142

In the 2nd case below, Python tries to look for a local variable. When it doesn't find one, why can't it look in the outer scope like it does for the 1st case?

This looks for x in the local scope, then outer scope:

def f1():
    x = 5
    def f2():
         print x

This gives local variable 'x' referenced before assignment error:

def f1():
    x = 5
    def f2():
        x+=1

I am not allowed to modify the signature of function f2() so I can not pass and return values of x. However, I do need a way to modify x. Is there a way to explicitly tell Python to look for a variable name in the outer scope (something similar to the global keyword)?

Python version: 2.7

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  • 5
    python 3 provides nonlocal for this purpose, but I think it's worth asking why you need to do this. This is a little safer than using global, but still doesn't feel right. Commented Aug 16, 2012 at 12:52
  • 4
    @mgilson there are many good reasons to do this, see e.g. the rationale section in pep-3104. Commented Aug 16, 2012 at 12:58
  • what version of python do you use? Commented Aug 16, 2012 at 13:00
  • @soulcheck I use Python 2.7, edited that in my question. Commented Aug 16, 2012 at 13:02
  • @Dhara sorry, don't know why the comment got posted. It was fairly obvious that you're on 2.7. Commented Aug 16, 2012 at 13:06

2 Answers 2

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208

In Python 3.x this is possible:

def f1():
        x = 5
        def f2():
                nonlocal x
                x+=1
        return f2

The problem and a solution to it, for Python 2.x as well, are given in this post. Additionally, please read PEP 3104 for more information on this subject.

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2 Comments

I just looked this up. nonlocal is the statement to use here.
@user1320237 global does not work instead of nonlocal -- that was precisely my question!
84 
def f1():
    x = { 'value': 5 }
    def f2():
        x['value'] += 1

Workaround is to use a mutable object and update members of that object. Name binding is tricky in Python, sometimes.

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4 Comments

This works, although I ended up using a list with a single element instead of a dictionary. Thanks!
I think it is more correct to use nonlocal or global
Global is not a good idea as it will pollute the global namespace. Nonlocal is a python 3 feature, and this question specified python 2.7.
global doesn't work in py2.7 for question purpose, however x['value'] is harder to write on parent scope than just value, so consider passing value to nested function, and then just returning value from nested scope to parent scope

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