Python swapping lists

Question

In python, when I assign a list to another, like:

a = [1,2,3]
b = a

Now b and a point to the same list. Now considering two lists,

a = [1,2,3]
b = [4,5,6]
a,b = b,a

Now how is it that they are swapped like any other data type and does not end up both pointing to the same list?

Answer 1

Looks like Python internally swaps the items. Check this program

a, b = [1, 2], [2, 3]

def func():
    a, b = b, a

import dis
dis.dis(func)

Output

  4           0 LOAD_FAST                0 (b)
              3 LOAD_FAST                1 (a)
              6 ROT_TWO             
              7 STORE_FAST               1 (a)
             10 STORE_FAST               0 (b)
             13 LOAD_CONST               0 (None)
             16 RETURN_VALUE

So, Python pushes references from b and a in the stack with LOAD_FAST. So, now the top most element is the reference pointed by a and the next one is the reference pointed by b. Then it uses ROT_TWO to swap the top two elements of the stack. So, now, the top most element is the reference pointed by b and the next one is the reference pointed by a and then assigns the top two elements of the stack to a and b respectively with STORE_FAST.

That's how sorting is happening in the assignment statement, when the number of items we deal with is lesser than 4.

If the number of items is greater than or equal to four, it builds a tuple and unpacks the values. Check this program

a, b, c, d = [1, 2], [2, 3], [4, 5], [5, 6]

def func():
    a, b, c, d  = d, c, b, a

import dis
dis.dis(func)

Output

  4           0 LOAD_FAST                0 (d)
              3 LOAD_FAST                1 (c)
              6 LOAD_FAST                2 (b)
              9 LOAD_FAST                3 (a)
             12 BUILD_TUPLE              4
             15 UNPACK_SEQUENCE          4
             18 STORE_FAST               3 (a)
             21 STORE_FAST               2 (b)
             24 STORE_FAST               1 (c)
             27 STORE_FAST               0 (d)
             30 LOAD_CONST               0 (None)
             33 RETURN_VALUE

Answer 2

Because Python assignment first evaluates the right-hand-side expression, then applies the result to the left-hand-side targets.

So, first, Python creates (<reference to list b>, <reference to list a>) as a tuple, then assigns the first item in that tuple to a, and the second item in that tuple to b. This swaps the references around neatly.

You could expand the assignment to read it like this:

tmp = b, a
a = tmp[0]
b = tmp[1]