Why does assigning a list to a new variable in Python not create a separate copy?

Unexpected List Mutation

When creating a list like v = [0,0,0,0,0,0,0,0,0], you might assume that assigning a new list to a variable creates a separate reference. However, code like the following can demonstrate unexpected behavior:

<code class="python">vec = v
vec[5] = 5</code>

Both vec and v now contain the value 5 at the index 5. Why does this happen?

Reference Assignment

In Python, lists are mutable objects. Assigning vec = v does not create a new copy of the list. Instead, it assigns a reference to v. Both vec and v point to the same underlying list object in memory.

Any modifications made to either vec or v will affect the original list because they are the same list. This is why when vec[5] is changed, v also changes.

Solution

To create a separate copy of the list, use the list() function:

<code class="python">vec = list(v)</code>

This creates a new list object that contains a copy of the elements from v. Any changes made to vec will not affect v, and vice versa.

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