如何从类体中获取当前类的引用?

6 投票
2 回答
2475 浏览
提问于 2025-04-15 18:49

我想在一个基类中保存一个字典,里面记录所有的子类(包括非直接子类),这样我就可以通过字符串来实例化它们。我这么做是因为CLSID是通过网页表单发送的,所以我想限制选择的范围,只能是那些从子类中设置的选项。(我不想用eval()或者globals()来处理类名)。

class BaseClass(object):
    CLSID = 'base'
    CLASSES = {}

    def from_string(str):
        return CLASSES[str]()

class Foo(BaseClass):
    CLSID = 'foo'
    BaseClass.CLASSES[CLSID] = Foo

class Bar(BaseClass):
    CLSID = 'bar'
    BaseClass.CLASSES[CLSID] = Bar

显然,这样做是不行的。但是有没有类似于@classmethod的东西可以用在初始化上?我的想法是,这个类方法在每个类被读取时只运行一次,并把这个类注册到基类中。这样的话,下面的代码就可以工作了:(这也能省去在FooBar中多写一行代码)

class BaseClass(object):
    CLSID = 'base'
    CLASSES = {}

    @classmethod
    def __init__(cls):
        BaseClass.CLASSES[cls.CLSID] = cls 

    def from_string(str):
        return CLASSES[str]()

我考虑过使用__subclasses__,然后在CLSID上用filter(),但那样只对直接子类有效。

所以,希望我能解释清楚我的目的,问题是如何让这个工作?或者我是不是走错了方向?

2 个回答

3

你可以尝试使用元类来完成这个任务,但我觉得一个更简单的办法就可以解决问题:

class BaseClass(object):
    CLASS_ID = None
    _CLASSES = {}

    @classmethod
    def create_from_id(cls, class_id):
        return CLASSES[class_id]()

    @classmethod
    def register(cls):
        assert cls.CLASS_ID is not None, "subclass %s must define a CLASS_ID" % cls
        cls._CLASSES[cls.CLASS_ID] = cls

然后要定义一个子类,只需使用:

class Foo(BaseClass):
    CLASS_ID = 'foo'

Foo.register()

最后,使用基类中的工厂方法来为你创建实例:

foo = BaseClass.create_from_id('foo')

在这个解决方案中,类定义之后,你必须调用注册类的方法来把子类注册到基类中。此外,默认的 CLASS_ID 是 None,这样可以避免如果用户忘记定义它时,覆盖掉注册表中的基类。

7

将这个与基类牢牢绑定在一起:

class AutoRegister(type):
  def __new__(mcs, name, bases, D):
    self = type.__new__(mcs, name, bases, D)
    if "ID" in D:  # only register if has ID attribute directly
      if self.ID in self._by_id:
        raise ValueError("duplicate ID: %r" % self.ID)
      self._by_id[self.ID] = self
    return self

class Base(object):
  __metaclass__ = AutoRegister
  _by_id = {}
  ID = "base"

  @classmethod
  def from_id(cls, id):
    return cls._by_id[id]()

class A(Base):
  ID = "A"

class B(Base):
  ID = "B"

print Base.from_id("A")
print Base.from_id("B")

或者将不同的关注点实际上分开:

class IDFactory(object):
  def __init__(self):
    self._by_id = {}
  def register(self, cls):
    self._by_id[cls.ID] = cls
    return cls

  def __call__(self, id, *args, **kwds):
    return self._by_id[id](*args, **kwds)
  # could use a from_id function instead, as above

factory = IDFactory()

@factory.register
class Base(object):
  ID = "base"

@factory.register
class A(Base):
  ID = "A"

@factory.register
class B(Base):
  ID = "B"

print factory("A")
print factory("B")

你可能已经注意到我更喜欢哪一种。将其与类的层次结构分开定义,你可以很容易地扩展和修改,比如可以用两个名字注册(使用ID属性只能允许一个):

class IDFactory(object):
  def __init__(self):
    self._by_id = {}

  def register(self, cls):
    self._by_id[cls.ID] = cls
    return cls

  def register_as(self, name):
    def wrapper(cls):
      self._by_id[name] = cls
      return cls
    return wrapper

  # ...

@factory.register_as("A")  # doesn't require ID anymore
@factory.register          # can still use ID, even mix and match
@factory.register_as("B")  # imagine we got rid of B,
class A(object):           #  and A fulfills that roll now
  ID = "A"

你还可以将工厂实例“放在”基类内部,同时保持它的独立性:

class IDFactory(object):
  #...

class Base(object):
  factory = IDFactory()

  @classmethod
  def register(cls, subclass):
    if subclass.ID in cls.factory:
      raise ValueError("duplicate ID: %r" % subclass.ID)
    cls.factory[subclass.ID] = subclass
    return subclass

@Base.factory.register  # still completely decoupled
                        # (it's an attribute of Base, but that can be easily
                        # changed without modifying the class A below)
@Base.register  # alternatively more coupled, but possibly desired
class A(Base):
  ID = "A"

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