Extending builtin classes in python

久未见 提交于 2019-11-28 19:12:37
S.Lott

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

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 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

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
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