Are Python Lists mutable?

Question

When I type following code,

x=[1,2,4]
print(x)
print("x",id(x))
x=[2,5,3]
print(x)
print("x",id(x))

it gives the output as

[1, 2, 4]
x 47606160
[2, 5, 3]
x 47578768

If lists are mutable then why it give 2 memory address when changing the list x?

Answer 1

You did not mutate (change) the list object referenced by x with this line:

x=[2,5,3]

Instead, that line creates a new list object and then reassigns the variable x to it. So, x now references the new object and id(x) gives a different number than before:

>>> x=[1,2,4]  # x references the list object [1,2,4]
>>> x
[1, 2, 4]
>>> x=[2,5,3]  # x now references an entirely new list object [2,5,3]
>>> x
[2, 5, 3]
>>>

Answer 2

You created a new list object and bound it to the same name, x. You never mutated the existing list object bound to x at the start.

Names in Python are just references. Assignment is binding a name to an object. When you assign to x again, you are pointing that reference to a different object. In your code, you simply created a whole new list object, then rebound x to that new object.

If you want to mutate a list, call methods on that object:

x.append(2)
x.extend([2, 3, 5])

or assign to indices or slices of the list:

x[2] = 42
x[:3] = [5, 6, 7]

Demo:

>>> x = [1, 2, 3]
>>> id(x)
4301563088
>>> x
[1, 2, 3]
>>> x[:2] = [42, 81]
>>> x
[42, 81, 3]
>>> id(x)
4301563088

We changed the list object (mutated it), but the id() of that list object did not change. It is still the same list object.

Perhaps this excellent presentation by Ned Batchelder on Python names and binding can help: Facts and myths about Python names and values.

Answer 3

You are not mutating the list, you are creating a new list and assigning it to the name x. That's why id is giving you different outputs. Your first list is gone and will be garbage-collected (unless there's another reference to it somewhere).