Скачать или смотреть Understanding the Mystery of Dictionaries and For Loops in Python 3

Discover why you see unexpected changes in your Python dictionary when using loops and how to properly manage references in lists.
---
This video is based on the question https://stackoverflow.com/q/62739017/ asked by the user 'ojacomarket' ( https://stackoverflow.com/u/13553379/ ) and on the answer https://stackoverflow.com/a/62739113/ provided by the user 'Leo Arad' ( https://stackoverflow.com/u/13101239/ ) at 'Stack Overflow' website. Thanks to these great users and Stackexchange community for their contributions.

Visit these links for original content and any more details, such as alternate solutions, latest updates/developments on topic, comments, revision history etc. For example, the original title of the Question was: Mystery of Dictionaries and For loop in Python 3

Also, Content (except music) licensed under CC BY-SA https://meta.stackexchange.com/help/l...
The original Question post is licensed under the 'CC BY-SA 4.0' ( https://creativecommons.org/licenses/... ) license, and the original Answer post is licensed under the 'CC BY-SA 4.0' ( https://creativecommons.org/licenses/... ) license.

If anything seems off to you, please feel free to write me at vlogize [AT] gmail [DOT] com.
---
Understanding the Mystery of Dictionaries and For Loops in Python 3

Python is a powerful programming language known for its simplicity and versatility. However, as you delve deeper, you may encounter some unexpected behavior, especially when working with dictionaries and loops. In this guide, we will explore the nuances of Python dictionaries, specifically how they behave when used in loops, to help clear up any confusion you might have.

The Problem: Unexpected Modifications in Lists

Imagine you have a dictionary representing a dog, stored as follows:

[[See Video to Reveal this Text or Code Snippet]]

You decide to create a list containing three instances of this dog. You write the following code:

[[See Video to Reveal this Text or Code Snippet]]

You might expect that modifying the first dog's color later on would only change that specific dog. However, when you run this code:

[[See Video to Reveal this Text or Code Snippet]]

You get this output:

[[See Video to Reveal this Text or Code Snippet]]

This result is due to a common pitfall in Python: reference sharing in mutable objects. Let’s break down why this happens.

Understanding the Behavior: Reference vs. Object

Single Reference to a Dictionary

In your first code snippet, when you append dog to the dogs list three times:

[[See Video to Reveal this Text or Code Snippet]]

All three entries in the dogs list are referencing the same dictionary object. Any change to that dictionary, like updating the color to 'white', reflects across all references in the list. This means when you’re modifying one dog, all three dogs seem to change.

Creating New Objects in a Loop

To achieve the desired outcome of having three distinct dog dictionaries, consider rewriting your code as follows:

[[See Video to Reveal this Text or Code Snippet]]

In this solution, you're creating a new dictionary object on each iteration of the loop. Now, when you change the color of the first dog, only that specific dictionary is modified:

[[See Video to Reveal this Text or Code Snippet]]

Conclusion

To summarize, the difference in behavior when working with lists of dictionaries in Python stems from how references work with mutable types. Understanding this will help you avoid unexpected changes in your code. Here’s what you need to remember:

Reference Sharing: Appending the same dictionary object results in all entries in the list reflecting changes to that single object.

Creating New Objects: Constructing a new dictionary within the loop ensures that each entry in the list is independent, allowing specific modifications.

By grasping these concepts, you can confidently work with dictionaries and loops in Python without running into these common issues. Happy coding!