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。

这个答案不如使用@unutbu提到的特殊内置__subclasses__()类方法好，所以我只是把它作为一个练习。subclasses()函数的定义返回一个字典，该字典将所有子类名称映射到子类本身。

def traced_subclass(baseclass):
    class _SubclassTracer(type):
        def __new__(cls, classname, bases, classdict):
            obj = type(classname, bases, classdict)
            if baseclass in bases: # sanity check
                attrname = '_%s__derived' % baseclass.__name__
                derived = getattr(baseclass, attrname, {})
                derived.update( {classname:obj} )
                setattr(baseclass, attrname, derived)
             return obj
    return _SubclassTracer

def subclasses(baseclass):
    attrname = '_%s__derived' % baseclass.__name__
    return getattr(baseclass, attrname, None)


class BaseClass(object):
    pass

class SubclassA(BaseClass):
    __metaclass__ = traced_subclass(BaseClass)

class SubclassB(BaseClass):
    __metaclass__ = traced_subclass(BaseClass)

print subclasses(BaseClass)

输出:

{'SubclassB': <class '__main__.SubclassB'>,
 'SubclassA': <class '__main__.SubclassA'>}

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__()]
        )
    )

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。

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], [])