在Python中寻找函数参数

5 投票
1 回答
1640 浏览
提问于 2025-04-16 02:06

我想知道一个类的 __init__ 方法有哪些参数。最简单的方法是这样做:

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

但是,如果这个类有装饰器的话,这个方法就不管用了。因为它会给出装饰器返回的函数的参数列表。我想要的是原始的 __init__ 方法和它的参数。如果有装饰器的话,装饰器函数会在装饰器返回的函数的闭包中找到:

cls.__init__.__func__.__closure__[0]

不过,如果闭包里还有其他东西,那就更复杂了,装饰器有时会这样做:

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>)

这时我就得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能有 *args**kwargs 作为参数。如果有多个装饰器,我又该如何选择我关心的那个呢?

那么,如何才能找到一个函数的参数,即使这个函数可能被装饰过?还有,怎样才能沿着装饰器的链条回到被装饰的函数呢?

更新:

这是我现在实际操作的方式(为了保护被指控者的身份,名字已经更改):

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 报告四个常量的值不太有趣,而实例的字典版本是空的。不过,如果你创建一个子类,你可以重写这四个常量中的任何一个,或者在 __init__ 中添加其他参数:

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}

1 个回答

3

考虑一下这个装饰器:

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)

在这里,rickroll 装饰器接收了 bar 方法,然后把它丢掉,换成一个新的函数。这个新函数不管原来参数的名字(甚至可能是数字!),而是直接返回一段经典歌曲的歌词。

之后再也没有提到过原来的函数,所以垃圾回收器随时可以把它清理掉。

在这种情况下,我看不出你怎么能找到参数名 p1 和 p2。根据我的理解,连 Python 的解释器自己都不知道它们原来叫什么。

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