Extending builtin classes in python

Question

How can I extend a builtin class in python? I would like to add a method to the str class.
I've done some searching but all I'm finding is older posts, I'm hoping someone knows of something newer.

Solution 1

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_method
ValueError: invalid literal for int() with base 10: 'Hi Mom'

>>> z = X("271828")
>>> z.lower()
'271828'
>>> z.my_method()
271828

Solution 2

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

Solution 3

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 update


class 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

Comments

[UnkwnTech]

I was hoping that I wouldn't have to subclass but after some more research I have come to the conclusion that it is not possible any other way, so I'll accept this answer.

[user1066101]

@Unkwntech: Can't imagine a better way than subclassing. Seriously. Whatever you had in mind (monkeypatching?) never seems to work out as well as a clear, precise, obvious subclass.

[orip]

@S.Lott - Ruby's open classes and C#'s extension methods come to mind.

[user1066101]

@orip: Aware of these. Find monkeypatching in all forms to be a management nightmare -- they effectively prevent reuse by injecting features in obscure ways.

[dagfr]

In this sample type(z.lower()) is str an not X. How to keep it an X ?

[GalacticRaph]

@dagfr by overriding the .lower() method in the X class and returning a new X object

[caram]

Python 3.9, __dict__ is no longer a dict but a mappingproxy, you have to use @MVP's solution instead.

[caram]

At last I can do this like in Ruby. Works like a charm. I prefer the second definition.