Negative index in python for loop [duplicate]

Answer 1

note: a[-1] refers to the last element of the list.

you assign to a[index] in every loop iteration over a

for a[index] in a:

a[index] will be assigned the values a[0], a[1], ..., a[-1] during the loop and end up being assigned to

a[index] = a[-1]

usually you try not to mess with the list you are iterating over:

for item in a:
    # do something with item

Answer 2

This is because in Python, this:

a = [0, 1, 2, 3]
for i in a:
    ... statements ...

can be understood as this:

a = [0, 1, 2, 3]
it = iter(a)
i = next(it)
... statements ...
i = next(it)
... statements ...
i = next(it)
... statements ...
i = next(it)
... statements ...

Now if you replace i by a[index], then you end up replacing a[index] by the last element you iterated over, which is the last element in your list.

Answer 3

It's somewhat intuitive if you refactor as a while loop

a = [0, 1, 2, 3]

for a[-1] in a: 
    print(a[-1])

becomes

a = [0, 1, 2, 3]
while i < len(a):
  e = a[i]
  a[-1] = e    # the last element is constantly being overwritten
  print(a[-1]) # same as a[i]

So the value of a over time is the following

[0, 1, 2, 3] # original
[0, 1, 2, 0] # prints 0
 |--------^
[0, 1, 2, 1] # prints 1
    |-----^
[0, 1, 2, 2] # prints 2
       |--^
[0, 1, 2, 2] # prints 2
         |^

Similarly in the second case

for a[0] in a:
    print(a[0])

becomes

a = [0, 1, 2, 3]
while i < len(a):
  e = a[i]
  a[0] = e    # the first element is constantly being overwritten
  print(a[0]) # same as a[i]

and the value of a is

[0, 1, 2, 3] # original
[0, 1, 2, 3] # prints 0
 ^|
[1, 1, 2, 3] # prints 1
 ^--|
[2, 1, 2, 2] # prints 2
 ^-----|
[3, 1, 2, 3] # prints 3
 ^--------|