Programmatically testing whether a variable is an optimoptions object

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

I don't have the toolbox so can't test, but don't they all derive from the same (possibly abstract) base class?
superclasses(obj)
will tell you the base classes (if any). isa on the base class would work.
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

"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

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)

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

Hi Luna,
That would work, but is pretty much the same as the provisional solution mentioned in my original post.

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Asked:

on 27 Feb 2020

Edited:

on 28 Feb 2020

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