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'>]

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'>]

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], [])

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。

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

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'>}