Consider these two functions:
---
def nok():
    a = None
    def f():
        if a:
            a = 1
    f()

def ok():
    a = None
    def f():
        if a:
            b = 1
    f()
---

Function ok() executes fine, but function nok() trigger an exception:

Traceback (most recent call last):
  File "pb.py", line 20, in <module>
    nok()
  File "pb.py", line 7, in nok
    f()
  File "pb.py", line 5, in f
    if a:
UnboundLocalError: local variable 'a' referenced before assignment

There is no reason for this to happen

Regards,
Rodrigo Ventura

The reason is that in nok Python sees the assignment to a (a = 1) and determines that the 'a' variable is local to the scope of f, and since the assignment comes after the "if a:" and at that point 'a' has no value, an error is raised.
In ok there's no assignment to 'a', so Python assume that 'a' refers to the 'a' variable defined in the outer scope.

See also http://docs.python.org/faq/programming.html#why-am-i-getting-an-unboundlocalerror-when-the-variable-has-a-value

Ezio,

thank you for the explanation.

Is it possible to access variable a in nok's scope from function f without using the global keyword in f? (so that variable a remains local to nok, rather than global to python)

Rodrigo

In 3.x, yes (the nonlocal keyword).  Not in 2.7, though.