Strange inline assignment

Question

I'm struggling with this strange behaviour in Python (2 and 3):

>>> a = [1, 2]
>>> a[a.index(1)], a[a.index(2)] = 2, 1

This results in:

>>> a
[1, 2]

But if you write

>>> a = [1, 2]
>>> a[a.index(1)], a[a.index(2)] = x, y

where x, y != 2, 1 (can be 1, 1, 2, 2 , 3, 5, etc.), this results in:

>>> a == [x, y]
True

As one would expect. Why doesn't a[a.index(1)], a[a.index(2)] = 2, 1 produce the result a == [2, 1]?

>>> a == [2, 1]
False

Answer 1

Because it actually gets interpreted like this:

>>> a = [1, 2]
>>> a
[1, 2]
>>> a[a.index(1)] = 2
>>> a
[2, 2]
>>> a[a.index(2)] = 1
>>> a
[1, 2]

To quote, per the standard rules for assignment (emphasis mine):

• If the target list is a comma-separated list of targets: The object must be an iterable with the same number of items as there are targets in the target list, and the items are assigned, from left to right, to the corresponding targets.

The assignment to a[a.index(1)] (i.e. a[0]) happens before the second assignment asks for a.index(2), by which time a.index(2) == 0.

You will see the same behaviour for any assignment:

foo = [a, b]
foo[foo.index(a)], foo[foo.index(b)] = x, y

where x == b (in this case, any assignment where the first value on the right-hand side is 2).