Understanding Python Namespace with Dictionary: Why Is the Output Different?

Explore the behavior of `Python` namespaces with dictionaries and understand why scope matters in your code.
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Understanding Python Namespace with Dictionary: Why Is the Output Different?

As a novice or intermediate Python programmer, you might have encountered situations where the output of your functions was not what you expected. One common source of confusion lies in how Python handles namespaces, especially when it comes to dictionaries. In this guide, we'll delve into a specific Python scenario regarding dictionary scope and function output. We'll explain why the output of a function may not change even when a variable seems to be redefined in a nested function. Let's break it down!

The Scenario

Consider the following dictionary definition in Python:

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Next, we define a function, fun1, which prints the value associated with the key 'a' from the dictionary r:

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Output:

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This works perfectly, and we receive the expected output. However, when we create another function, fun2, we redefine the dictionary r within that function:

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Output:

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The Issue: Why Isn't the Output [10, 11, 12]?

You may be scratching your head at this point. Why does fun2 still print [1, 2, 3] even after redefining r inside it? The key to understanding this behavior lies in the concept of scope and how namespaces work in Python.

Understanding Scope

In Python, scope determines the visibility of variables. When r is defined outside of any function, it resides in the global namespace. The fun1 function accesses this global variable r, which is why it prints [1, 2, 3] within fun1 — it's referencing the original dictionary defined at the beginning.

When you define r inside fun2, you create a new, local variable r that only exists within the scope of fun2. This local r does not affect the r defined globally. Hence, when fun1() is called within fun2(), it still refers to the original r from the global namespace.

Solutions to Access the New Dictionary

If you want fun1 to print the updated dictionary from fun2, there are a couple of ways to achieve this:

1. Define fun1 Inside fun2

By defining fun1 inside fun2, you allow it access to the local r:

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2. Pass r as a Parameter

Passing r as a parameter to fun1 is another effective and straightforward solution:

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With these methods, fun1 will correctly use the local r, and you’ll see the output as [10, 11, 12].

Conclusion

Understanding how namespaces and scopes work in Python is fundamental to writing bug-free code. It's essential to grasp that when you define a variable in a function, it creates a new scope that does not affect the outer scopes unless explicitly stated. By defining functions appropriately or passing parameters, you can control which version of a variable your code uses.

Happy coding! If you have any questions or further dilemmas regarding Python programming, feel free to leave a comment below.