Python share global variable only for functions inside of function

Question

I have a function which will recursively execute another function inside and I want to share variable for all execution of that function.

Something like that:

def testglobal():
  x = 0
  def incx():
    global x
    x += 2
  incx()
  return x
testglobal() # should return 2

However, I'm getting error NameError: name 'x' is not defined

There is hacky solution to make list and use first value of that list as x. But this is so ugly.

So how can I share x with incx function ? Or should I use completely different approach ?

In python 3 there is a new keyword, nonlocal, that does exactly what you want. Keep in mind that this is a closure so you have access to x without alterations, but assigning (e.g. x = 1) inside incx will make x local to incx and therefore not refer to the same variable. nonlocal achieves this. — alexpeits
– alexpeits, Commented Sep 23, 2016 at 14:28

Duncan · Accepted Answer · 2016-09-23 14:27:34Z

3

This will work unless you are still using Python 2.x:

def testglobal():
  x = 0
  def incx():
    nonlocal x
    x += 2
  incx()
  return x

testglobal() # should return 2

Possible a cleaner solution though would be to define a class to store your state between method calls.

answered Sep 23, 2016 at 14:27

Duncan

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Comments

linusg · Accepted Answer · 2016-09-23 14:28:28Z

2

Use the nonlocal statement, so incx will use the x variable from testglobal:

def testglobal():
    x = 0
    def incx():
        nonlocal x
        x += 2
    incx()
    return x

testglobal()

answered Sep 23, 2016 at 14:28

linusg

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Comments

chepner · Accepted Answer · 2016-09-23 15:00:22Z

1

You want to use the nonlocal statement to access x, which is not global but local to testglobal.

def testglobal():
  x = 0
  def incx():
    nonlocal x
    x += 2
  incx()
  return x
assert 2 == testglobal()

The closest you can come to doing this in Python 2 is to replace x with a mutable value, similar to the argument hack you mentioned in your question.

def testglobal():
  x = [0]
  def incx():
    x[0] += 2
  incx()
  return x[0]
assert 2 == testglobal()

Here's an example using a function attribute instead of a list, an alternative that you might find more attractive.

def testglobal():
  def incx():
    incx.x += 2
  incx.x = 0
  incx()
  return inc.x
assert 2 == testglobal()

edited Sep 23, 2016 at 15:00

answered Sep 23, 2016 at 14:27

chepner

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2 Comments

Oles Savluk Over a year ago

But this was introduced only in python 3, right ? But this concept looks so common for, how to do the same in python 2 ?

chepner Over a year ago

You can't, short of your mutable-argument hack. Closures only capture the value of a variable for reading; you can't modify the value nonlocally (which is why Python 3 added the nonlocal keyword).

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