If I run the script:
z=1:1000;
parfor i=1:numel(z)
zfun(z(i))
end
which calls the function:
function zfun(z)
for j=1:50
for k=1:100
myz=z*10+myz*k;
end
end
save(fullfile('filepath', num2str(myz)), 'myz');
end
I get an error of Undefined function or variable 'myz' for the save command.
Any ideas how to fix this?

2 commentaires

Walter Roberson
Walter Roberson le 20 Jan 2017
Which MATLAB version?
My spidy-senses are suggesting you might be using R2015b or R2016a but not R2016b ?
JohnDylon
JohnDylon le 20 Jan 2017
R2016b trial for Linux. (No error for 2014a on PC though!)

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 Réponse acceptée

Walter Roberson
Walter Roberson le 20 Jan 2017

0 votes

You do not initialize the variable in the zfun loop.

5 commentaires

JohnDylon
JohnDylon le 21 Jan 2017
Initializating "myz=[]" outside of the zfun loop and passing it through the function as zfun(z(i), myz) done the job. Thanks again!
Walter Roberson
Walter Roberson le 21 Jan 2017
There should not be a need for that; you should be able to just initialize it to 1 inside zfun.
Initializing to [] is going to result in a lot of multiplying with the empty matrix, which will not be productive.
Then you say
function zfun(z)
myz=1;
for j=1:50
for k=1:100
myz=z*10+myz*k;
end
end
save(fullfile('filepath', num2str(myz)), 'myz');
end
is better to do?
Walter Roberson
Walter Roberson le 21 Jan 2017
Yes, that would have less overhead. However, due to finite precision of floating point numbers, the result of the above is going to be -inf below about -1/17 and +inf above it. The zero is at about -0.058197670686932642438500200510901 . If you were trying to find the zero, it would be much more effective to use the symbolic toolbox.
JohnDylon
JohnDylon le 22 Jan 2017
Thank you for all valuable guidance.

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