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

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

for i in mylist:
    result += [i, mylist.pop()]

Note:

Beware: Just like @Tadhg McDonald-Jensen has said (see the comment below) it'll destroy half of original list object.

Answer 2

Not terribly different from some of the other answers, but it avoids a conditional expression for determining the sign of the index.

a = range(10)
b = [a[i // (2*(-1)**(i&1))] for i in a]

i & 1 alternates between 0 and 1. This causes the exponent to alternate between 1 and -1. This causes the index divisor to alternate between 2 and -2, which causes the index to alternate from end to end as i increases. The sequence is a[0], a[-1], a[1], a[-2], a[2], a[-3], etc.

(I iterate i over a since in this case each value of a is equal to its index. In general, iterate over range(len(a)).)

Answer 3

Two versions not seen yet:

b = list(sum(zip(a, a[::-1]), ())[:len(a)])

and

import itertools as it

b = [a[j] for j in it.accumulate(i*(-1)**i for i in range(len(a)))]