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
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'>]
其他回答
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。
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], [])
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'>]
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__()]
)
)