如何以编程方式更改python装饰器中函数的argspec？

Question

Given a function:

def func(f1, kw='default'):
    pass
bare_argspec = inspect.getargspec(func)

@decorator
def func2(f1, kw='default'):
    pass
decorated_argspec = inspect.getargspec(func2)

How can I create a decorator such that bare_argspec == decorated_argspec ?

(As to why, the framework that calls the decorated function does argspec inspection to choose what to pass in, so the decorator has to retain the same argspec in order to play nice. When I posed this question on #python, I got a long speech about why the framework sucks, which is not what I'm looking for; I have to solve the problem here. Also, I'm just interested in the answer, too)

Answer 1

Michele Simionato's decorator module has a decorator called decorator which preserves function argspecs.

import inspect
import decorator

def func(f1, kw='default'):
    pass
bare_argspec = inspect.getargspec(func)
print(bare_argspec)
# ArgSpec(args=['f1', 'kw'], varargs=None, keywords=None, defaults=('default',))

@decorator.decorator
def mydecorator(func,*args,**kw):
    result=func(*args,**kw)
    return result

@mydecorator
def func2(f1, kw='default'):
    pass
decorated_argspec = inspect.getargspec(func2)
print(decorated_argspec)
# ArgSpec(args=['f1', 'kw'], varargs=None, keywords=None, defaults=('default',))

assert(bare_argspec==decorated_argspec)

Answer 2

There's the decorator module:

from decorator import decorator
@decorator
def trace(func, *args, **kw):
    print 'calling', func, 'with', args, kw
    return func(*args, **kw)

That makes trace a decorator with the same argspecs as the decorated function. Example:

>>> @trace
... def f(x, y=1, z=2, *args, **kw):
...     pass

>>> f(0, 3)
calling f with (0, 3, 2), {}

>>> from inspect import getargspec
>>> print getargspec(f)
ArgSpec(args=['x', 'y', 'z'], varargs='args', keywords='kw', defaults=(1, 2))

Answer 3

functools.update_wrapper()和/或functools.wraps()足够好？