Question

I am wondering if there is a standard library function in Python which will rearrange the elements of a list like below:

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

function(a)

print a

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

It should get one element from beginning of original list, then one from end, then second from beginning and so on. Then rearrange the list.

Regards,

Answer 1

You could build a fast, memory efficient generator using itertools which does what you want:

from itertools import chain, izip

def reorder(a):
    gen = chain.from_iterable(izip(a, reversed(a)))
    for _ in a:
        yield next(gen)

>>> list(reorder(a))
<<< [1, 7, 2, 6, 3, 5, 4]

You’ll find that itertools has a great collection of building blocks for creating your own efficient iterators. A slightly more succinct solution:

>>> list(chain.from_iterable(izip(a, reversed(a))))[:len(a)]
<<< [1, 7, 2, 6, 3, 5, 4]

List comprehensions are another really concise way to build lists:

>>> [x for t in zip(a, reversed(a)) for x in t][:len(a)]
<<< [1, 7, 2, 6, 3, 5, 4]

Finally here’s a short one-liner just for fun:

>>> sum(zip(a, a[::-1]), ())[:len(a)]
<<< (1, 7, 2, 6, 3, 5, 4)