Get name of current class?

Question

How do I get the name of the class I am currently in?

Example:

def get_input(class_name):
    [do things]
    return class_name_result

class foo():
    input = get_input([class name goes here])

Due to the nature of the program I am interfacing with (vistrails), I cannot use __init__() to initialize input.

Answer 1

PEP 3155 introduced __qualname__, which was implemented in Python 3.3.

For top-level functions and classes, the __qualname__ attribute is equal to the __name__ attribute. For nested classes, methods, and nested functions, the __qualname__ attribute contains a dotted path leading to the object from the module top-level.

It is accessible from within the very definition of a class or a function, so for instance:

class Foo:
    print(__qualname__)

will effectively print Foo. You'll get the fully qualified name (excluding the module's name), so you might want to split it on the . character.

However, there is no way to get an actual handle on the class being defined.

>>> class Foo:
...     print('Foo' in globals())
... 
False

Answer 2

You can access it by the class' private attributes:

cls_name = self.__class__.__name__

EDIT:

As said by Ned Batcheler, this wouldn't work in the class body, but it would in a method.

Answer 3

I think, it should be like this:

    class foo():
        input = get_input(__qualname__)

Answer 4

obj.__class__.__name__ will get you any objects name, so you can do this:

class Clazz():
    def getName(self):
        return self.__class__.__name__

Usage:

>>> c = Clazz()
>>> c.getName()
'Clazz'

Answer 5

Within the body of a class, the class name isn't defined yet, so it is not available. Can you not simply type the name of the class? Maybe you need to say more about the problem so we can find a solution for you.

I would create a metaclass to do this work for you. It's invoked at class creation time (conceptually at the very end of the class: block), and can manipulate the class being created. I haven't tested this:

class InputAssigningMetaclass(type):
    def __new__(cls, name, bases, attrs):
        cls.input = get_input(name)
        return super(MyType, cls).__new__(cls, name, bases, newattrs)

class MyBaseFoo(object):
    __metaclass__ = InputAssigningMetaclass

class foo(MyBaseFoo):
    # etc, no need to create 'input'

class foo2(MyBaseFoo):
    # etc, no need to create 'input'

Answer 6

EDIT: Yes, you can; but you have to cheat: The currently running class name is present on the call stack, and the traceback module allows you to access the stack.

>>> import traceback
>>> def get_input(class_name):
...     return class_name.encode('rot13')
... 
>>> class foo(object):
...      _name = traceback.extract_stack()[-1][2]
...     input = get_input(_name)
... 
>>> 
>>> foo.input
'sbb'

However, I wouldn't do this; My original answer is still my own preference as a solution. Original answer:

probably the very simplest solution is to use a decorator, which is similar to Ned's answer involving metaclasses, but less powerful (decorators are capable of black magic, but metaclasses are capable of ancient, occult black magic)

>>> def get_input(class_name):
...     return class_name.encode('rot13')
... 
>>> def inputize(cls):
...     cls.input = get_input(cls.__name__)
...     return cls
... 
>>> @inputize
... class foo(object):
...     pass
... 
>>> foo.input
'sbb'
>>>