Is it possible to modify a variable in python that is in an outer (enclosing), but not global, scope?

Question:

Consider this example:

def A():
    b = 1
    def B():
        # I can access 'b' from here.
        print(b)
        # But can i modify 'b' here?
    B()
A()

For the code in the B function, the variable b is in a non-global, enclosing (outer) scope. How can I modify b from within B? I get an UnboundLocalError if I try it directly, and using global does not fix the problem since b is not global.


Python implements lexical, not dynamic scope – like almost all modern languages. The techniques here will not allow access to the caller’s variables – unless the caller also happens to be an enclosing function – because the caller is not in scope. For more on this problem, see How can I access variables from the caller, even if it isn't an enclosing scope (i.e., implement dynamic scoping)?.

Asked By: grigoryvp

||

Answers:

No you cannot, at least in this way.

Because the “set operation” will create a new name in the current scope, which covers the outer one.

Answered By: zchenah

On Python 3, use the nonlocal keyword:

The nonlocal statement causes the listed identifiers to refer to previously bound variables in the nearest enclosing scope excluding globals. This is important because the default behavior for binding is to search the local namespace first. The statement allows encapsulated code to rebind variables outside of the local scope besides the global (module) scope.

def foo():
    a = 1
    def bar():
        nonlocal a
        a = 2
    bar()
    print(a)  # Output: 2

On Python 2, use a mutable object (like a list, or dict) and mutate the value instead of reassigning a variable:

def foo():
    a = []
    def bar():
        a.append(1)
    bar()
    bar()
    print a

foo()

Outputs:

[1, 1]
Answered By: Adam Wagner

I’m a little new to Python, but I’ve read a bit about this. I believe the best you’re going to get is similar to the Java work-around, which is to wrap your outer variable in a list.

def A():
   b = [1]
   def B():
      b[0] = 2
   B()
   print(b[0])

# The output is '2'

Edit: I guess this was probably true before Python 3. Looks like nonlocal is your answer.

Answered By: Mike Edwards

I don’t think you should want to do this. Functions that can alter things in their enclosing context are dangerous, as that context may be written without the knowledge of the function.

You could make it explicit, either by making B a public method and C a private method in a class (the best way probably); or by using a mutable type such as a list and passing it explicitly to C:

def A():
    x = [0]
    def B(var): 
        var[0] = 1
    B(x)
    print x

A()
Answered By: Sideshow Bob

You can, but you’ll have to use the global statment (not a really good solution as always when using global variables, but it works):

def A():
    global b
    b = 1

    def B():
      global b
      print( b )
      b = 2

    B()
A()
Answered By: Cédric Julien

I don’t know if there is an attribute of a function that gives the __dict__ of the outer space of the function when this outer space isn’t the global space == the module, which is the case when the function is a nested function, in Python 3.

But in Python 2, as far as I know, there isn’t such an attribute.

So the only possibilities to do what you want is:

1) using a mutable object, as said by others

2)

def A() :
    b = 1
    print 'b before B() ==', b

    def B() :
        b = 10
        print 'b ==', b
        return b

    b = B()
    print 'b after B() ==', b

A()

result

b before B() == 1
b == 10
b after B() == 10

.

Nota

The solution of C├ędric Julien has a drawback:

def A() :
    global b # N1
    b = 1
    print '   b in function B before executing C() :', b

    def B() :
        global b # N2
        print '     b in function B before assigning b = 2 :', b
        b = 2
        print '     b in function B after  assigning b = 2 :', b

    B()
    print '   b in function A , after execution of B()', b

b = 450
print 'global b , before execution of A() :', b
A()
print 'global b , after execution of A() :', b

result

global b , before execution of A() : 450
   b in function B before executing B() : 1
     b in function B before assigning b = 2 : 1
     b in function B after  assigning b = 2 : 2
   b in function A , after execution of B() 2
global b , after execution of A() : 2

The global b after execution of A() has been modified and it may be not whished so

That’s the case only if there is an object with identifier b in the global namespace

Answered By: eyquem

You can use an empty class to hold a temporary scope. It’s like the mutable but a bit prettier.

def outer_fn():
   class FnScope:
     b = 5
     c = 6
   def inner_fn():
      FnScope.b += 1
      FnScope.c += FnScope.b
   inner_fn()
   inner_fn()
   inner_fn()

This yields the following interactive output:

>>> outer_fn()
8 27
>>> fs = FnScope()
NameError: name 'FnScope' is not defined
Answered By: chrisk

For anyone looking at this much later on a safer but heavier workaround is. Without a need to pass variables as parameters.

def outer():
    a = [1]
    def inner(a=a):
        a[0] += 1
    inner()
    return a[0]
Answered By: Michael Giba

The short answer that will just work automagically

I created a python library for solving this specific problem. It is released under the unlisence so use it however you wish. You can install it with pip install seapie or check out the home page here https://github.com/hirsimaki-markus/SEAPIE

[email protected]:home$ pip install seapie

from seapie import Seapie as seapie
def A():
    b = 1

    def B():
        seapie(1, "b=2")
        print(b)

    B()
A()

outputs

2

the arguments have following meaning:

  • The first argument is execution scope. 0 would mean local B(), 1 means parent A() and 2 would mean grandparent <module> aka global
  • The second argument is a string or code object you want to execute in the given scope
  • You can also call it without arguments for interactive shell inside your program

The long answer

This is more complicated. Seapie works by editing the frames in call stack using CPython api. CPython is the de facto standard so most people don’t have to worry about it.

The magic words you are probably most likely interesed in if you are reading this are the following:

frame = sys._getframe(1)          # 1 stands for previous frame
parent_locals = frame.f_locals    # true dictionary of parent locals
parent_globals = frame.f_globals  # true dictionary of parent globals

exec(codeblock, parent_globals, parent_locals)

ctypes.pythonapi.PyFrame_LocalsToFast(ctypes.py_object(frame),ctypes.c_int(1))
# the magic value 1 stands for ability to introduce new variables. 0 for update-only

The latter will force updates to pass into local scope. local scopes are however optimized differently than global scope so intoducing new objects has some problems when you try to call them directly if they are not initialized in any way. I will copy few ways to circumvent these problems from the github page

  • Assingn, import and define your objects beforehand
  • Assingn placeholder to your objects beforehand
  • Reassign object to itself in main program to update symbol table: x = locals()[“x”]
  • Use exec() in main program instead of directly calling to avoid optimization. Instead of calling x do: exec(“x”)

If you are feeling that using exec() is not something you want to go with you can
emulate the behaviour by updating the the true local dictionary (not the one returned by locals()). I will copy an example from https://faster-cpython.readthedocs.io/mutable.html

import sys
import ctypes

def hack():
    # Get the frame object of the caller
    frame = sys._getframe(1)
    frame.f_locals['x'] = "hack!"
    # Force an update of locals array from locals dict
    ctypes.pythonapi.PyFrame_LocalsToFast(ctypes.py_object(frame),
                                          ctypes.c_int(0))

def func():
    x = 1
    hack()
    print(x)

func()

Output:

hack!
Answered By: Markus Hirsimäki
Categories: questions Tags: ,
Answers are sorted by their score. The answer accepted by the question owner as the best is marked with
at the top-right corner.