Why this gives error?
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I downloaded the attached codes " hPSO1.m", and "hPSOoptions.m" and tried to run it with my fitness function"myfitness.m" (also in the attachment) but it gives the following error:
Unrecognized function or variable 'options'.
Error in main (line 3)
[x,fval,gfx,output] = hPSO1(@main3, 4, options, varargin{:})
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Answers (1)
Jan
on 22 Nov 2022
Edited: Jan
on 22 Nov 2022
This is the failing function:
function [x,fval,gfx,output] = main(varargin)
[x,fval,gfx,output] = hPSO1(@main3, 4, options, varargin{:})
end
As the error message tells clearly, the variable options is not defined. What do you expect it to be?
Of course you can use defined variables only. Because this is such fundamental for programming, I recommend to read the tutorial again: https://www.mathworks.com/learn/tutorials/matlab-onramp.html
In the next step the function hPSO1 will fail:
function [x,fval,gfx,output]=hPSO1(main3,4,hPSOoptions,varargin)
% ^ ???
Providing a constant in the list of inputs is not meaningful. How should Matlab handle this? It could consider the value 4 as defined?
In the function myfitness a comment character is missing in line 6:
K=3; 3rd constant
You see a corresponding warning in the editor. Consider and fix these warnings.
Although you got instructions how to improve this in two other of your questions already, you still use the slow and complicated form:
abc=0.0;
for m1=1:M*N
abc=abc+(abs(yo(m1,1)-ye(m1,1))).^2;
end
abc=abc/(M*N);
e=abc;
% Faster:
e = mean((yo - ye).^2);
3 Comments
Jan
on 22 Nov 2022
Edited: Jan
on 22 Nov 2022
I told you already in my answer, that it is not meaningful to use a constant in the definition of a function. Now you do this again:
function [x,fval,gfx,output] = main(4)
% ^ ???
What should be set to the value 4 here?
Which Matlab version are you using? In older versions you cannot insert the definition of a function inside a script.
Did you solve the Onramp tutorial already? It is explained there exhaustively, how inputs arguments are defined for functions.
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