Given a class C with a function or method f, I use inspect.ismethod(obj.f) (where obj is an instance of C) to find out if f is bound method or not. Is there a way to do the same directly at the class level (without creating an object)?
inspect.ismethod does not work as this:
class C(object):
@staticmethod
def st(x):
pass
def me(self):
pass
obj = C()
results in this (in Python 3):
>>> inspect.ismethod(C.st)
False
>>> inspect.ismethod(C.me)
False
>>> inspect.ismethod(obj.st)
False
>>> inspect.ismethod(obj.me)
True
I guess I need to check if the function/method is member of a class and not static but I was not able to do it easily. I guess it could be done using classify_class_attrs as shown here
How would you determine where each property and method of a Python class is defined?
but I was hoping there was another more direct way.
There are no unbound methods in Python 3, so you cannot detect them either. All you have is regular functions. At most you can see if they have a qualified name with a dot, indicating that they are nested, and their first argument name is self:
if '.' in method.__qualname__ and inspect.getargspec(method).args[0] == 'self':
# regular method. *Probably*
This of course fails entirely for static methods and nested functions that happen to have self as a first argument, as well as regular methods that do not use self as a first argument (flying in the face of convention).
For static methods and class methods, you'd have to look at the class dictionary instead:
>>> isinstance(vars(C)['st'], staticmethod)
True
That's because C.__dict__['st'] is the actual staticmethod instance, before binding to the class.