Extract elements of list at odd positions

So I want to create a list which is a sublist of some existing list.

For example,

L = [1, 2, 3, 4, 5, 6, 7], I want to create a sublist li such that li contains all the elements in L at odd positions.

While I can do it by

L = [1, 2, 3, 4, 5, 6, 7]
li = []
count = 0
for i in L:
    if count % 2 == 1:
        li.append(i)
    count += 1

But I want to know if there is another way to do the same efficiently and in fewer number of steps.

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

For the odd positions, you probably want:

>>>> list_ = list(range(10))
>>>> print list_[1::2]
[1, 3, 5, 7, 9]
>>>>

I like List comprehensions because of their Math (Set) syntax. So how about this:

L = [1, 2, 3, 4, 5, 6, 7]
odd_numbers = [y for x,y in enumerate(L) if x%2 != 0]
even_numbers = [y for x,y in enumerate(L) if x%2 == 0]

Basically, if you enumerate over a list, you'll get the index x and the value y. What I'm doing here is putting the value y into the output list (even or odd) and using the index x to find out if that point is odd (x%2 != 0).