The Python Oracle

Adding a method to an existing object instance

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Chapters
00:00 Adding A Method To An Existing Object Instance
00:19 Answer 1 Score 32
00:46 Accepted Answer Score 1172
02:20 Answer 3 Score 105
03:06 Answer 4 Score 42
05:05 Thank you

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Full question
https://stackoverflow.com/questions/972/...

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Content licensed under CC BY-SA
https://meta.stackexchange.com/help/lice...

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Tags
#python #oop #methods #monkeypatching

#avk47



ACCEPTED ANSWER

Score 1172


In Python, there is a difference between functions and bound methods.

>>> def foo():
...     print "foo"
...
>>> class A:
...     def bar( self ):
...         print "bar"
...
>>> a = A()
>>> foo
<function foo at 0x00A98D70>
>>> a.bar
<bound method A.bar of <__main__.A instance at 0x00A9BC88>>
>>>

Bound methods have been "bound" (how descriptive) to an instance, and that instance will be passed as the first argument whenever the method is called.

Callables that are attributes of a class (as opposed to an instance) are still unbound, though, so you can modify the class definition whenever you want:

>>> def fooFighters( self ):
...     print "fooFighters"
...
>>> A.fooFighters = fooFighters
>>> a2 = A()
>>> a2.fooFighters
<bound method A.fooFighters of <__main__.A instance at 0x00A9BEB8>>
>>> a2.fooFighters()
fooFighters

Previously defined instances are updated as well (as long as they haven't overridden the attribute themselves):

>>> a.fooFighters()
fooFighters

The problem comes when you want to attach a method to a single instance:

>>> def barFighters( self ):
...     print "barFighters"
...
>>> a.barFighters = barFighters
>>> a.barFighters()
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: barFighters() takes exactly 1 argument (0 given)

The function is not automatically bound when it's attached directly to an instance:

>>> a.barFighters
<function barFighters at 0x00A98EF0>

To bind it, we can use the MethodType function in the types module:

>>> import types
>>> a.barFighters = types.MethodType( barFighters, a )
>>> a.barFighters
<bound method ?.barFighters of <__main__.A instance at 0x00A9BC88>>
>>> a.barFighters()
barFighters

This time other instances of the class have not been affected:

>>> a2.barFighters()
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
AttributeError: A instance has no attribute 'barFighters'

More information can be found by reading about descriptors and metaclass programming.




ANSWER 2

Score 105


Module new is deprecated since python 2.6 and removed in 3.0, use types

see http://docs.python.org/library/new.html

In the example below I've deliberately removed return value from patch_me() function. I think that giving return value may make one believe that patch returns a new object, which is not true - it modifies the incoming one. Probably this can facilitate a more disciplined use of monkeypatching.

import types

class A(object):#but seems to work for old style objects too
    pass

def patch_me(target):
    def method(target,x):
        print "x=",x
        print "called from", target
    target.method = types.MethodType(method,target)
    #add more if needed

a = A()
print a
#out: <__main__.A object at 0x2b73ac88bfd0>  
patch_me(a)    #patch instance
a.method(5)
#out: x= 5
#out: called from <__main__.A object at 0x2b73ac88bfd0>
patch_me(A)
A.method(6)        #can patch class too
#out: x= 6
#out: called from <class '__main__.A'>



ANSWER 3

Score 42


I think that the above answers missed the key point.

Let's have a class with a method:

class A(object):
    def m(self):
        pass

Now, let's play with it in ipython:

In [2]: A.m
Out[2]: <unbound method A.m>

Ok, so m() somehow becomes an unbound method of A. But is it really like that?

In [5]: A.__dict__['m']
Out[5]: <function m at 0xa66b8b4>

It turns out that m() is just a function, reference to which is added to A class dictionary - there's no magic. Then why A.m gives us an unbound method? It's because the dot is not translated to a simple dictionary lookup. It's de facto a call of A.__class__.__getattribute__(A, 'm'):

In [11]: class MetaA(type):
   ....:     def __getattribute__(self, attr_name):
   ....:         print str(self), '-', attr_name

In [12]: class A(object):
   ....:     __metaclass__ = MetaA

In [23]: A.m
<class '__main__.A'> - m
<class '__main__.A'> - m

Now, I'm not sure out of the top of my head why the last line is printed twice, but still it's clear what's going on there.

Now, what the default __getattribute__ does is that it checks if the attribute is a so-called descriptor or not, i.e. if it implements a special __get__ method. If it implements that method, then what is returned is the result of calling that __get__ method. Going back to the first version of our A class, this is what we have:

In [28]: A.__dict__['m'].__get__(None, A)
Out[28]: <unbound method A.m>

And because Python functions implement the descriptor protocol, if they are called on behalf of an object, they bind themselves to that object in their __get__ method.

Ok, so how to add a method to an existing object? Assuming you don't mind patching class, it's as simple as:

B.m = m

Then B.m "becomes" an unbound method, thanks to the descriptor magic.

And if you want to add a method just to a single object, then you have to emulate the machinery yourself, by using types.MethodType:

b.m = types.MethodType(m, b)

By the way:

In [2]: A.m
Out[2]: <unbound method A.m>

In [59]: type(A.m)
Out[59]: <type 'instancemethod'>

In [60]: type(b.m)
Out[60]: <type 'instancemethod'>

In [61]: types.MethodType
Out[61]: <type 'instancemethod'>



ANSWER 4

Score 32


In Python monkeypatching generally works by overwriting a class or function's signature with your own. Below is an example from the Zope Wiki:

from SomeOtherProduct.SomeModule import SomeClass
def speak(self):
   return "ook ook eee eee eee!"
SomeClass.speak = speak

This code will overwrite/create a method called speak in the class. In Jeff Atwood's recent post on monkey patching, he showed an example in C# 3.0 which is the current language I use for work.