Reputation: 6693
Is it possible to get the name of a subclass? For example:
class Foo:
def bar(self):
print type(self)
class SubFoo(Foo):
pass
SubFoo().bar()
will print: < type 'instance' >
I'm looking for a way to get "SubFoo"
.
I know you can do isinstance
, but I don't know the name of the class a priori, so that doesn't work for me.
Upvotes: 24
Views: 20037
Reputation: 923
Just a reminder that this issue doesn't exist in Python 3.x
and
print(type(self))
will give the more informative <class '__main__.SubFoo'>
instead of bad old < type 'instance' >
.
This is because object
is subclassed automatically in Python 3.x
, making every class a new-style
class.
More discussion at What is the purpose of subclassing the class "object" in Python?.
Like others pointed out, to just get the subclass name do print(type(self).__name__)
.
Upvotes: 3
Reputation: 1698
SubFoo.__name__
And parents: [cls.__name__ for cls in SubFoo.__bases__]
Upvotes: 5
Reputation: 9005
#!/usr/bin/python
class Foo(object):
def bar(self):
print type(self)
class SubFoo(Foo):
pass
SubFoo().bar()
Subclassing from object
gives you new-style classes (which are not so new any more - python 2.2!) Anytime you want to work with the self attribute a lot you will get a lot more for your buck if you subclass from object. Python's docs ... new style classes. Historically Python left the old-style way Foo()
for backward compatibility. But, this was a long time ago. There is not much reason anymore not to subclass from object.
Upvotes: 13
Reputation: 19905
you can use
SubFoo().__class__.__name__
which might be off-topic, since it gives you a class name :)
Upvotes: 16
Reputation: 798606
It works a lot better when you use new-style classes.
class Foo(object):
....
Upvotes: 3