Extending builtin classes in python

Just subclass the type

>>> class X(str):...     def my_method(self):...         return int(self)...>>> s = X("Hi Mom")>>> s.lower()'hi mom'>>> s.my_method()Traceback (most recent call last):  File "<stdin>", line 1, in <module>  File "<stdin>", line 3, in my_methodValueError: invalid literal for int() with base 10: 'Hi Mom'>>> z = X("271828")>>> z.lower()'271828'>>> z.my_method()271828

One way could be to use the "class reopening" concept (natively existing in Ruby) that can be implemented in Python using a class decorator.An exemple is given in this page:http://www.ianbicking.org/blog/2007/08/opening-python-classes.html

I quote:

I think with class decorators you could do this:

@extend(SomeClassThatAlreadyExists)class SomeClassThatAlreadyExists:    def some_method(self, blahblahblah):        stuff

Implemented like this:

def extend(class_to_extend):    def decorator(extending_class):        class_to_extend.__dict__.update(extending_class.__dict__)        return class_to_extend    return decorator

Assuming that you can not change builtin classes.To simulate a "class reopening" like Ruby in Python3 where __dict__ is an mappingproxy object and not dict object :

def open(cls):  def update(extension):    for k,v in extension.__dict__.items():      if k != '__dict__':        setattr(cls,k,v)    return cls  return updateclass A(object):  def hello(self):    print('Hello!')A().hello()   #=> Hello!#reopen class A@open(A)class A(object):  def hello(self):    print('New hello!')  def bye(self):    print('Bye bye')A().hello()   #=> New hello!A().bye()     #=> Bye bye

In Python2 I could also write a decorator function 'open' as well:

def open(cls):  def update(extension):    namespace = dict(cls.__dict__)    namespace.update(dict(extension.__dict__))    return type(cls.__name__,cls.__bases__,namespace)  return update