Pythonic way to mix two lists

Question

I have two lists of length n and n+1:

[a_1, a_2, ..., a_n]
[b_1, b_2, ..., b_(n+1)]

I want a function giving as a result a list with alternate elements from the two, that is

[b_1, a_1, ..., b_n, a_n, b_(n+1)]

The following works, but does not look smart:

def list_mixing(list_long,list_short):
    list_res = []
    for i in range(len(list_short)):
        list_res.extend([list_long[i], list_short[i]])
    list_res.append(list_long[-1])
    return list_res

Can anyone suggest a more pythonic way of doing this? Thanks!

Answer 1

>>> import itertools
>>> a
['1', '2', '3', '4', '5', '6']
>>> b
['a', 'b', 'c', 'd', 'e', 'f']
>>> list(itertools.chain.from_iterable(zip(a,b)))
['1', 'a', '2', 'b', '3', 'c', '4', 'd', '5', 'e', '6', 'f']

zip() produces a iterable with the length of shortest argument. You can either append a[-1] to the result, or use itertools.zip_longest(izip_longest for Python 2.x) with a fill value and delete that value afterwards.

And you can use more than two input sequences with this solution.

For not appending the last value, you can try this dirty approach, but I don't really recommend it, it isn't clear:

>>> a
[1, 2, 3, 4, 5]
>>> b
['a', 'b', 'c', 'd', 'e', 'f']
>>> [a[i//2] if i%2 else b[i//2] for i in range(len(a)*2+1)]
['a', 1, 'b', 2, 'c', 3, 'd', 4, 'e', 5, 'f']

(For Python 2.x, use single /)

Answer 2

more-itertools has roundrobin which exactly does the job:

from more_itertools import roundrobin

l1 = [1, 3, 5]
l2 = [2, 4, 6, 8, 10]

print(list(roundrobin(l1,l2)))
# [1, 2, 1, 3, 4, 3, 5, 6, 5, 8, 10]