Programmatically testing whether a variable is an optimoptions object

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I was wondering if there was something like an isoptimoptions function that could be used to test in general whether a variable was generated by optimoptions(). The isa command alone will not seem to do it, because optimoptions objects occupy separate classes, depending on which solver the object was generated for.
>> opts1=optimoptions('lsqnonlin');
>> opts2=optimoptions('linprog');
>> whos opts1 opts2
Name Size Bytes Class Attributes
opts1 1x1 29913 optim.options.Lsqnonlin
opts2 1x1 14775 optim.options.Linprog
The following user-defined command seems to work, but I'm wondering if there are more direct alternatives,
function bool=isaoptimoptions(opts)
bool = startsWith(class(opts),'optim.options')
end
  4 Comments
Guillaume
Guillaume on 28 Feb 2020
superclasses(obj)
will tell you the base classes (if any). isa on the base class would work.
Matt J
Matt J on 28 Feb 2020
Theyhave several parent classes, but those I've checked all seem to have 'optim.options.SolverOptions' as a parent in common.
>> superclasses(optimoptions('linprog'))
Superclasses for class optim.options.Linprog:
optim.options.MultiAlgorithm
optim.options.SolverOptions
matlab.mixin.CustomDisplay
>> superclasses(optimoptions('intlinprog'))
Superclasses for class optim.options.Intlinprog:
optim.options.SingleAlgorithm
optim.options.SolverOptions
matlab.mixin.CustomDisplay
>> superclasses(optimoptions('lsqlin'))
Superclasses for class optim.options.Lsqlin:
optim.options.MultiAlgorithm
optim.options.SolverOptions
matlab.mixin.CustomDisplay

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Accepted Answer

Guillaume
Guillaume on 28 Feb 2020
Edited: Guillaume on 28 Feb 2020
"those I've checked all seem to have 'optim.options.SolverOptions' as a parent in common."
In that case:
isa(opts, 'optim.options.SolverOptions')
edit:
Strangely enough the base class optim.options.SolverOptions is part of base matlab (or a toolbox I've got even though I've haven't got the optimisation toolbox) and available as an m file.
The comment in the m file does say that this is the "base class for Optimization Toolbox options", so yes, the above is what you want.
  1 Comment
Matt J
Matt J on 28 Feb 2020
Edited: Matt J on 28 Feb 2020
Yeah, I guess that makes sense. It would be nice if it were documented in the toolbox that this will always be the root class ...

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More Answers (1)

Luna
Luna on 28 Feb 2020
Hi,
Maybe something like this helps?
metaClass = metaclass(opts1);
if contains(metaClass.Name,'optim.options')
disp('This is a optim.options object.')
else
end
  1 Comment
Matt J
Matt J on 28 Feb 2020
Hi Luna,
That would work, but is pretty much the same as the provisional solution mentioned in my original post.

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