Adding method to lambda in Python

Question

I want to do something like this in python:

(lambda: ... some code ...).valueHandling(...some parameters)

That is, I want to extend the type function with a new method called valueHandling. I tried different thing like:

1)

def valueHandling(self):
  ...

function.valueHandling=valueHandling

2)

function.__class__.__dict__['valueHandling']=valueHandling

3) etc

None worked... Is there a way to do it? To add a method to function so all 'lambdas' can respond to that method?

Thanks

I'm really confused as to what you want to happen here. What does *respond to* mean in your final sentence? — Gareth Latty, Oct 16 '12 at 23:00
@Lattyware he wants to add a new method to all function objects — Johanna Larsson, Oct 16 '12 at 23:01
You can't extend function in a way that affects the core `function` type. See this: http://stackoverflow.com/questions/192649/can-you-monkey-patch-methods-on-core-types-in-python — Ian Clelland, Oct 16 '12 at 23:06
@JoranBeasley He wants to alter all `lambda`s so that they all have a `.valueHandling()` method/function — TankorSmash, Oct 16 '12 at 23:06
could you write it as: `valueHandling(lambda x,y,z: ... some code ..., ...some other parameters)` — jfs, Oct 16 '12 at 23:43
`self` parameter is only for unbound methods (methods defined on the class). You are setting an attribute directly on the object. It should not have a `self` parameter — newacct, Oct 17 '12 at 06:16

score 0 · Answer 1 · answered Oct 16 '12 at 23:05

CPython doesn't allow this:

TypeError: can't set attributes of built-in/extension type 'function'

(and if you try to use the .__dict__ method...)

TypeError: 'dictproxy' object does not support item assignment

This is because these types are implemented in C and trade flexibility for speed.

Karoly Horvath · Answer 2 · 2012-10-16T23:23:40.080

0

You can set properties of individual (function/lambda/etc..) objects:

l = lambda x: x*2
def fun():
  return 1
l.f = fun
l.f()
1

You cannot do this for all functions - you have to manually add it to each of them.

You could write a short wrapper, and then mylambda(lambda x: ...) :

def mylambda(x):
  x.f = fun
  return x

edited Oct 16 '12 at 23:23

answered Oct 16 '12 at 23:17

Karoly Horvath

Thank you all! It is a pity Python does not allow you to add functionality to core classes like function... I wanted to do that to implement exception just using objects and without the try/except sintax. I just did it as Karoly suggested. Thanks! – Hernan Oct 17 '12 at 02:27
From the documentation: ""Function objects also support getting and setting arbitrary attributes, which can be used, for example, to attach metadata to functions. Regular attribute dot-notation is used to get and set such attributes. Note that the current implementation only supports function attributes on user-defined functions. Function attributes on built-in functions may be supported in the future"" – Bakuriu Oct 17 '12 at 13:05

2 Answers2