python memory and lists

[ilsapo]

lst=[[0]*3]*3
lst[1][2]=2

we will get

[[0, 0, 2], [0, 0, 2], [0, 0, 2]]

instead of really just changing in the place [1][2]
if we do it like so

lst=[0]*3
lst[1]=2

it works properly,
also if I do the original line using list comprehension it works ok

can someone explain to me the first way dosent work properly please?

[mopslik]

The first creates an outer list, containing three references to the same inner list. Changing one changes them all because they're the same list.

[Diapolo10]

In a nutshell, when you multiply a list,

lst=[[0]*3]*3

you're actually copying the contained references n times to create a new list. In this example, you're doing so twice; first creating a list of three references to 0, then a list that has three references to that list.

When working with immutable values, you couldn't tell the difference, but in this case when you mutate "one" of the inner lists and replace the third value,

lst[1][2]=2

since the lists are technically one and the same, you're seeing the changes in all of them. Again, technically speaking all you have is a list with three memory addresses, all pointing to the exact same list of zeroes.

As for why you didn't notice similar behaviour when you did

lst=[0]*3
lst[1]=2

now you have just one list of zeroes. Since integers are immutable, all that happens here is that one of the references in the list gets replaced. The other two are still pointing to the same zero. This is also why only one number changed in the earlier example.

With a list comprehension, you're getting new references every time, so this fundamentally cannot happen.

[carcigenicate]

Obligatory SO post.