In Python, how do you make a subclass from a superclass?
A heroic little example:
class SuperHero(object): #superclass, inherits from default object
def getName(self):
raise NotImplementedError #you want to override this on the child classes
class SuperMan(SuperHero): #subclass, inherits from SuperHero
def getName(self):
return "Clark Kent"
class SuperManII(SuperHero): #another subclass
def getName(self):
return "Clark Kent, Jr."
if __name__ == "__main__":
sm = SuperMan()
print sm.getName()
sm2 = SuperManII()
print sm2.getName()
# Initialize using Parent
#
class MySubClass(MySuperClass):
def __init__(self):
MySuperClass.__init__(self)
Or, even better, the use of Python's built-in function, super()
(see the Python 2/Python 3 documentation for it) may be a slightly better method of calling the parent for initialization:
# Better initialize using Parent (less redundant).
#
class MySubClassBetter(MySuperClass):
def __init__(self):
super(MySubClassBetter, self).__init__()
Or, same exact thing as just above, except using the zero argument form of super()
, which only works inside a class definition:
class MySubClassBetter(MySuperClass):
def __init__(self):
super().__init__()
There is another way to make subclasses in python dynamically with a function type()
:
SubClass = type('SubClass', (BaseClass,), {'set_x': set_x}) # Methods can be set, including __init__()
You usually want to use this method when working with metaclasses. When you want to do some lower level automations, that alters way how python creates class. Most likely you will not ever need to do it in this way, but when you do, than you already will know what you are doing.
class Class1(object):
pass
class Class2(Class1):
pass
Class2 is a sub-class of Class1
You use:
class DerivedClassName(BaseClassName):
For details, see the Python docs, section 9.5.