List value swapping: What's the correct order and why?

Question

For a list of integers, such as A = [2, 10, -5], I get the error

Traceback (most recent call last):
  File "so.py", line 6, in <module>
    v, A[v-1] = A[v-1], v
IndexError: list assignment index out of range

Code:

for i, v in enumerate(A):
    while 1<=v<=len(A) and v != A[v-1]:
        v, A[v-1] = A[v-1], v

but this works:

for i, v in enumerate(A):
    while 1<=v<=len(A) and v != A[v-1]:
        A[v-1], v = v, A[v-1]

Why the order of the swapping elements matters here? v is always being checked to be in bound.

Weirdly enough cannot reproduce a smaller example. But, A = [6, 5, 4, 3, 2] becomes an infinite loop.

You should refer to this [post](https://stackoverflow.com/questions/21047524/how-does-swapping-of-members-in-the-python-tuples-a-b-b-a-work-internally). — Oladimeji Mudele, Nov 15 '18 at 17:41
`i` is the index. `v` the value. `A[i]` makes sense. `A[v]` does not. — hpaulj, Nov 15 '18 at 17:58
Note that your inner loop is a `while`, not an `if`. Your list [6, 5, 4, 3, 2] gets stuck in the `while`. — Prune, Nov 15 '18 at 18:10
@Prune: I got that `while` gets stuck here. And explanation comes from your ans below. Thanks. — kaiserasif, Nov 15 '18 at 18:19
@kaiserasif Can you describe in words what you want your code to do? Maybe give an example? — Elias Riedel Gårding, Nov 15 '18 at 20:31
@Elias: put every integer in the corresponding position if the value falls in the valid range, otherwise skip. However, Prune's example clarifies what was bothering me, and what Tim had answered previously. — kaiserasif, Nov 15 '18 at 21:55

score 7 · Answer 1 · answered Nov 15 '18 at 17:35

7

Python swaps the variables in the order provided, so v is assigned the value at A[v-1], and then tries to reassign A[v-1] - but since v has been modified to be a list element, v-1 is out of range of A.

answered Nov 15 '18 at 17:35

Tim

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Then, why the opposite doesn't cause the problem, i.e. `A[v-1]` set to `v` first, then `v` still gets the previous `A[v-1]` value from before the update. – kaiserasif Nov 15 '18 at 17:44
1

Because you have an if statement ensuring that `v` is less than the length of the array, so `A[v-1]` is always set to a value that can index `A` in the next step. – Tim Nov 15 '18 at 17:57

Prune · Accepted Answer · 2018-11-15T18:06:38.473

I reproduced this with [2, 10, -5]. The detailed sequence of operations is

i, v = 0, 2
1 <= 2 ?     OK
2 != A[1] ?  OK ... stay in loop
    v, A[v-1] = 10, 2
    # This assignment breaks into ...
    v = 10
    A[10-1] = 2
    # ... and this second assignment is out of range.

If you switch the assignment order:

    A[v-1], v = 2, 10
    # This assignment breaks into ...
    A[2-1] = 2
    v = 10

In this case, your while conditions have properly guarded a legal assignment.

Note that you are not swapping list values; v is a local variable; it is not a reference to A[i]. For instance, in the above example, you do get v=10, but this does not affect the value of A[0].

score 0 · Answer 3 · answered Nov 15 '18 at 18:07

A = [6, 5, 4, 3, 2]

for i, v in enumerate(A):
    while 1<=v<=len(A) and v != A[v-1]:
        v, A[v-1] = A[v-1], v

You need to try running out your algorithm in your head:

Start:

i = 0, v = 6

v(6) is not between 1 and 5: next iteration

i = 1, v = 5, A[5-1] = 2

5 is between 1 and 5; v is not equal to 2: swap -> v = 2; A[4] = 2

i = 1, v = 2, A[2-1] = 5

2 is between 1 and 5; v is not equal to 5: swap -> v = 5; A[1] = 5

i = 1, v = 5, A[5-1] = 2

5 is between 1 and 5; v is not equal to 2: swap -> v = 2; A[4] = 2

i = 1, v = 2, A[2-1] = 5

2 is between 1 and 5; v is not equal to 5: swap -> v = 5; A[1] = 5

... and on and on

I don't think your algorithm makes sense. It's unclear why you are using the values in your list to index the list during your loop. I think this confusion about the index and values is at the root of your problem.

List value swapping: What's the correct order and why?

3 Answers3