# LaVOZs

The World’s Largest Online Community for Developers

'; python - strange result when removing item from a list - LavOzs.Com

# strange result when removing item from a list [duplicate]

You're modifying the list while you iterate over it. That means that the first time through the loop, `i == 1`, so 1 is removed from the list. Then the `for` loop goes to the second item in the list, which is not 2, but 3! Then that's removed from the list, and then the `for` loop goes on to the third item in the list, which is now 5. And so on. Perhaps it's easier to visualize like so, with a ^ pointing to the value of `i`:

``````[1, 2, 3, 4, 5, 6...]
^
``````

That's the state of the list initially; then 1 is removed and the loop goes to the second item in the list:

``````[2, 3, 4, 5, 6...]
^
[2, 4, 5, 6...]
^
``````

And so on.

There's no good way to alter a list's length while iterating over it. The best you can do is something like this:

``````numbers = [n for n in numbers if n >= 20]
``````

or this, for in-place alteration (the thing in parens is a generator expression, which is implicitly converted into a tuple before slice-assignment):

``````numbers[:] = (n for in in numbers if n >= 20)
``````

If you want to perform an operation on n before removing it, one trick you could try is this:

``````for i, n in enumerate(numbers):
if n < 20 :
print "do something"
numbers[i] = None
numbers = [n for n in numbers if n is not None]
``````

@senderle's answer is the way to go!

Having said that to further illustrate even a bit more your problem, if you think about it, you will always want to remove the index 0 twenty times:

``````[1,2,3,4,5............50]
^
[2,3,4,5............50]
^
[3,4,5............50]
^
``````

So you could actually go with something like this:

``````aList = range(50)
i = 0
while i < 20:
aList.pop(0)
i += 1

print aList #[21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49]
``````

I hope it helps.

The ones below are not bad practices AFAIK.

EDIT (Some more):

``````lis = range(50)
lis = lis[20:]
``````

Will do the job also.

EDIT2 (I'm bored):

``````functional = filter(lambda x: x> 20, range(50))
``````

Removing items from a list is simple: begin at the list's end:

``````li = range(1,15)
print li,'\n'

for i in xrange(len(li)-1,-1,-1):
if li[i] < 6:
del li[i]

print li
``````

result

``````[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14]

[6, 7, 8, 9, 10, 11, 12, 13, 14]
``````

I really like the answer @senderle provided. The only thing I want to add (I'd have commented, but can't get proper code formatting) is that you can also get away with side processing for your list comprehensions if you declare your functions ahead of time like so:

``````def checkRemoval(item):
# some processing here
return passed_inspection # Should be True / False.

newList = [item for item in oldList if checkRemoval(item)]
``````