我需要一种工作方法来获取从Python中的基类继承的所有类。
新类型(即从object
子类化,这是Python 3中的默认类型)具有返回子类的__subclasses__
方法:
class Foo(object): pass
class Bar(Foo): pass
class Baz(Foo): pass
class Bing(Bar): pass
以下是子类的名称:
print([cls.__name__ for cls in Foo.__subclasses__()])
# ['Bar', 'Baz']
以下是子类本身:
print(Foo.__subclasses__())
# [<class '__main__.Bar'>, <class '__main__.Baz'>]
确认子类确实将Foo
列为其基数:
for cls in Foo.__subclasses__():
print(cls.__base__)
# <class '__main__.Foo'>
# <class '__main__.Foo'>
请注意,如果您想要子类,则必须递归:
def all_subclasses(cls):
return set(cls.__subclasses__()).union(
[s for c in cls.__subclasses__() for s in all_subclasses(c)])
print(all_subclasses(Foo))
# {<class '__main__.Bar'>, <class '__main__.Baz'>, <class '__main__.Bing'>}
请注意,如果尚未执行子类的类定义 - 例如,如果尚未导入子类的模块 - 那么该子类尚不存在,并且__subclasses__
将找不到它。
你提到“给它的名字”。由于Python类是第一类对象,因此您不需要使用具有类名称的字符串来代替类或类似的类。您可以直接使用该类,您可能应该这样做。
如果你有一个表示类名称的字符串,并且想要找到该类的子类,那么有两个步骤:找到给定其名称的类,然后如上所述找到__subclasses__
的子类。
如何从名称中找到该类取决于您期望找到它的位置。如果您希望在与尝试定位类的代码相同的模块中找到它,那么
cls = globals()[name]
会做这个工作,或者在你希望在当地人找到它的不太可能的情况下,
cls = locals()[name]
如果类可以在任何模块中,那么您的名称字符串应该包含完全限定名称 - 类似于'pkg.module.Foo'
而不仅仅是'Foo'
。使用importlib
加载类的模块,然后检索相应的属性:
import importlib
modname, _, clsname = name.rpartition('.')
mod = importlib.import_module(modname)
cls = getattr(mod, clsname)
但是,如果找到该类,cls.__subclasses__()
将返回其子类列表。
我无法想象它的真实世界用例,但是一种强大的方式(即使在Python 2旧样式类中)也将扫描全局命名空间:
def has_children(cls):
g = globals().copy() # use a copy to make sure it will not change during iteration
g.update(locals()) # add local symbols
for k, v in g.items(): # iterate over all globals object
try:
if (v is not cls) and issubclass(v, cls): # found a strict sub class?
return True
except TypeError: # issubclass raises a TypeError if arg is not a class...
pass
return False
它适用于Python 2新样式类和Python 3类以及Python 2经典类
如果你只想要直接子类,那么.__subclasses__()
工作正常。如果您想要所有子类,子类的子类等,您将需要一个函数来为您完成。
这是一个简单易读的函数,以递归方式查找给定类的所有子类:
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
一般形式的最简单的解决方案:
def get_subclasses(cls):
for subclass in cls.__subclasses__():
yield from get_subclasses(subclass)
yield subclass
还有一个类方法,以防你有一个继承自的类:
@classmethod
def get_subclasses(cls):
for subclass in cls.__subclasses__():
yield from subclass.get_subclasses()
yield subclass
__init_subclass__
正如其他答案所提到的,您可以检查__subclasses__
属性以获取子类列表,因为python 3.6您可以通过覆盖__init_subclass__
方法来修改此属性创建。
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,如果您知道自己在做什么,就可以覆盖__subclasses__
的行为,并从此列表中省略/添加子类。
FWIW,这就是我所说的@unutbu's answer只使用本地定义的类 - 并且使用eval()
而不是vars()
将使它适用于任何可访问的类,而不仅仅是当前范围中定义的类。
对于那些不喜欢使用eval()
的人来说,也有一种方法可以避免它。
首先,这是一个具体的例子,展示了使用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')
向下移动到定义的函数来改进,这使得使用它更容易,而不会失去使用eval()
获得的额外通用性,这与vars()
不同于上下文敏感:
# 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,在某些情况下,出于安全原因避免使用eval()
是可能的,甚至可能是重要的,所以这里是没有它的版本:
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'>]
用于获取所有子类列表的更短版本:
from itertools import chain
def subclasses(cls):
return list(
chain.from_iterable(
[list(chain.from_iterable([[x], subclasses(x)])) for x in cls.__subclasses__()]
)
)
这不像使用@unutbu提到的特殊内置__subclasses__()
class方法那样好,所以我只是把它作为练习来呈现。 Thesubclasses()
function定义返回一个字典,它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,因为它是通过将尾递归实现转换为循环生成器而创建的:
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], [])
如何根据名称查找类的所有子类?
在访问对象本身的情况下,我们当然可以轻松地做到这一点,是的。
简单地给它的名字是个坏主意,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answer创建了一个实现,因为它回答了这个问题并且比这里的其他解决方案更优雅,这里是:
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'>]