How do I find out the name of the class used to create an instance of an object in Python?
I'm not sure if I should use the inspect
module or parse the __class__
attribute.
Have you tried the __name__
attribute 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:
... def whoami(self):
... print(type(self).__name__)
...
>>>
>>> class B(A):
... pass
...
>>>
>>>
>>> o = B()
>>> o.whoami()
'B'
>>>
self.__class__.__name__
instead of type(self).__name__
to get the same behaviour. Unless there is something the type()
function does that I am not aware of?
type(item)
on a list item the result will be <type 'instance'>
while item.__class__.__name__
holds the class name.
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.
In Python 2,
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'
>>>
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'
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
Apart from grabbing the special __name__
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.
__qualname__
is for Python 3.3+
__qualname__
vs __name__
: stackoverflow.com/questions/58108488/what-is-qualname-in-python
You can simply use __qualname__
which stands for qualified name of a function or class
Example:
>>> class C:
... class D:
... def meth(self):
... pass
...
>>> C.__qualname__
'C'
>>> C.D.__qualname__
'C.D'
>>> C.D.meth.__qualname__
'C.D.meth'
documentation link qualname
To get instance classname:
type(instance).__name__
or
instance.__class__.__name__
both are the same
__class__
, or in old style classes (which are obsolete)
Success story sharing
dir(x.__class__)
does not list it?__class__
over thetype
method? Like so:type(x).__name__
. Isn't calling double underscore members directly discouraged? I can't see a way around using__name__
, though.__class__
directly to be compatible with old-style classes, since their type is justinstance
.def typename(x): return type(x).__name__