How to elegantly interleave two lists of uneven length in python?

Question

I want to merge two lists in python, with the lists being of different lengths, so that the elements of the shorter list are as equally spaced within the final list as possi

Answer 1

Borrowing heavily from Jon Clements' solution, you could write a function that takes an arbitrary number of sequences and returns merged sequence of evenly-spaced items:

import itertools as IT

def evenly_spaced(*iterables):
    """
    >>> evenly_spaced(range(10), list('abc'))
    [0, 1, 'a', 2, 3, 4, 'b', 5, 6, 7, 'c', 8, 9]
    """
    return [item[1] for item in
            sorted(IT.chain.from_iterable(
            zip(IT.count(start=1.0 / (len(seq) + 1), 
                         step=1.0 / (len(seq) + 1)), seq)
            for seq in iterables))]

iterables = [
    ['X']*2,
    range(1, 11),
    ['a']*3
    ]

print(evenly_spaced(*iterables))

yields

[1, 2, 'a', 3, 'X', 4, 5, 'a', 6, 7, 'X', 8, 'a', 9, 10]

Answer 2

if a is the longer list and b is the shorter

from itertools import groupby

len_ab = len(a) + len(b)
groups = groupby(((a[len(a)*i//len_ab], b[len(b)*i//len_ab]) for i in range(len_ab)),
                 key=lambda x:x[0])
[j[i] for k,g in groups for i,j in enumerate(g)]

eg

>>> a = range(8)
>>> b = list("abc")
>>> len_ab = len(a) + len(b)
>>> groups = groupby(((a[len(a)*i//len_ab], b[len(b)*i//len_ab]) for i in range(len_ab)), key=lambda x:x[0])
>>> [j[i] for k,g in groups for i,j in enumerate(g)]
[0, 'a', 1, 2, 'b', 3, 4, 5, 'c', 6, 7]

You can use this trick to make sure a is longer than b

b, a = sorted((a, b), key=len)