Given a method, how do I return the class it belongs to in Python 3.3 onward?

问题

Given x = C.f after:

class C:
    def f(self):
        pass

What do I call on x that will return C?

The best I could do is execing a parsed portion of x.__qualname__, which is ugly:

exec('d = ' + ".".join(x.__qualname__.split('.')[:-1]))

For a use case, imagine that I want a decorator that adds a super call to any method it's applied to. How can that decorator, which is only given the function object, get the class to super (the ??? below)?

def ensure_finished(iterator):
    try:
        next(iterator)
    except StopIteration:
        return
    else:
        raise RuntimeError

def derived_generator(method):
    def new_method(self, *args, **kwargs):
        x = method(self, *args, **kwargs)
        y = getattr(super(???, self), method.__name__)\
            (*args, **kwargs)

        for a, b in zip(x, y):
            assert a is None and b is None
            yield

        ensure_finished(x)
        ensure_finished(y)

    return new_method

回答1:

If your aim is to get rid of the exec statement, but are willing to use the __qualname__ attribute, even though you are still required to manually parse it, then at least for simple cases the following seems to work:

x.__globals__[x.__qualname__.rsplit('.', 1)[0]]

or:

getattr(inspect.getmodule(x), x.__qualname__.rsplit('.', 1)[0])

I'm not a Python expert, but I think the second solution is better, considering the following documentation excerpts:

• from What's new in Python 3.3:

  Functions and class objects have a new __qualname__ attribute representing the “path” from the module top-level to their definition. For global functions and classes, this is the same as __name__. For other functions and classes, it provides better information about where they were actually defined, and how they might be accessible from the global scope.

• from __qualname__'s description in PEP 3155:

  For nested classed, methods, and nested functions, the __qualname__ attribute contains a dotted path leading to the object from the module top-level.

EDIT:

1. As noted in the comments by @eryksun, parsing __qualname__ like this goes beyond its intended usage and is extremely fragile considering how __qualname__ reflects closures. A more robust approach needs to exclude closure namespaces of the form name.<locals>. For example:

   >>> class C:
   ...     f = (lambda x: lambda s: x)(1)
   ... 
   >>> x = C.f
   >>> x
   <function C.<lambda>.<locals>.<lambda> at 0x7f13b58df730>
   >>> x.__qualname__
   'C.<lambda>.<locals>.<lambda>'
   >>> getattr(inspect.getmodule(x), x.__qualname__.rsplit('.', 1)[0])
   Traceback (most recent call last):
     File "<stdin>", line 1, in <module>
   AttributeError: 'module' object has no attribute 'C.<lambda>.<locals>'
   

   This specific case can be handled in the following manner:

   >>> getattr(inspect.getmodule(x),
   ...         x.__qualname__.split('.<locals>', 1)[0].rsplit('.', 1)[0])
   <class '__main__.C'>
   

   Nonetheless, it's unclear what other corner cases exist now or may come up in future releases.

2. As noted in the comment by @MichaelPetch, this answer is relevant only for Python 3.3 onward, as only then the __qualname__ attribute was introduced into the language.

   • However, according to @WouterBolsterlee, github.com/wbolster/qualname provides an equivalent for older Python versions.

3. For a complete solution that handles bound methods as well, please refer to this answer.

来源：https://stackoverflow.com/questions/25921450/given-a-method-how-do-i-return-the-class-it-belongs-to-in-python-3-3-onward