Yeah, I was stuck in a rut so that it took me until I stepped away for a
bit to realize the same conclusion!
I didn't realize I could just add the is_commutative flag to
UndefinedFunction, so I like your solution better than mine, where I had
ended up subclassing AppliedUndef (as that is the subclass of Function that
has the is_commutative property). Since I'm unlikely to deal with
non-commutative symbols, your solution should serve me fine.
Thanks!
Duane
On Monday, November 23, 2015 at 3:29:41 PM UTC-6, Aaron Meurer wrote:
>
> Function("P") is just syntactic sugar. As you can see from the
> constructor, it's the same as UndefinedFunction("P"). If you want to
> customize the behavior, you should subclass UndefinedFunction.
>
> However, in this case, it looks like UndefinedFunction("P",
> is_commutative=True) works.
>
> Aaron Meurer
>
>
> On Mon, Nov 23, 2015 at 2:32 PM, Duane Nykamp <dqny...@comcast.net
> <javascript:>> wrote:
> > I'm trying to create a subclass of Symbol that returns a subclass of
> > Function when called. The only reason is to customize the
> is_commutative
> > property so that an expression like P(A)/P(Eq(x,1)) will output with the
> > P(x=1) in the denominator rather than as P^-1(x=1), which might confuse
> my
> > students.
> >
> > However, if I create a unmodified Function subclass, the resulting
> function
> > is not callable.
> >
> > In [7]: class Function2(Function):
> > ...: pass
> > ...:
> >
> > In [8]: P=Function2('P')
> >
> > In [9]: P(x)
> >
> ---------------------------------------------------------------------------
> > TypeError Traceback (most recent call
> last)
> > <ipython-input-9-49600f79dd99> in <module>()
> > ----> 1 P(x)
> >
> > TypeError: 'Function2' object is not callable
> >
> >
> > The problem is due to __new__ of Function
> >
> > def __new__(cls, *args, **options):
> > # Handle calls like Function('f')
> > if cls is Function:
> > return UndefinedFunction(*args, **options)
> >
> > but, I can't change the condition to issubclass(cls,Function), as that
> makes
> > it always true and messes up the function of Function.
> >
> > Or, the specific question I'm actually struggling with is:
> > Can I create a
> > P=Function('P')
> > so that
> >
> > In [4]: P(A)/P(Eq(x,1))
> > Out[4]:
> > -1
> > P(A)⋅P (x = 1)
> >
> > outputs with latex with the P(x=1) in the denomiator.
> >
> > Thanks,
> > Duane
> >
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