Python and the self parameter

Posted by Svend on Stack Overflow See other posts from Stack Overflow or by Svend
Published on 2010-12-31T01:56:50Z Indexed on 2010/12/31 2:54 UTC
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I'm having some issues with the self parameter, and some seemingly inconsistent behavior in Python is annoying me, so I figure I better ask some people in the know. I have a class, Foo. This class will have a bunch of methods, m1, through mN. For some of these, I will use a standard definition, like in the case of m1 below. But for others, it's more convinient to just assign the method name directly, like I've done with m2 and m3.

import os

def myfun(x, y):
    return x + y

class Foo():
    def m1(self, y, z):
        return y + z + 42

    m2 = os.access
    m3 = myfun

f = Foo()
print f.m1(1, 2)
print f.m2("/", os.R_OK)
print f.m3(3, 4)

Now, I know that os.access does not take a self parameter (seemingly). And it still has no issues with this type of assignment. However, I cannot do the same for my own modules (imagine myfun defined off in mymodule.myfun). Running the above code yields the following output:

3
True
Traceback (most recent call last):
  File "foo.py", line 16, in <module>
    print f.m3(3, 4)
TypeError: myfun() takes exactly 2 arguments (3 given)

The problem is that, due to the framework I work in, I cannot avoid having a class Foo at least. But I'd like to avoid having my mymodule stuff in a dummy class. In order to do this, I need to do something ala

def m3(self,a1, a2):
    return mymodule.myfun(a1,a2)

Which is hugely redundant when you have like 20 of them. So, the question is, either how do I do this in a totally different and obviously much smarter way, or how can I make my own modules behave like the built-in ones, so it does not complain about receiving 1 argument too many.

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