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
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'>]
其他回答
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'>]
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__()]
)
)
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'>}
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'>]
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