfor i=1:iLen
jVal=2:0.5:2.5;
jLen=length(jVal);
dummy=zeros(1,jLen);
j=1:jLen
dummy(j)=jVal(j); %/// This line is in ERROR
matrix(i,:)=dummy;
end
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