python self-less

this works in the desired way:

class d:
    def __init__(self,arg):
        self.a = arg
    def p(self):
        print "a= ",self.a

x = d(1)
y = d(2)
x.p()
y.p()

yielding

a=  1
a=  2

i've tried eliminating the "self"s and using a global statement in __init__

class d:
    def __init__(self,arg):
        global a
        a = arg
    def p(self):
        print "a= ",a

x = d(1)
y = d(2)
x.p()
y.p()

yielding, undesirably:

a=  2
a=  2

is there a way to write it without having to use "self"?

Solution

Python methods are just functions that are bound to the class or instance of a class. The only difference is that a method (aka bound function) expects the instance object as the first argument. Additionally when you invoke a method from an instance, it automatically passes the instance as the first argument. So by defining self in a method, you're telling it the namespace to work with.

This way when you specify self.a the method knows you're modifying the instance variable a that is part of the instance namespace.

Python scoping works from the inside out, so each function (or method) has its own namespace. If you create a variable a locally from within the method p (these names suck BTW), it is distinct from that of self.a. Example using your code:

class d:
    def __init__(self,arg):
        self.a = arg
    def p(self):
        a = self.a - 99
        print "my a= ", a
        print "instance a= ",self.a


x = d(1)
y = d(2)
x.p()
y.p()

Which yields:

my a=  -98
instance a=  1
my a=  -97
instance a=  2

Lastly, you don't have to call the first variable self. You could call it whatever you want, although you really shouldn't. It's convention to define and reference self from within methods, so if you care at all about other people reading your code without wanting to kill you, stick to the convention!