Question

a = [0,1,2,3,4,5]
for b in a:
  print ":"+str(b)
  a.pop(0)

Thinking that this would work in order going through the entire list and all its items I ran this code and expected this.

:0
0
:1
1
:2
2
:3
3
:4
4
:5
5

Instead I got this:

:0
0
:2
1
:4
2

Now I understand why this happened but is this an error in python? Shouldn’t it still go through all of the original objects instead of the length of the current list? And why did this not throw and out of bounds error?
IE: Shouldn’t it still have done:

:0
0
:1
2
:2
4
:3
Error
:4
Error
:5
Error

Answer 1

You are looping over a list and altering it at the same time. By using .pop() you are shortening the list, but the iterator pointer is not updated.

Use a copy instead:

for b in list(a):

or

for b in a[:]:

where the [:] slice notation returns a list copy.

Another approach would be to use a while loop instead:

while a:
    print a.pop(0)

because an empty list tests as boolean False.

The python for loop uses it’s argument as an iterator, it does not itself keep an index. There is no way for the for loop to ‘know’ you removed elements. Instead, it’s the list() iterator that keeps that pointer:

>>> a = [0,1,2,3,4,5]
>>> itera = iter(a)
>>> itera.next()  # index 0 -> a[0] is 0
0
>>> a.pop(0)
0
>>> a
[1,2,3,4,5]
>>> itera.next()  # index 1 -> a[1] is 2
2

That iterator keeps a counter, and every time you call next() on the iterator it’ll give you the value at the next index, whatever that value may be, until the counter is equal to the current lenght of the list.