how to get derived class name from base class

Question

I have a base class Person and derived classes Manager and Employee. Now, what I would like to know is the object created is Man

Answer 1

Python objects provide a __class__ attribute which stores the type used to make that object. This in turns provides a __name__ attribute which can be used to get the name of the type as a string. So, in the simple case:

class A(object):
    pass
class B(A):
    pass

b = B()
print b.__class__.__name__

Would give:

'B'

So, if I follow your question correctly you would do:

m = Manager()
print m.__class__.__name__
'Manager'

Answer 2

In your example you do not need to know the class, you just call the method by referring to the class instance:

# if the derived class is Employee then i would like go to the method title 
# of employee and if its a Manager then go to the title method of Manager
person['post'] = object.title()

But do not use object as a variable name, you hide the built-in name.

Answer 3

Not exactly sure that I understand what you are asking, but you can use x.__class__.__name__ to retrieve the class name as a string, e.g.

class Person:
    pass

class Manager(Person):
    pass

class Employee(Person):
    pass

def get_class_name(instance):
    return instance.__class__.__name__

>>> m = Manager()
>>> print get_class_name(m)
Manager
>>> print get_class_name(Employee())
Employee

Or, you could use isinstance to check for different types:

>>> print isinstance(m, Person)
True
>>> print isinstance(m, Manager)
True
>>> print isinstance(m, Employee)
False

So you could do something like this:

def handle_person(person):
    if isinstance(person, Manager):
        person.read_paper()     # method of Manager class only
    elif isinstance(person, Employee):
        person.work_hard()      # method of Employee class only
    elif isinstance(person, Person):
        person.blah()           # method of the base class
    else:
        print "Not a person"

Answer 4

The best way to "do this" is to not do it. Instead, create methods on Person that are overridden on Manager or Employee, or give the subclasses their own methods that extend the base class.

class Person(object):
    def doYourStuff(self):
        print "I'm just a person, I don't have anything to do"

class Manager(object):
    def doYourStuff(self):
        print "I hereby manage you"

class Employee(object):
    def doYourStuff(self):
        print "I work long hours"

If you find yourself needing to know in the base class which subclass is being instantiated, your program probably has a design error. What will you do if someone else later extends Person to add a new subclass called Contractor? What will Person do when the subclass isn't any of the hard-coded alternatives it knows about?

Answer 5

I don't know if this is what you want, and the way you'd like it implemented, but here's a try:

>>> class Person(object):
    def _type(self):
        return self.__class__.__name__


>>> p = Person()
>>> p._type()
'Person'
>>> class Manager(Person):
    pass

>>> m = Manager()
>>> m._type()
'Manager'
>>> 

Pros: only one definition of the _type method.

Answer 6

Would you be looking for something like this?

>>> class Employee:
...     pass
... 
>>> class Manager(Employee):
...     pass
... 
>>> e = Employee()
>>> m = Manager()
>>> print e.__class__.__name__
Employee
>>> print m.__class__.__name__
Manager
>>> e.__class__.__name__ == 'Manager'
False
>>> e.__class__.__name__ == 'Employee'
True