On Thu, Dec 06, 2012 at 01:13:09AM +0000, Dima Pasechnik wrote:
> > Here's a related question: suppose I have an object G in sage. Is there a
> > "correct" way to ask G if is it a CombinatorialFreeModule? I can check for
> >
> > if hasattr(G,'_basis_keys'): ...
> >
> > but I would have thought that there was a better way to do this...
> isinstance?
>
>
> sage: F = CombinatorialFreeModule(QQ, ['a','b','c'])
> sage: isinstance(F,CombinatorialFreeModule)
> True
> sage: isinstance([],CombinatorialFreeModule)
> False
Depending on what you want to test exactly, you might or not want to do
sage: F in ModulesWithBasis
which will eventually include QQ^3, QQ['x'], ...
Cheers,
Nicolas
--
Nicolas M. Thiéry "Isil" <nthi...@users.sf.net>
http://Nicolas.Thiery.name/
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