Question

我希望能够向班级的__init__方法询问它的参数是什么。直截了当的方法如下：

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

但是，如果类有任何装饰器，那将无效。它将给出装饰器返回的函数的参数列表。我想了解原始__init__方法并获取原始参数。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]

然而，如果闭包中有其他东西，装饰者可能不时会这样做，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>)

然后我必须决定是否要从装饰器的参数或原始函数的参数。装饰器返回的函数可以为其参数设置*args和**kwargs。如果有多个装饰者，我必须决定哪个是我关心的呢？

那么即使功能可以被装饰，找到函数参数的最佳方法是什么？另外，将一系列装饰器放回装饰函数的最佳方法是什么？

更新

这实际上是我现在正在做的事情（名称已被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报告了四个容器的无趣值，并且dict版本的实例为空。但是，如果您是子类，则可以覆盖四个中的任何一个，或者您可以在__init__中包含其他参数：

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}

Answer 1

考虑这个装饰者：

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bar方法，丢弃它，用一个新函数替换它，忽略其命名不同（可能编号！）的参数，而是从经典歌曲返回一行。

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，我看不到你如何找到参数名称p1和p2。根据我的理解，即使Python解释器本身也不知道它们曾经被称为什么。

在Python中查找函数的参数

1 个答案: