Python: can a decorator determine if a function is

问题:

I'm writing a decorator, and for various annoying reasons[0] it would be expedient to check if the function it is wrapping is being defined stand-alone or as part of a class (and further which classes that new class is subclassing).

For example:

def my_decorator(f):
    defined_in_class = ??
    print "%r: %s" %(f, defined_in_class)

@my_decorator
def foo(): pass

class Bar(object):
    @my_decorator
    def bar(self): pass

Should print:

<function foo …>: False
<function bar …>: True

Also, please note:

• At the point decorators are applied the function will still be a function, not an unbound method, so testing for instance/unbound method (using typeof or inspect) will not work.
• Please only offer suggestions that solve this problem — I'm aware that there are many similar ways to accomplish this end (ex, using a class decorator), but I would like them to happen at decoration time, not later.

[0]: specifically, I'm writing a decorator that will make it easy to do parameterized testing with nose. However, nose will not run test generators on subclasses of unittest.TestCase, so I would like my decorator to be able to determine if it's being used inside a subclass of TestCase and fail with an appropriate error. The obvious solution - using isinstance(self, TestCase) before calling the wrapped function doesn't work, because the wrapped function needs to be a generator, which doesn't get executed at all.

回答1:

Take a look at the output of inspect.stack() when you wrap a method. When your decorator's execution is underway, the current stack frame is the function call to your decorator; the next stack frame down is the @ wrapping action that is being applied to the new method; and the third frame will be the class definition itself, which merits a separate stack frame because the class definition is its own namespace (that is wrapped up to create a class when it is done executing).

I suggest, therefore:

defined_in_class = (len(frames) > 2 and
                    frames[2][4][0].strip().startswith('class '))

If all of those crazy indexes look unmaintainable, then you can be more explicit by taking the frame apart piece by piece, like this:

import inspect
frames = inspect.stack()
defined_in_class = False
if len(frames) > 2:
    maybe_class_frame = frames[2]
    statement_list = maybe_class_frame[4]
    first_statment = statement_list[0]
    if first_statment.strip().startswith('class '):
        defined_in_class = True

Note that I do not see any way to ask Python about the class name or inheritance hierarchy at the moment your wrapper runs; that point is "too early" in the processing steps, since the class creation is not yet finished. Either parse the line that begins with class yourself and then look in that frame's globals to find the superclass, or else poke around the frames[1] code object to see what you can learn — it appears that the class name winds up being frames[1][0].f_code.co_name in the above code, but I cannot find any way to learn what superclasses will be attached when the class creation finishes up.

回答2:

Some hacky solution that I've got:

import inspect

def my_decorator(f):
    args = inspect.getargspec(f).args
    defined_in_class = bool(args and args[0] == 'self')
    print "%r: %s" %(f, defined_in_class)

But it relays on the presence of self argument in function.

回答3:

A little late to the party here, but this has proven to be a reliable means of determining if a decorator is being used on a function defined in a class:

frames = inspect.stack()

className = None
for frame in frames[1:]:
    if frame[3] == "<module>":
        # At module level, go no further
        break
    elif '__module__' in frame[0].f_code.co_names:
        className = frame[0].f_code.co_name
        break

The advantage of this method over the accepted answer is that it works with e.g. py2exe.

回答4:

You could check if the decorator itself is being called at the module level or nested within something else.

defined_in_class = inspect.currentframe().f_back.f_code.co_name != "<module>"

回答5:

I think the functions in the inspect module will do what you want, particularly isfunction and ismethod:

>>> import inspect
>>> def foo(): pass
... 
>>> inspect.isfunction(foo)
True
>>> inspect.ismethod(foo)
False
>>> class C(object):
...     def foo(self):
...             pass
... 
>>> inspect.isfunction(C.foo)
False
>>> inspect.ismethod(C.foo)
True
>>> inspect.isfunction(C().foo)
False
>>> inspect.ismethod(C().foo)
True

You can then follow the Types and Members table to access the function inside the bound or unbound method:

>>> C.foo.im_func
<function foo at 0x1062dfaa0>
>>> inspect.isfunction(C.foo.im_func)
True
>>> inspect.ismethod(C.foo.im_func)
False