Why Does Modifying a Nested List in Python Lead to Unexpected Results?

Nested List Indices Revisited

In Python, nested lists offer a convenient way to organize data. However, working with indices for these lists can sometimes lead to unexpected results. Let's explore a common issue related to nested list indices with a specific code example:

Consider the following code snippet:

<code class="python">some_list = 4 * [(4 * [0])]
for i in range(3):
    for j in range(3):
        some_list[i+1][j+1] = 1
for i in range(4):
    print(some_list[i])</code>

The intended objective is to create a 2D list where the values at each index (excluding the first row and first column) are set to 1. However, the code fails to achieve this. Instead, it produces a list where all rows and columns are filled with 1s.

Explanation:

The root cause of this issue lies in Python's method of passing lists by reference. Unlike basic value types, lists are passed around as references to the same memory location. This means that any changes made to one list variable will be reflected in all other variables referencing the same list.

In the given code, the line some_list = 4 * [(4 * [0])] multiplies a list containing a single 0 value four times. However, it does not create four distinct lists. Instead, it creates a single list with four references pointing to the same 0 value.

As a result, when the code updates the value at [i 1][j 1] for the first iteration of the loop, it modifies the value that all other references point to. This explains why all elements end up with the value 1.

Solution:

To resolve this issue and create four distinct lists, each with its own set of values, it's necessary to avoid the second multiplication. One way to do this is to use a loop to initialize each row of the list separately:

<code class="python">some_list = [(4 * [0]) for _ in range(4)]</code>

By looping over range(4), this code creates four distinct lists, each with a separate set of values. As a result, subsequent modifications to the list values will not affect other rows.

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