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

Nested List Indices

When working with nested lists in Python, it's important to understand the concept of list references and how it affects list indices.

List References

Unlike many programming languages that pass variables by value, Python passes lists by reference. This means that assigning a list to another variable creates a reference to the same underlying list object in memory.

Nested List Indices

In the code provided, the problem arises from the use of list multiplication. The line some_list = 4 * [(4 * [0])] creates four references to the same list object, meaning that changes to one of the sublists will affect all four.

Expected Results

The expected results are:

[0, 0, 0, 0]
[0, 1, 1, 1]
[0, 1, 1, 1]
[0, 1, 1, 1]

Actual Results

The actual results are:

[0, 1, 1, 1]
[0, 1, 1, 1]
[0, 1, 1, 1]
[0, 1, 1, 1]

Solution

To avoid this issue, use a loop to create a nested list without creating multiple references to the same list. The corrected code is:

some_list = [(4 * [0]) for _ in range(4)]

This loop creates four independent lists, each with four zeros, ensuring that the expected behavior is achieved.

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