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

The basic principle behind your question is a so-called roundrobin algorithm. The itertools-documentation-page contains a possible implementation of it:

from itertools import cycle, islice

def roundrobin(*iterables):
    """This function is taken from the python documentation!
    roundrobin('ABC', 'D', 'EF') --> A D E B F C
    Recipe credited to George Sakkis"""
    pending = len(iterables)
    nexts = cycle(iter(it).__next__ for it in iterables) # next instead of __next__ for py2
    while pending:
        try:
            for next in nexts:
                yield next()
        except StopIteration:
            pending -= 1
            nexts = cycle(islice(nexts, pending))

so all you have to do is split your list into two sublists one starting from the left end and one from the right end:

import math
mid = math.ceil(len(a)/2) # Just so that the next line doesn't need to calculate it twice

list(roundrobin(a[:mid], a[:mid-1:-1]))
# Gives you the desired result: [0, 9, 1, 8, 2, 7, 3, 6, 4, 5]

alternatively you could create a longer list (containing alternating items from sequence going from left to right and the items of the complete sequence going right to left) and only take the relevant elements:

list(roundrobin(a, reversed(a)))[:len(a)]

or using it as explicit generator with next:

rr = roundrobin(a, reversed(a))
[next(rr) for _ in range(len(a))]

or the speedy variant suggested by @Tadhg McDonald-Jensen (thank you!):

list(islice(roundrobin(a,reversed(a)),len(a)))