How to call nested function independently in Python script

Question

In the below Python code, I am trying to call the nested function (at step2) independently but getting an error. Condition is step1 should execute before step2, this is to ensure that c1 gets the same value of p1

import datetime

def neighbor():

    n1 = datetime.datetime.now()
    print("n1",n1)
    return n1

def parent():

    p1 = neighbor()
    print("p1",p1)

    def child():
        nonlocal p1
        print("c1",p1)


parent()    # step1

child()     # step2     

Answer 1

The nested function is only available to the function in whose scope you defined it, just like if you declare a local variable inside a function (and indeed, a function is really just a variable whose value is a "callable").

One way to enforce this is to put the definitions in a class.

class Parent:
    def __init__(self):
        self.p1 = neighbor()

    def child(self):
        print("c1", self.p1)

Now, you will be able to call the child method from outside the class scope, but only when you have created a Parent object instance. (The __init__ method gets called when you create the instance.)

instance = Parent()  # create instance
... # maybe other code here
instance.child()

An additional benefit is that you can have multiple Parent instances if you want to; each has its own state (the set of attributes and their values, like here the p1 attribute which you access via instance.p1 or inside the class via self.p1 ).

Learning object-oriented programming concepts doesn't end here, but this is actually a pretty simple and gentle way to get started.