Make dictionary function behave like class method

Question

I have a dictionary of methods:

def name(self):
   print self.name

def age(self):
   print self.age

display = {
   'name': name,
   'age': age
}

And I have a class Person where I want display['name']() and display['age']() to work as methods:

class Person:
   def __init__(self):
      self.name = 'Bob'
      self.age = 22
      self.showName = display['name']
      self.showAge = display['age']

person = Person()

The problem is that when I call person.showName() I get the error

TypeError: showName() takes exactly 1 argument (0 given)

How can I make the function showName act like a method, and implicitly pass self as the first argument?

possible duplicate of [add method to loaded class/module in python](http://stackoverflow.com/questions/7749914/add-method-to-loaded-class-module-in-python) — Björn Pollex, Jan 18 '12 at 13:34
Sorry, misread the question - it's not a duplicate of that one. — Björn Pollex, Jan 18 '12 at 13:34
Are you aware of the difference between a class-method and an instance-method? Is the wording in the title deliberate, or are you confusing something? — Björn Pollex, Jan 18 '12 at 13:36
I guess it is a duplicate of [this one](http://stackoverflow.com/questions/972/adding-a-method-to-an-existing-object) instead. — Björn Pollex, Jan 18 '12 at 13:36
@BjörnPollex: I'm probably confused with something. :) Python is very new to me. Feel free to edit the title, I'll accept the change. — Randomblue, Jan 18 '12 at 13:36

Roman Bodnarchuk · Accepted Answer · 2012-01-18T14:33:25.510

4

Easy way is to move definitions of your functions outside of __init__:

>>> class Person:
...     showName = display['name']
...     showAge = display['age']
...     def __init__(self):
...             self.name = 'Bob'
...             self.age = 22
...
>>> Person().showAge()
22

Harder way is to understand why does this happens and use MethodType:

>>> class Person:
...     def __init__(self):
...             self.name = 'Bob'
...             self.age = 22
...             self.showName = types.MethodType(display['name'], self)
...             self.showAge = types.MethodType(display['age'], self)
...
>>> Person().showAge()
22

In short, a function should be a method to be threaten a bit special while performing object.attribute lookup. This threatening also includes passing of self parameter, which your functions expect. MethodType takes function and binds it to some object thus creating a method.

First example works since functions defined along with class definition are threaten as methods automatically.

edited Jan 18 '12 at 14:33

answered Jan 18 '12 at 13:45

Roman Bodnarchuk

29,461
12
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Didn't know about methodtype, very nice. – Bite code Jan 18 '12 at 13:51
An alternative to `types.MethodType(display['name'], self)` is `display['name'].__get__(self)`. – Sven Marnach Jan 18 '12 at 13:57
I now have a context problem. The context of `showName` does not contain functions in the context of `Person`. – Randomblue Jan 18 '12 at 14:00
@Randomblue you should be able to access all `Person` functions using `self` parameter in `name` or `age` function. – Roman Bodnarchuk Jan 18 '12 at 14:03

score 1 · Answer 2 · answered Jan 18 '12 at 13:37

1

from functools import partial
class Person:
   def __init__(self):
        self.showName = partial(display['name'], self=self)

partial allows better support for passing arguments to the encapsulated function than lambda, if you'll have any need for more arguments in the future.

answered Jan 18 '12 at 13:37

dorsh

23,750
2
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2

While this works, there is a [better solution](http://stackoverflow.com/questions/972/adding-a-method-to-an-existing-object). – Björn Pollex Jan 18 '12 at 13:39

score 0 · Answer 3 · answered Jan 18 '12 at 13:37

0

You can use something like:

self.showName = lambda : display['name'](self)

answered Jan 18 '12 at 13:37

karolx

508
4
9

1

This a horrible solution and a very bad misuse of lambdas. – Björn Pollex Jan 18 '12 at 13:38

score -2 · Answer 4 · answered Jan 18 '12 at 13:34

-2

Do this:

display['name']()

Should work.

answered Jan 18 '12 at 13:34

TyrantWave

4,583
2
22
25

1

He writes in his question that he already knows this. He wants normal method-call syntax. – Björn Pollex Jan 18 '12 at 13:35

Make dictionary function behave like class method

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