Cycle a list from alternating sides

Question

Given a list

a = [0,1,2,3,4,5,6,7,8,9]

how can I get

b = [0,9,1,8,2,7,3,6,4,5]

That is, produce a new lis

Answer 1

cycle between getting items from the forward iter and the reversed one. Just make sure you stop at len(a) with islice.

from itertools import islice, cycle

iters = cycle((iter(a), reversed(a)))
b = [next(it) for it in islice(iters, len(a))]

>>> b
[0, 9, 1, 8, 2, 7, 3, 6, 4, 5]

This can easily be put into a single line but then it becomes much more difficult to read:

[next(it) for it in islice(cycle((iter(a),reversed(a))),len(a))]

Putting it in one line would also prevent you from using the other half of the iterators if you wanted to:

>>> iters = cycle((iter(a), reversed(a)))
>>> [next(it) for it in islice(iters, len(a))]
[0, 9, 1, 8, 2, 7, 3, 6, 4, 5]
>>> [next(it) for it in islice(iters, len(a))]
[5, 4, 6, 3, 7, 2, 8, 1, 9, 0]

Answer 2

>>> [a[-i//2] if i % 2 else a[i//2] for i in range(len(a))]
[0, 9, 1, 8, 2, 7, 3, 6, 4, 5]

Explanation:
This code picks numbers from the beginning (a[i//2]) and from the end (a[-i//2]) of a, alternatingly (if i%2 else). A total of len(a) numbers are picked, so this produces no ill effects even if len(a) is odd.
[-i//2 for i in range(len(a))] yields 0, -1, -1, -2, -2, -3, -3, -4, -4, -5,
[ i//2 for i in range(len(a))] yields 0, 0, 1, 1, 2, 2, 3, 3, 4, 4,
and i%2 alternates between False and True,
so the indices we extract from a are: 0, -1, 1, -2, 2, -3, 3, -4, 4, -5.

My assessment of pythonicness:
The nice thing about this one-liner is that it's short and shows symmetry (+i//2 and -i//2).
The bad thing, though, is that this symmetry is deceptive:
One might think that -i//2 were the same as i//2 with the sign flipped. But in Python, integer division returns the floor of the result instead of truncating towards zero. So -1//2 == -1.
Also, I find accessing list elements by index less pythonic than iteration.

Answer 3

For fun, here is an itertools variant:

>>> a = [0,1,2,3,4,5,6,7,8,9]
>>> list(chain.from_iterable(izip(islice(a, len(a)//2), reversed(a))))
[0, 9, 1, 8, 2, 7, 3, 6, 4, 5]

This works where len(a) is even. It would need a special code for odd-lengthened input.

Enjoy!

Answer 4

One way to do this for even-sized lists (inspired by this post):

a = range(10)

b = [val for pair in zip(a[:5], a[5:][::-1]) for val in pair]

Answer 5

You can partition the list into two parts about the middle, reverse the second half and zip the two partitions, like so:

a = [0,1,2,3,4,5,6,7,8,9]
mid = len(a)//2
l = []
for x, y in zip(a[:mid], a[:mid-1:-1]):
    l.append(x)
    l.append(y)
# if the length is odd
if len(a) % 2 == 1:
    l.append(a[mid])
print(l)

Output:

[0, 9, 1, 8, 2, 7, 3, 6, 4, 5]

Answer 6

Not at all elegant, but it is a clumsy one-liner:

a = range(10)
[val for pair in zip(a[:len(a)//2],a[-1:(len(a)//2-1):-1]) for val in pair]

Note that it assumes you are doing this for a list of even length. If that breaks, then this breaks (it drops the middle term). Note that I got some of the idea from here.