Python pop() vs pop(0)

Question

So the following is confusing me.

#!/usr/bin/python

test = [0, 0, 0, 1, 2, 3, 4, 5, 6]
test1 = [0, 0, 0, 1, 2, 3, 4, 5, 6]

for _dummy in test:
    if(_dummy == 0):
        test.pop()
for _dummy in test1:
    if(_dummy == 0):
        test1.pop(0)

print test
print test1

Results

ubuntu-vm:~/sandbox$ ./test.py 
[0, 0, 0, 1, 2, 3]
[0, 1, 2, 3, 4, 5, 6]

Perhaps, I'm fundamentally misunderstanding how pop is implemented. But my understanding is that it removes the item at the given index in the list, and returns it. If no index is specified, it defaults to the last item. So it would seem that in the first loop it should remove 3 items from the left of the list, and in the second loop it should remove 3 items from the end of the list.

Answer 1

The first test isn't surprising; three elements are removed off the end.

The second test is a bit surprising. Only two elements are removed. Why?

List iteration in Python essentially consists of an incrementing index into the list. When you delete an element you shift all the elements on the right over. This may cause the index to point to a different element.

Illustratively:

start of loop
[0,0,0,1,2,3,4,5,6]
 ^   <-- position of index

delete first element (since current element = 0)
[0,0,1,2,3,4,5,6]
 ^

next iteration
[0,0,1,2,3,4,5,6]
   ^

delete first element (since current element = 0)
[0,1,2,3,4,5,6]
   ^

and from now on no zeros are encountered, so no more elements are deleted.

---

To avoid confusion in the future, try not to modify lists while you're iterating over them. While Python won't complain (unlike dictionaries, which cannot be modified during iteration), it will result in weird and usually counterintuitive situations like this one.