Is it possible to modify variable in python that is in outer, but not global, scope?

Question

Given following code:

def A() :
    b = 1

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


        

Answer 1

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 Cédric 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

Answer 2

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

user@pc: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!

Answer 3

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()