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From: Jeff Nowak on 9 Aug 2010 08:27
I'm new to parfor and was hoping someone might be able to tell me what
is wrong with this code. Matlab gives the following error.

??? Error: The variable lEmax in a parfor cannot be classified.
See Parallel for Loops in MATLAB, "Overview".

Error in ==> testparfor at 18
parfor iHit=1:nHits


I think lEmax fits the description of a sliced output variable.
Below is the test code. Any help is appreciated.

Jeff


nPool = 4;
if matlabpool('size') == 0
matlabpool('open','local',nPool)
end

nfft = 214;
freqBinsInd = 1:nfft;
fs=2500000;

nEmaxRows = 12;
nHits = 50;

timevals = 1:nHits;

lEmax = [];
lEmax(nEmaxRows,nHits) = sqrt(-1);

parfor iHit=1:nHits
ii = iHit;
freqval = (-nfft/2+freqBinsInd(ii)-1)*fs/nfft;
it = complex(randn(8,1),randn(8,1));

lEmax(1,iHit) = it(1);
lEmax(2,iHit) = it(2);
lEmax(3,iHit) = it(3);
lEmax(4,iHit) = it(4);
lEmax(5,iHit) = it(5);
lEmax(6,iHit) = it(6);
lEmax(7,iHit) = it(7);
lEmax(8,iHit) = it(8);

azel = [cos(rand(1));sin(rand(1))]*pi/180;

lEmax(9,iHit) = mod(90-azel(1),360);
lEmax(10,iHit) = azel(2);
lEmax(11,iHit) = timevals(iHit);
lEmax(12,iHit) = freqval;
end
From: Steven_Lord on 9 Aug 2010 09:15


"Jeff Nowak" <nowak(a)ll.mit.edu> wrote in message
news:RuS7o.1586$oF2.260(a)llnews...
> I'm new to parfor and was hoping someone might be able to tell me what is
> wrong with this code. Matlab gives the following error.
>
> ??? Error: The variable lEmax in a parfor cannot be classified.
> See Parallel for Loops in MATLAB, "Overview".
>
> Error in ==> testparfor at 18
> parfor iHit=1:nHits
>
>
> I think lEmax fits the description of a sliced output variable.
> Below is the test code. Any help is appreciated.
>
> Jeff
>
>
> nPool = 4;
> if matlabpool('size') == 0
> matlabpool('open','local',nPool)
> end
>
> nfft = 214;
> freqBinsInd = 1:nfft;
> fs=2500000;
>
> nEmaxRows = 12;
> nHits = 50;
>
> timevals = 1:nHits;
>
> lEmax = [];
> lEmax(nEmaxRows,nHits) = sqrt(-1);
>
> parfor iHit=1:nHits
> ii = iHit;
> freqval = (-nfft/2+freqBinsInd(ii)-1)*fs/nfft;
> it = complex(randn(8,1),randn(8,1));
>
> lEmax(1,iHit) = it(1);
> lEmax(2,iHit) = it(2);

*snip*

As written, IEmax is not a sliced variable. The criteria that a variable
must satisfy to be sliced are:

http://www.mathworks.com/access/helpdesk/help/toolbox/distcomp/brdqtjj-1.html#bq_tcng-1

IEmax satisfies the "Type of First-Level Indexing", "Form of Indexing", and
"Shape of Array" criteria. Where it fails is in the "Fixed Index Listing"
criterion, as you have one instance of IEmax being indexed with indices (1,
iHit) and another instance with (2, iHit).

Move the construction of azel up above the assignment to IEmax and construct
the vector of values that you want to assign into IEmax as a temporary
variable vectorRHS, then use:

IEmax(:, iHit) = vectorRHS;
% or
IEmax(1:nEmaxRows, iHit) = vectorRHS;

as the only assignment to IEmax in the loop and it should satisfy all four
of those criteria.

--
Steve Lord
slord(a)mathworks.com
comp.soft-sys.matlab (CSSM) FAQ: http://matlabwiki.mathworks.com/MATLAB_FAQ
To contact Technical Support use the Contact Us link on
http://www.mathworks.com

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