nonlocal keyword in Python 2.x

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Question :

nonlocal keyword in Python 2.x

I’m trying to implement a closure in Python 2.6 and I need to access a nonlocal variable but it seems like this keyword is not available in python 2.x. How should one access nonlocal variables in closures in these versions of python?

Asked By: adinsa

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Answer #1:

Inner functions can read nonlocal variables in 2.x, just not rebind them. This is annoying, but you can work around it. Just create a dictionary, and store your data as elements therein. Inner functions are not prohibited from mutating the objects that nonlocal variables refer to.

To use the example from Wikipedia:

def outer():
    d = {'y' : 0}
    def inner():
        d['y'] += 1
        return d['y']
    return inner

f = outer()
print(f(), f(), f()) #prints 1 2 3
Answered By: Chris B.

Answer #2:

The following solution is inspired by the answer by Elias Zamaria, but contrary to that answer does handle multiple calls of the outer function correctly. The “variable” inner.y is local to the current call of outer. Only it isn’t a variable, since that is forbidden, but an object attribute (the object being the function inner itself). This is very ugly (note that the attribute can only be created after the inner function is defined) but seems effective.

def outer():
    def inner():
        inner.y += 1
        return inner.y
    inner.y = 0
    return inner

f = outer()
g = outer()
print(f(), f(), g(), f(), g()) #prints (1, 2, 1, 3, 2)
Answered By: Marc van Leeuwen

Answer #3:

Rather than a dictionary, there’s less clutter to a nonlocal class. Modifying @ChrisB’s example:

def outer():
    class context:
        y = 0
    def inner():
        context.y += 1
        return context.y
    return inner

Then

f = outer()
assert f() == 1
assert f() == 2
assert f() == 3
assert f() == 4

Each outer() call creates a new and distinct class called context (not merely a new instance). So it avoids @Nathaniel’s beware about shared context.

g = outer()
assert g() == 1
assert g() == 2

assert f() == 5
Answered By: Bob Stein

Answer #4:

I think the key here is what you mean by “access”. There should be no issue with reading a variable outside of the closure scope, e.g.,

x = 3
def outer():
    def inner():
        print x
    inner()
outer()

should work as expected (printing 3). However, overriding the value of x does not work, e.g.,

x = 3
def outer():
    def inner():
        x = 5
    inner()
outer()
print x

will still print 3. From my understanding of PEP-3104 this is what the nonlocal keyword is meant to cover. As mentioned in the PEP, you can use a class to accomplish the same thing (kind of messy):

class Namespace(object): pass
ns = Namespace()
ns.x = 3
def outer():
    def inner():
        ns.x = 5
    inner()
outer()
print ns.x
Answered By: ig0774

Answer #5:

There is another way to implement nonlocal variables in Python 2, in case any of the answers here are undesirable for whatever reason:

def outer():
    outer.y = 0
    def inner():
        outer.y += 1
        return outer.y
    return inner

f = outer()
print(f(), f(), f()) #prints 1 2 3

It is redundant to use the name of the function in the assignment statement of the variable, but it looks simpler and cleaner to me than putting the variable in a dictionary. The value is remembered from one call to another, just like in Chris B.’s answer.

Answered By: Elias Zamaria

Answer #6:

Here’s something inspired by a suggestion Alois Mahdal made in a comment regarding another answer:

class Nonlocal(object):
    """ Helper to implement nonlocal names in Python 2.x """
    def __init__(self, **kwargs):
        self.__dict__.update(kwargs)


def outer():
    nl = Nonlocal(y=0)
    def inner():
        nl.y += 1
        return nl.y
    return inner

f = outer()
print(f(), f(), f()) # -> (1 2 3)

Update

After looking back at this recently, I was struck by how decorator-like it was—when it dawned on me that implementing it as one would make it more generic & useful (although doing so arguably degrades its readability to some degree).

# Implemented as a decorator.

class Nonlocal(object):
    """ Decorator class to help implement nonlocal names in Python 2.x """
    def __init__(self, **kwargs):
        self._vars = kwargs

    def __call__(self, func):
        for k, v in self._vars.items():
            setattr(func, k, v)
        return func


@Nonlocal(y=0)
def outer():
    def inner():
        outer.y += 1
        return outer.y
    return inner


f = outer()
print(f(), f(), f()) # -> (1 2 3)

Note that both versions work in both Python 2 and 3.

Answered By: martineau

Answer #7:

There is a wart in python’s scoping rules – assignment makes a variable local to its immediately enclosing function scope. For a global variable, you would solve this with the global keyword.

The solution is to introduce an object which is shared between the two scopes, which contains mutable variables, but is itself referenced through a variable which is not assigned.

def outer(v):
    def inner(container = [v]):
        container[0] += 1
        return container[0]
    return inner

An alternative is some scopes hackery:

def outer(v):
    def inner(varname = 'v', scope = locals()):
        scope[varname] += 1
        return scope[varname]
    return inner

You might be able to figure out some trickery to get the name of the parameter to outer, and then pass it as varname, but without relying on the name outer you would like need to use a Y combinator.

Answered By: Marcin

Answer #8:

Another way to do it (although it’s too verbose):

import ctypes

def outer():
    y = 0
    def inner():
        ctypes.pythonapi.PyCell_Set(id(inner.func_closure[0]), id(y + 1))
        return y
    return inner

x = outer()
x()
>> 1
x()
>> 2
y = outer()
y()
>> 1
x()
>> 3
Answered By: Ezer Fernandes

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