Assigning method to object at runtime in Python

Question

I'm trying to do the Javascript equivalent in Python:

a.new_func = function(arg1, arg2) {
    var diff = arg1 - arg2;
    return diff * diff;
}

Right now, the way I'm doing this is by defining the method first, and then assigning it, but my question is whether or not Python allows a shorthand to do the assigning and the defining part in the same line. Something like this:

a.new_func = def new_func(arg1, arg2):
    diff = arg1 - arg2
    return diff * diff

Instead of this:

def new_func(arg1, arg2):
    diff = arg1 - arg2
    return diff * diff
a.new_func = new_func

I realize the difference is not major, but am still interested to know whether or not it's possible.

user2357112 · Accepted Answer · 2014-01-17 03:29:47Z

9

Python supports no such syntax.

I suppose if you wanted, you could write a decorator. It might look a bit nicer:

def method_of(instance):
    def method_adder(function):
        setattr(instance, function.__name__, function)
        return function
    return method_adder

@method_of(a)
def new_func(arg1, arg2):
    stuff()

Or if you want the method to have access to self:

def method_of(instance):
    def method_adder(function):
        setattr(instance, function.__name__, function.__get__(instance))
        return function
    return method_adder

@method_of(a)
def new_func(self, arg1, arg2):
    stuff()

edited Jan 17, 2014 at 3:29

answered Jan 17, 2014 at 3:24

user2357112

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1 Comment

user2357112 Over a year ago

This actually turned out much neater than I expected. If I ever need to dynamically add methods to objects, I'll probably use a decorator. I wonder how many times this has been reinvented.

inspectorG4dget · Accepted Answer · 2014-01-17 03:27:11Z

6

The closest thing to what you're looking for are lambda expressions, which are not as easy to use as properly written functions:

a.new_function = lambda arg1,arg2 : (arg1-arg2)**2

However, in almost all cases, defining a function, and assigning it, the way you have done in your example is the way to go

answered Jan 17, 2014 at 3:27

inspectorG4dget

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Comments

finiteautomata · Accepted Answer · 2014-01-17 03:54:47Z

You should notice that an instance method is not a lambda.

For instance, lets do a simple experiment in IPython

In [12]: class A:
   ....:     def f(self):
   ....:         return 1
   ....:     

In [13]: A.f.__class__
Out[13]: instancemethod

In [14]: another_f = lambda self: 1

In [15]: another_f.__class__
Out[15]: function

Trying to bind an attribute to a lambda will fail miserably when calling it.

In [27]: an_instance = A()

In [28]: an_instance.g = lambda self: 2

In [29]: an_instance.g()
---------------------------------------------------------------------------
TypeError                                 Traceback (most recent call last)
<ipython-input-29-5122c91d3e8f> in <module>()
----> 1 an_instance.g()

TypeError: <lambda>() takes exactly 1 argument (0 given)

What you should do, instead, is wrapping the lambda with types.MethodType

In [31]: an_instance.g = types.MethodType(lambda self: 2, an_instance)

In [32]: an_instance.g()
Out[32]: 2

There is some weird magic happening behind called descriptors. In my opinion, this is not a quite OO solution, but... Well, if you want to know more about it, here is a link http://www.cafepy.com/article/python_attributes_and_methods/python_attributes_and_methods.html

vibhor · Accepted Answer · 2014-01-17 03:39:43Z

0

So I am assuming you want to know that how to define the function runtime.

May be something similar in the link create function dynamically

answered Jan 17, 2014 at 3:39

vibhor

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