Re: [sympy] node.func.__name__ on UndefinedFunctions?

[Aaron Meurer]

It works in that case too

In [62]: u = Function('u')

In [63]: u.__name__
Out[63]: 'u'

The important thing to note is that Function('u') creates *class*, not
an object. Roughly speaking, Function('u') is syntatic sugar for

class u(Function):
pass

(except it also sets the metaclass as UndefinedFunction)

In other words, if you have a = u(x), then a.func is u.

Aaron Meurer


On Tue, Apr 5, 2016 at 2:04 PM, Nico <nico.sc...@gmail.com> wrote:
> I need to get the name (as a string) of certain given functions, and I've
> typically done it like
> ```
> node.func.__name__
> ```
> That works, except for UndefinedFunctions. Is that an oversight? How to get
> the 'hello' from u in
> ```
> u = sympy.Function('hello')
> ```
> ?
>
> Cheers,
> Nico
>
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[Nico]

Great, thanks!