## Solution

Yes, you can:

``l = L[1::2]``

And this is all. The result will contain the elements placed on the following positions (`0`-based, so first element is at position `0`, second at `1` etc.):

``1, 3, 5``

so the result (actual numbers) will be:

``2, 4, 6``

## Explanation

The `[1::2]` at the end is just a notation for list slicing. Usually it is in the following form:

``some_list[start:stop:step]``

If we omitted `start`, the default (`0`) would be used. So the first element (at position `0`, because the indexes are `0`-based) would be selected. In this case the second element will be selected.

Because the second element is omitted, the default is being used (the end of the list). So the list is being iterated from the second element to the end.

We also provided third argument (`step`) which is `2`. Which means that one element will be selected, the next will be skipped, and so on...

So, to sum up, in this case `[1::2]` means:

1. take the second element (which, by the way, is an odd element, if you judge from the index),
2. skip one element (because we have `step=2`, so we are skipping one, as a contrary to `step=1` which is default),
3. take the next element,
4. Repeat steps 2.-3. until the end of the list is reached,

EDIT : @PreetKukreti gave a link for another explanation on Python's list slicing notation. See here: Explain Python's slice notation

## Extras - replacing counter with `enumerate()`

In your code, you explicitly create and increase the counter. In Python this is not necessary, as you can enumerate through some iterable using `enumerate()`:

``````for count, i in enumerate(L):
if count % 2 == 1:
l.append(i)``````

The above serves exactly the same purpose as the code you were using:

``````count = 0
for i in L:
if count % 2 == 1:
l.append(i)
count += 1``````

More on emulating `for` loops with counter in Python: Accessing the index in Python 'for' loops