NameSpaces in Python (2/2)

Please check part-1 to get an understanding on the available scopes in python. Now let’s check how these various scopes work together in tandem

Global Scope vs Local Scope

Fig 1. A global variable ‘a’ accessed inside the function f1()

Since ‘a’ can be accessed inside the function, lets change its value

Fig 2. Error in accessing ‘global’ variable ‘a’ inside the function f1()

As per the Global scope rules, variable ‘a’ should be accessible inside the function f1(). Infact we are able to print its value in this snippet (Fig 1). However the moment, there is a new value assigned to the variable, python compiler throws ‘local variable referenced before assignment’

This is happening because scope hierarchy is coupled with assignment rules

The moment we try to assign a value to ‘a’ inside the function, Python interpreter treats it as ‘local’ variable and since its previous value is not available (Python: No seperate declaration and only value assignment is available), it throws error

How to fix this? Use ‘global’ keyword to declare the variable inside a function

Global variable ‘a’ accessed and assigned to a new value inside the function

In the above code snippet, the value of ‘a’ is changed inside the function

Creating a new global variable inside a function

Python also gives provision to create a new global variable from inside a function. ‘global’ keyword can be used for this purpose as well

Initially there is no variable named ‘b’ available in global scope

Created a global variable inside the function using ‘global’ keyword

Now accessing ‘b’ would work without error

Local vs Enclosed Space

Understanding local, enclosed spaces

In the above code snippet:

func1(): ‘a’ is a local variable

nest1(): Python interpreter sees the assignment of ‘a’ without any tag. Which makes it to treat it ‘a’ as local variable for the function nest1() than treating it as the variable in the enclosed space. That is the reason even after calling nest1(), value of ‘a’ in func1() is still 10, rather than 20

How to change this behavior?

Accessing and Updating Variables inside nesting functions

By adding the ‘nonlocal’ keyword before ‘a’ enables python interpreter to search in the enclosing space rather than the function local space

Can we create a new nonlocal variable (similar to global variable creation)?

Creating a non-local variable inside nested function throws error

Unlike ‘global’ keyword, ‘nonlocal’ keyword cannot be used to create new non-local variables. This keyword can only be used to access the already available variables in the enclosing space. Why?

That’s how Python defined non-local and global :)

Global vs Enclosed

‘global’ keyword in enclosing space doesn’t impact the operations inside nested function

In the above code snippet, ‘global’ scope for ‘a’ is defined in func1(). When nest1() encounters assignment of ‘a’, it will be treated as local variable for the function nest1() and its scope will be over once the control exits this function. Hence the reason for ‘a’ value not getting updated. How to fix this?

Use ‘global’ keyword inside the nested function rather than in the enclosing space

Global vs Enclosed vs Local Space

Using ‘global’ word inside functions to modify its value

Over-riding built-in scope

Over-riding built in scope

What can be tried further?

1. Inside func1(): Try using ‘nonlocal’ rather than ‘global’ and check the behaviour
2. Inside func1(): Create a local variable with same name as global variable and try to modify its value inside nested function using ‘global’, ‘nonlocal’ keywords
3. How to get back the built-in functionality after over-riding

Please also let me know, if you find any other behaviours apart from the things we have discussed in the two parts. ~~