如何查找用于在Python中创建对象实例的类的名称?
我不确定应该使用inspect模块还是解析__class__属性。
如何查找用于在Python中创建对象实例的类的名称?
我不确定应该使用inspect模块还是解析__class__属性。
当前回答
在Python 2中,
type(instance).__name__ != instance.__class__.__name__
# if class A is defined like
class A():
...
type(instance) == instance.__class__
# if class A is defined like
class A(object):
...
例子:
>>> class aclass(object):
... pass
...
>>> a = aclass()
>>> type(a)
<class '__main__.aclass'>
>>> a.__class__
<class '__main__.aclass'>
>>>
>>> type(a).__name__
'aclass'
>>>
>>> a.__class__.__name__
'aclass'
>>>
>>> class bclass():
... pass
...
>>> b = bclass()
>>>
>>> type(b)
<type 'instance'>
>>> b.__class__
<class __main__.bclass at 0xb765047c>
>>> type(b).__name__
'instance'
>>>
>>> b.__class__.__name__
'bclass'
>>>
其他回答
问得好。
下面是一个基于GHZ的简单示例,可能会对某人有所帮助:
>>> class person(object):
def init(self,name):
self.name=name
def info(self)
print "My name is {0}, I am a {1}".format(self.name,self.__class__.__name__)
>>> bob = person(name='Robert')
>>> bob.info()
My name is Robert, I am a person
类型()?
>>> class A:
... def whoami(self):
... print(type(self).__name__)
...
>>>
>>> class B(A):
... pass
...
>>>
>>>
>>> o = B()
>>> o.whoami()
'B'
>>>
或者,您可以使用classmethoddecorator:
class A:
@classmethod
def get_classname(cls):
return cls.__name__
def use_classname(self):
return self.get_classname()
用法:
>>> A.get_classname()
'A'
>>> a = A()
>>> a.get_classname()
'A'
>>> a.use_classname()
'A'
是否要将类的名称作为字符串?
instance.__class__.__name__
在Python 2中,
type(instance).__name__ != instance.__class__.__name__
# if class A is defined like
class A():
...
type(instance) == instance.__class__
# if class A is defined like
class A(object):
...
例子:
>>> class aclass(object):
... pass
...
>>> a = aclass()
>>> type(a)
<class '__main__.aclass'>
>>> a.__class__
<class '__main__.aclass'>
>>>
>>> type(a).__name__
'aclass'
>>>
>>> a.__class__.__name__
'aclass'
>>>
>>> class bclass():
... pass
...
>>> b = bclass()
>>>
>>> type(b)
<type 'instance'>
>>> b.__class__
<class __main__.bclass at 0xb765047c>
>>> type(b).__name__
'instance'
>>>
>>> b.__class__.__name__
'bclass'
>>>