获取实例的类名？

How do I find out a name of class that created an instance of an object in Python if the function I am doing this from is the base class of which the class of the instance has been derived?

Was thinking maybe the inspect module might have helped me out here, but it doesn’t seem to give me what I want. And short of parsing the __class__ member, I’m not sure how to get at this information.

Have you tried the of the class? ie type(x).__name__ will give you the name of the class, which I think is what you want.

>>> import itertools
>>> x = itertools.count(0)
>>> type(x).__name__
'count'

If you’re still using Python 2, note that the above method works with new-style classes only (in Python 3+ all classes are “new-style” classes). Your code might use some old-style classes. The following works for both:

x.__class__.__name__

Do you want the name of the class as a string?

instance.__class__.__name__

type() ?

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

class A:
  pass

a = A()
str(a.__class__)

The sample code above (when input in the interactive interpreter) will produce '__main__.A' as opposed to 'A' which is produced if the __name__ attribute is invoked. By simply passing the result of A.__class__ to the str constructor the parsing is handled for you. However, you could also use the following code if you want something more explicit.

"{0}.{1}".format(a.__class__.__module__,a.__class__.__name__)

This behavior can be preferable if you have classes with the same name defined in separate modules.

The sample code provided above was tested in Python 2.7.5.

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):
  ...

Example:

>>> 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'
>>>

Good question.

Here’s a simple example based on GHZ’s which might help someone:

>>> 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

Alternatively you can use the classmethod decorator:

class A:
    @classmethod
    def get_classname(cls):
        return cls.__name__

    def use_classname(self):
        return self.get_classname()

Usage:

>>> A.get_classname()
'A'
>>> a = A()
>>> a.get_classname()
'A'
>>> a.use_classname()
'A'

Apart from grabbing the special attribute, you might find yourself in need of the qualified name for a given class/function. This is done by grabbing the types __qualname__.

In most cases, these will be exactly the same, but, when dealing with nested classes/methods these differ in the output you get. For example:

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'

Since introspection is what you’re after, this is always you might want to consider.

To get instance classname:

type(instance).__name__

or

instance.__class__.__name__

both are the same