Question

def method(self, *args):
   def function(*args):
      #can this access all method's variables, data, etc.

Answer 1

Yes you can,because python follows the following look-up rule while finding a variable:

LEGB:

L:local
E:enclosing
G:global
B:built-in

So, in your case it’s E:

python 2.x:

in python 2.x you can’t modify those variables in the func

class A:
    def meth(self):
        foo=1
        bar=2
        def func():
            foo=2     # if you place this statement below the print statement then you'll get
                      # UnboundLocalError: local variable 'foo' referenced before assignment
            print foo,bar
        func()    
        print (foo) #meth's foo is unchanged
a=A()
a.meth()

output:

2 2 
1

python 3.x:
use nonlocal to even modify the variables:

class A:
    def meth(self):
        foo=1
        bar=2
        def func():
            nonlocal foo,bar             
            print (foo,bar)
            foo=2               #changes meth's foo to 2
        func()    
        print (foo)
a=A()
a.meth()

output:

1 2
2