我需要一种工作方法来获取从Python基类继承的所有类。


当前回答

下面是一个没有递归的版本:

def get_subclasses_gen(cls):

    def _subclasses(classes, seen):
        while True:
            subclasses = sum((x.__subclasses__() for x in classes), [])
            yield from classes
            yield from seen
            found = []
            if not subclasses:
                return

            classes = subclasses
            seen = found

    return _subclasses([cls], [])

这与其他实现的不同之处在于它返回原始类。 这是因为它使代码更简单,并且:

class Ham(object):
    pass

assert(issubclass(Ham, Ham)) # True

如果get_subclasses_gen看起来有点奇怪,那是因为它是通过将尾递归实现转换为循环生成器创建的:

def get_subclasses(cls):

    def _subclasses(classes, seen):
        subclasses = sum(*(frozenset(x.__subclasses__()) for x in classes))
        found = classes + seen
        if not subclasses:
            return found

        return _subclasses(subclasses, found)

    return _subclasses([cls], [])

其他回答

Python 3.6 - __init_subclass__

正如其他回答提到的,你可以检查__subclasses__属性来获得子类列表,因为python 3.6你可以通过重写__init_subclass__方法来修改这个属性的创建。

class PluginBase:
    subclasses = []

    def __init_subclass__(cls, **kwargs):
        super().__init_subclass__(**kwargs)
        cls.subclasses.append(cls)

class Plugin1(PluginBase):
    pass

class Plugin2(PluginBase):
    pass

这样,如果你知道你在做什么,你可以重写__subclasses__的行为,并从这个列表中省略/添加子类。

虽然我非常倾向于__init_subclass__方法,这将保留定义顺序,并避免组合增长顺序,如果你有一个非常密集的层次结构,到处都有多个继承:

def descendents(cls):
    '''Does not return the class itself'''
    R = {}
    def visit(cls):
        for subCls in cls.__subclasses__():
            R[subCls] = True
            visit(subCls)
    visit(cls)
    return list(R.keys())

这是因为字典会记住键的插入顺序。列表方法也会起作用。

获取所有子类列表的一个更短的版本:

from itertools import chain

def subclasses(cls):
    return list(
        chain.from_iterable(
            [list(chain.from_iterable([[x], subclasses(x)])) for x in cls.__subclasses__()]
        )
    )

一般形式的最简单解:

def get_subclasses(cls):
    for subclass in cls.__subclasses__():
        yield from get_subclasses(subclass)
        yield subclass

和类方法,如果你有一个单一的类,你继承:

@classmethod
def get_subclasses(cls):
    for subclass in cls.__subclasses__():
        yield from subclass.get_subclasses()
        yield subclass

下面是一个简单但有效的代码版本:

def get_all_subclasses(cls):
    subclass_list = []

    def recurse(klass):
        for subclass in klass.__subclasses__():
            subclass_list.append(subclass)
            recurse(subclass)

    recurse(cls)

    return set(subclass_list)

它的时间复杂度是O(n)如果没有多重继承,n是所有子类的数目。 它比递归地创建列表或使用生成器生成类的函数更有效,后者的复杂度可能是(1)O(nlogn)当类层次结构是平衡树时,或(2)O(n²)当类层次结构是有偏树时。