类型（）？

>>> class A:
...     def whoami(self):
...         print(type(self).__name__)
...
>>>
>>> class B(A):
...     pass
...
>>>
>>>
>>> o = B()
>>> o.whoami()
'B'
>>>

如果您希望为对象列表（或可迭代集合）解决此问题，以下是我的解决方法：

from operator import attrgetter

# Will use a few data types to show a point
my_list = [1, "2", 3.0, [4], object(), type, None]

# I specifically want to create a generator
my_class_names = list(map(attrgetter("__name__"), map(type, my_list))))

# Result:
['int', 'str', 'float', 'list', 'object', 'type', 'NoneType']


# Alternatively, use a lambda
my_class_names = list(map(lambda x: type(x).__name__, my_list))

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'
>>>

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

除了获取特殊的__name__属性之外，您可能会发现自己需要给定类/函数的限定名称。这是通过获取__qualiname__类型来完成的。

在大多数情况下，它们将完全相同，但在处理嵌套类/方法时，它们在输出方面有所不同。例如：

class Spam:
    def meth(self):
        pass
    class Bar:
        pass

>>> s = Spam()
>>> type(s).__name__ 
'Spam'
>>> type(s).__qualname__
'Spam'
>>> type(s).Bar.__name__       # type not needed here
'Bar'
>>> type(s).Bar.__qualname__   # type not needed here 
'Spam.Bar'
>>> type(s).meth.__name__
'meth'
>>> type(s).meth.__qualname__
'Spam.meth'

因为内省是你追求的，所以这是你可能需要考虑的。

要获取实例类名：

type(instance).__name__

or

instance.__class__.__name__

两者都一样