Why Does Modifying a Nested List Element in Python Affect All Its Copies?

Mary-Kate Olsen
Release: 2024-10-31 01:44:29
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Why Does Modifying a Nested List Element in Python Affect All Its Copies?

Nested List Indices: Interpreting Python's Reference-Based Behavior

In Python, lists are treated as mutable, reference-based data structures. This characteristic poses a potential pitfall while working with nested lists, as demonstrated by the following code snippet:

<code class="python">some_list = 4 * [(4 * [0])]</code>
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Creating a nested list like this creates four references to the same underlying list. As a result, any modification made to one of the references affects all the others due to their shared nature. This behavior is evident in the provided code, where the expected output:

<code class="python">[0, 0, 0, 0]
[0, 1, 1, 1]
[0, 1, 1, 1]
[0, 1, 1, 1]</code>
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Conflicts with the actual output:

<code class="python">[0, 1, 1, 1]
[0, 1, 1, 1]
[0, 1, 1, 1]
[0, 1, 1, 1]</code>
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To avoid this gotcha, it's recommended to create a new list instance for each sublist using a comprehension approach:

<code class="python">some_list = [(4 * [0]) for _ in range(4)]</code>
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This approach ensures that each sublist is independent, resolving the issue and producing the intended output.

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