python - 在 Python 中查找函数的参数

Question

我希望能够询问类的 __init__ 方法它的参数是什么。直接的方法如下:

cls.__init__.__func__.__code__.co_varnames[:code.co_argcount]

但是，如果类有任何装饰器，那将不起作用。它将给出装饰器返回的函数的参数列表。我想深入了解原始的 __init__ 方法并获取那些原始参数。在装饰器的情况下，装饰器函数将在装饰器返回的函数的闭包中找到:

cls.__init__.__func__.__closure__[0]

但是，如果闭包中还有其他东西，装饰器可能会不时做一些事情，那就更复杂了:

def Something(test):
    def decorator(func):
        def newfunc(self):
            stuff = test
            return func(self)
        return newfunc
    return decorator

def test():
    class Test(object):
        @Something(4)
        def something(self):
            print Test
    return Test

test().something.__func__.__closure__
(<cell at 0xb7ce7584: int object at 0x81b208c>, <cell at 0xb7ce7614: function object at 0xb7ce6994>)

然后我必须决定是要装饰器的参数还是原始函数的参数。装饰器返回的函数可以有 *args 和 **kwargs 作为参数。如果有多个装饰器，我必须决定我关心哪一个怎么办？

那么即使函数可能被修饰，查找函数参数的最佳方法是什么？此外，沿着装饰器链返回装饰函数的最佳方法是什么？

更新:

以下是我现在的做法(姓名已更改以保护被告的身份):

import abc
import collections

IGNORED_PARAMS = ("self",)
DEFAULT_PARAM_MAPPING = {}
DEFAULT_DEFAULT_PARAMS = {}

class DICT_MAPPING_Placeholder(object):
    def __get__(self, obj, type):
        DICT_MAPPING = {}
        for key in type.PARAMS:
            DICT_MAPPING[key] = None
        for cls in type.mro():
            if "__init__" in cls.__dict__:
                cls.DICT_MAPPING = DICT_MAPPING
                break
        return DICT_MAPPING

class PARAM_MAPPING_Placeholder(object):
    def __get__(self, obj, type):
        for cls in type.mro():
            if "__init__" in cls.__dict__:
                cls.PARAM_MAPPING = DEFAULT_PARAM_MAPPING
                break
        return DEFAULT_PARAM_MAPPING

class DEFAULT_PARAMS_Placeholder(object):
    def __get__(self, obj, type):
        for cls in type.mro():
            if "__init__" in cls.__dict__:
                cls.DEFAULT_PARAMS = DEFAULT_DEFAULT_PARAMS
                break
        return DEFAULT_DEFAULT_PARAMS

class PARAMS_Placeholder(object):
    def __get__(self, obj, type):
        func = type.__init__.__func__
        # unwrap decorators here
        code = func.__code__
        keys = list(code.co_varnames[:code.co_argcount])
        for name in IGNORED_PARAMS:
            try: keys.remove(name)
            except ValueError: pass
        for cls in type.mro():
            if "__init__" in cls.__dict__:
                cls.PARAMS = tuple(keys)
                break
        return tuple(keys)

class BaseMeta(abc.ABCMeta):
    def __init__(self, name, bases, dict):
        super(BaseMeta, self).__init__(name, bases, dict)
        if "__init__" not in dict:
            return
        if "PARAMS" not in dict:
            self.PARAMS = PARAMS_Placeholder()
        if "DEFAULT_PARAMS" not in dict:
            self.DEFAULT_PARAMS = DEFAULT_PARAMS_Placeholder()
        if "PARAM_MAPPING" not in dict:
            self.PARAM_MAPPING = PARAM_MAPPING_Placeholder()
        if "DICT_MAPPING" not in dict:
            self.DICT_MAPPING = DICT_MAPPING_Placeholder()


class Base(collections.Mapping):
    __metaclass__ = BaseMeta
    """
    Dict-like class that uses its __init__ params for default keys.

    Override PARAMS, DEFAULT_PARAMS, PARAM_MAPPING, and DICT_MAPPING
    in the subclass definition to give non-default behavior.

    """
    def __init__(self):
        pass
    def __nonzero__(self):
        """Handle bool casting instead of __len__."""
        return True
    def __getitem__(self, key):
        action = self.DICT_MAPPING[key]
        if action is None:
            return getattr(self, key)
        try:
            return action(self)
        except AttributeError:
            return getattr(self, action)
    def __iter__(self):
        return iter(self.DICT_MAPPING)
    def __len__(self):
        return len(self.DICT_MAPPING)

print Base.PARAMS
# ()
print dict(Base())
# {}

此时 Base 报告了四个常量的无趣值，实例的字典版本为空。但是，如果您创建子类，您可以重写这四个中的任何一个，或者您可以将其他参数包含在 __init__ 中:

class Sub1(Base):
    def __init__(self, one, two):
        super(Sub1, self).__init__()
        self.one = one
        self.two = two

Sub1.PARAMS
# ("one", "two")
dict(Sub1(1,2))
# {"one": 1, "two": 2}

class Sub2(Base):
    PARAMS = ("first", "second")
    def __init__(self, one, two):
        super(Sub2, self).__init__()
        self.first = one
        self.second = two

Sub2.PARAMS
# ("first", "second")
dict(Sub2(1,2))
# {"first": 1, "second": 2}

Answer 1

考虑这个装饰器:

def rickroll(old_function):
    return lambda junk, junk1, junk2: "Never Going To Give You Up"

class Foo(object):
    @rickroll
    def bar(self, p1, p2):
        return p1 * p2

print Foo().bar(1, 2)

在其中，rickroll 装饰器采用了 bar 方法，将其丢弃，并用忽略其不同名称(可能还有编号!)参数的新函数替换它，而是返回经典歌曲中的一行。

没有对原始函数的进一步引用，垃圾收集器可以随时将其删除。

在这种情况下，我看不出您如何找到参数名称 p1 和 p2。据我了解，即使是 Python 解释器本身也不知道它们过去被称为什么。