Cant remove all elements via pop method

Question

When I'm trying to remove elements through the loop like this:

numbers = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 18, 12, 122]
for z in numbers:
    a = numbers.pop(0)
print(numbers)

That's what I'm getting:

[8, 9, 10, 18, 12, 122]

why it doesnt remove all elements? why is that so? thanks!

@KlausD.: probably a better way is to range over the `len(..)` since this requires less memory. — Willem Van Onsem, Feb 13 '17 at 13:21
@WillemVanOnsem True, but that would be another concept, since it iterates over a `range` object instead of a class. — Klaus D., Feb 13 '17 at 13:28
@KlausD.: indeed: you cannot access the element. Fair point. — Willem Van Onsem, Feb 13 '17 at 13:29

Willem Van Onsem · Accepted Answer · 2017-06-18T22:13:35.117

4

Your code will remove half. Because when you use a for loop, it will iterate over the elements. Now you remove elements at the other side of the list, so the two pointers will reach each other midway and so the loop will stop.

So you start with the list:

[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 18, 12, 122]
#^

*(the caret shows where the for loop pointer is located), next you pop and advance the caret:

[2, 3, 4, 5, 6, 7, 8, 9, 10, 18, 12, 122]
#   ^

and so on:

[4, 5, 6, 7, 8, 9, 10, 18, 12, 122]
#      ^
[5, 6, 7, 8, 9, 10, 18, 12, 122]
#         ^
[6, 7, 8, 9, 10, 18, 12, 122]
#            ^
[7, 8, 9, 10, 18, 12, 122]
#                 ^
[8, 9, 10, 18, 12, 122]
#                      ^

Now the index pointer has reached a point out of the range so the loop terminates. But it has not iterated over all elements: since each time you pop an element, it has jumped two paces at once so to speak.

Some general advice about looping over lists:

You better do not modify the list while you iterate over it.

A better way to do this is:

for _ in range(len(numbers)):
    a = numbers.pop(0)
print(numbers)

or, as @Matthias proposes:

while numbers:
    a = numbers.pop(0)
print(numbers)

Since the bool(..) of a list is true of it contains at least one element. So here each time we check whether there is at least still an element in numbers.

edited Jun 18 '17 at 22:13

answered Feb 13 '17 at 13:17

Willem Van Onsem

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Actually the OP is removing from the start (`.pop(0)`), not from the end... – bruno desthuilliers Feb 13 '17 at 13:19
@brunodesthuilliers: I just noticed. Editing... – Willem Van Onsem Feb 13 '17 at 13:19
`list.pop(0)` is O(n), I don't think it's good advice to use it in this way at all. Also the core of this question is clearly a dupe – Chris_Rands Feb 13 '17 at 13:24
@Chris_Rands: indeed deleting from the front of a list is never a good idea. Nevertheless it is possible and of course in a real application a programmer will do something with the list. – Willem Van Onsem Feb 13 '17 at 13:25
Alternative: `while numbers: a = numbers.pop(0)`. – Matthias Feb 13 '17 at 14:14
@Matthias: that's indeed a great idea. I've added it to the answer. – Willem Van Onsem Feb 13 '17 at 14:16
@WillemVanOnsem: Thank you for the additional explanation. I was in a little bit of a hurry. – Matthias Feb 13 '17 at 14:30

Cant remove all elements via pop method

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