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'>}

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'>}

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。

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

我怎么能找到一个类的所有子类给它的名字?

我们当然可以很容易地做到这一点，只要能访问对象本身。

仅仅给出它的名字是一个糟糕的想法，因为可以有多个同名的类，甚至在同一个模块中定义。

我为另一个答案创建了一个实现，因为它回答了这个问题，而且它比这里的其他解决方案更优雅，下面是:

def get_subclasses(cls):
    """returns all subclasses of argument, cls"""
    if issubclass(cls, type):
        subclasses = cls.__subclasses__(cls)
    else:
        subclasses = cls.__subclasses__()
    for subclass in subclasses:
        subclasses.extend(get_subclasses(subclass))
    return subclasses

用法:

>>> import pprint
>>> list_of_classes = get_subclasses(int)
>>> pprint.pprint(list_of_classes)
[<class 'bool'>,
 <enum 'IntEnum'>,
 <enum 'IntFlag'>,
 <class 'sre_constants._NamedIntConstant'>,
 <class 'subprocess.Handle'>,
 <enum '_ParameterKind'>,
 <enum 'Signals'>,
 <enum 'Handlers'>,
 <enum 'RegexFlag'>]