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__
Answered 2023-09-21 08:08:02
dir(x.__class__)
does not list it? - anyone __class__
over the type
method? Like so: type(x).__name__
. Isn't calling double underscore members directly discouraged? I can't see a way around using __name__
, though. - anyone __class__
directly to be compatible with old-style classes, since their type is just instance
. - anyone def typename(x): return type(x).__name__
- anyone Do you want the name of the class as a string?
instance.__class__.__name__
Answered 2023-09-21 08:08:02
type()
?
>>> class A:
... def whoami(self):
... print(type(self).__name__)
...
>>>
>>> class B(A):
... pass
...
>>>
>>>
>>> o = B()
>>> o.whoami()
'B'
>>>
Answered 2023-09-21 08:08:02
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? - anyone type(item)
on a list item the result will be <type 'instance'>
while item.__class__.__name__
holds the class name. - anyone 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.
Answered 2023-09-21 08:08:02
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'
>>>
Answered 2023-09-21 08:08:02
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'
Answered 2023-09-21 08:08:02
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
Answered 2023-09-21 08:08:02
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.
Answered 2023-09-21 08:08:02
__qualname__
is for Python 3.3+ - anyone __qualname__
vs __name__
: stackoverflow.com/questions/58108488/what-is-qualname-in-python - anyone 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
Answered 2023-09-21 08:08:02
To get instance classname:
type(instance).__name__
or
instance.__class__.__name__
both are the same
Answered 2023-09-21 08:08:02
__class__
, or in old style classes (which are obsolete) - anyone You can first use type
and then str
to extract class name from it.
class foo:pass;
bar:foo=foo();
print(str(type(bar))[8:-2][len(str(type(bar).__module__))+1:]);
foo
Answered 2023-09-21 08:08:02
If you're looking to solve this for a list (or iterable collection) of objects, here's how I would solve:
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))
Answered 2023-09-21 08:08:02