How can I programmatically change the argspec of a function in a python decorator?

Question

Given a function:

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

@decorator
def func2(f1, kw=\'default\'):
    pass
decorate         


        

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

Are functools.update_wrapper() and/or functools.wraps() good enough?

Answer 3

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))