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。

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。

注意:我看到有人(不是@unutbu)改变了引用的答案，使它不再使用vars()['Foo'] -所以我的帖子的主要观点不再适用。

FWIW，这是我的意思是@unutbu的答案只与本地定义的类一起工作-使用eval()而不是vars()将使它与任何可访问的类一起工作，而不仅仅是那些在当前范围内定义的类。

对于那些不喜欢使用eval()的人，还展示了一种避免使用它的方法。

首先，这里有一个具体的例子，演示了使用vars()的潜在问题:

class Foo(object): pass
class Bar(Foo): pass
class Baz(Foo): pass
class Bing(Bar): pass

# unutbu's approach
def all_subclasses(cls):
    return cls.__subclasses__() + [g for s in cls.__subclasses__()
                                       for g in all_subclasses(s)]

print(all_subclasses(vars()['Foo']))  # Fine because  Foo is in scope
# -> [<class '__main__.Bar'>, <class '__main__.Baz'>, <class '__main__.Bing'>]

def func():  # won't work because Foo class is not locally defined
    print(all_subclasses(vars()['Foo']))

try:
    func()  # not OK because Foo is not local to func()
except Exception as e:
    print('calling func() raised exception: {!r}'.format(e))
    # -> calling func() raised exception: KeyError('Foo',)

print(all_subclasses(eval('Foo')))  # OK
# -> [<class '__main__.Bar'>, <class '__main__.Baz'>, <class '__main__.Bing'>]

# using eval('xxx') instead of vars()['xxx']
def func2():
    print(all_subclasses(eval('Foo')))

func2()  # Works
# -> [<class '__main__.Bar'>, <class '__main__.Baz'>, <class '__main__.Bing'>]

这可以通过将eval('ClassName')移动到定义的函数中来改进，这使得它更容易使用，而不会失去使用eval()所获得的额外通用性，这与vars()不同，eval()不是上下文敏感的:

# easier to use version
def all_subclasses2(classname):
    direct_subclasses = eval(classname).__subclasses__()
    return direct_subclasses + [g for s in direct_subclasses
                                    for g in all_subclasses2(s.__name__)]

# pass 'xxx' instead of eval('xxx')
def func_ez():
    print(all_subclasses2('Foo'))  # simpler

func_ez()
# -> [<class '__main__.Bar'>, <class '__main__.Baz'>, <class '__main__.Bing'>]

最后，出于安全考虑，避免使用eval()是可能的，在某些情况下甚至是重要的，所以这里有一个不使用它的版本:

def get_all_subclasses(cls):
    """ Generator of all a class's subclasses. """
    try:
        for subclass in cls.__subclasses__():
            yield subclass
            for subclass in get_all_subclasses(subclass):
                yield subclass
    except TypeError:
        return

def all_subclasses3(classname):
    for cls in get_all_subclasses(object):  # object is base of all new-style classes.
        if cls.__name__.split('.')[-1] == classname:
            break
    else:
        raise ValueError('class %s not found' % classname)
    direct_subclasses = cls.__subclasses__()
    return direct_subclasses + [g for s in direct_subclasses
                                    for g in all_subclasses3(s.__name__)]

# no eval('xxx')
def func3():
    print(all_subclasses3('Foo'))

func3()  # Also works
# -> [<class '__main__.Bar'>, <class '__main__.Baz'>, <class '__main__.Bing'>]

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。

如果你只想要直接的子类，那么.__subclasses__()就可以了。如果你想要所有的子类，子类的子类等等，你需要一个函数来为你做这些。

下面是一个简单易读的函数，它可以递归地找到给定类的所有子类:

def get_all_subclasses(cls):
    all_subclasses = []

    for subclass in cls.__subclasses__():
        all_subclasses.append(subclass)
        all_subclasses.extend(get_all_subclasses(subclass))

    return all_subclasses

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__()]
        )
    )