python - Pythonic way to combine two lists in an alternating fashion?

Question

I have two lists, the first of which is guaranteed to contain exactly one more item than the second. I would like to know the most Pythonic way to create a new list whose even-index values come from the first list and whose odd-index values come from the second list.

# example inputs
list1 = ['f', 'o', 'o']
list2 = ['hello', 'world']

# desired output
['f', 'hello', 'o', 'world', 'o']

This works, but isn't pretty:

list3 = []
while True:
    try:
        list3.append(list1.pop(0))
        list3.append(list2.pop(0))
    except IndexError:
        break

How else can this be achieved? What's the most Pythonic approach?

Answer 1

Here's one way to do it by slicing:

>>> list1 = ['f', 'o', 'o']
>>> list2 = ['hello', 'world']
>>> result = [None]*(len(list1)+len(list2))
>>> result[::2] = list1
>>> result[1::2] = list2
>>> result
['f', 'hello', 'o', 'world', 'o']

Answer 2

There's a recipe for this in the itertools documentation:

from itertools import cycle, islice

def roundrobin(*iterables):
    "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)
    while pending:
        try:
            for next in nexts:
                yield next()
        except StopIteration:
            pending -= 1
            nexts = cycle(islice(nexts, pending))

EDIT:

For python's version greater than 3:

from itertools import cycle, islice

def roundrobin(*iterables):
    "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)
    while pending:
        try:
            for next in nexts:
                yield next()
        except StopIteration:
            pending -= 1
            nexts = cycle(islice(nexts, pending))

Answer 3

import itertools
print [x for x in itertools.chain.from_iterable(itertools.izip_longest(list1,list2)) if x]

I think this is the most pythonic way of doing it.

Answer 4

In Python 2, this should do what you want:

>>> iters = [iter(list1), iter(list2)]
>>> print list(it.next() for it in itertools.cycle(iters))
['f', 'hello', 'o', 'world', 'o']

Answer 5

Without itertools and assuming l1 is 1 item longer than l2:

>>> sum(zip(l1, l2+[0]), ())[:-1]
('f', 'hello', 'o', 'world', 'o')

In python 2, using itertools and assuming that lists don't contain None:

>>> filter(None, sum(itertools.izip_longest(l1, l2), ()))
('f', 'hello', 'o', 'world', 'o')

Answer 6

I know the questions asks about two lists with one having one item more than the other, but I figured I would put this for others who may find this question.

Here is Duncan's solution adapted to work with two lists of different sizes.

list1 = ['f', 'o', 'o', 'b', 'a', 'r']
list2 = ['hello', 'world']
num = min(len(list1), len(list2))
result = [None]*(num*2)
result[::2] = list1[:num]
result[1::2] = list2[:num]
result.extend(list1[num:])
result.extend(list2[num:])
result

This outputs:

['f', 'hello', 'o', 'world', 'o', 'b', 'a', 'r'] 

Answer 7

If both lists have equal length, you can do:

[x for y in zip(list1, list2) for x in y]

As the first list has one more element, you can add it post hoc:

[x for y in zip(list1, list2) for x in y] + [list1[-1]]

Answer 8

Here's a one liner that does it:

list3 = [ item for pair in zip(list1, list2 + [0]) for item in pair][:-1]

Answer 9

Here's a one liner using list comprehensions, w/o other libraries:

list3 = [sub[i] for i in range(len(list2)) for sub in [list1, list2]] + [list1[-1]]

Here is another approach, if you allow alteration of your initial list1 by side effect:

[list1.insert((i+1)*2-1, list2[i]) for i in range(len(list2))]

Answer 10

This one is based on Carlos Valiente's contribution above with an option to alternate groups of multiple items and make sure that all items are present in the output :

A=["a","b","c","d"]
B=[1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16]

def cyclemix(xs, ys, n=1):
    for p in range(0,int((len(ys)+len(xs))/n)):
        for g in range(0,min(len(ys),n)):
            yield ys[0]
            ys.append(ys.pop(0))
        for g in range(0,min(len(xs),n)):
            yield xs[0]
            xs.append(xs.pop(0))

print [x for x in cyclemix(A, B, 3)]

This will interlace lists A and B by groups of 3 values each:

['a', 'b', 'c', 1, 2, 3, 'd', 'a', 'b', 4, 5, 6, 'c', 'd', 'a', 7, 8, 9, 'b', 'c', 'd', 10, 11, 12, 'a', 'b', 'c', 13, 14, 15]

Answer 11

Might be a bit late buy yet another python one-liner. This works when the two lists have equal or unequal size. One thing worth nothing is it will modify a and b. If it's an issue, you need to use other solutions.

a = ['f', 'o', 'o']
b = ['hello', 'world']
sum([[a.pop(0), b.pop(0)] for i in range(min(len(a), len(b)))],[])+a+b
['f', 'hello', 'o', 'world', 'o']

Answer 12

My take:

a = "hlowrd"
b = "el ol"

def func(xs, ys):
    ys = iter(ys)
    for x in xs:
        yield x
        yield ys.next()

print [x for x in func(a, b)]

Answer 13

def combine(list1, list2):
    lst = []
    len1 = len(list1)
    len2 = len(list2)

    for index in range( max(len1, len2) ):
        if index+1 <= len1:
            lst += [list1[index]]

        if index+1 <= len2:
            lst += [list2[index]]

    return lst

Answer 14

from itertools import chain
list(chain(*zip('abc', 'def')))  # Note: this only works for lists of equal length
['a', 'd', 'b', 'e', 'c', 'f']

Answer 15

Stops on the shortest:

def interlace(*iters, next = next) -> collections.Iterable:
    """
    interlace(i1, i2, ..., in) -> (
        i1-0, i2-0, ..., in-0,
        i1-1, i2-1, ..., in-1,
        .
        .
        .
        i1-n, i2-n, ..., in-n,
    )
    """
    return map(next, cycle([iter(x) for x in iters]))

Sure, resolving the next/__next__ method may be faster.

Answer 16

This is nasty but works no matter the size of the lists:

list3 = [element for element in list(itertools.chain.from_iterable([val for val in itertools.izip_longest(list1, list2)])) if element != None]

Answer 17

Multiple one-liners inspired by answers to another question:

import itertools

list(itertools.chain.from_iterable(itertools.izip_longest(list1, list2, fillvalue=object)))[:-1]

[i for l in itertools.izip_longest(list1, list2, fillvalue=object) for i in l if i is not object]

[item for sublist in map(None, list1, list2) for item in sublist][:-1]

Answer 18

How about numpy? It works with strings as well:

import numpy as np

np.array([[a,b] for a,b in zip([1,2,3],[2,3,4,5,6])]).ravel()

Result:

array([1, 2, 2, 3, 3, 4])

Answer 19

An alternative in a functional & immutable way (Python 3):

from itertools import zip_longest
from functools import reduce

reduce(lambda lst, zipped: [*lst, *zipped] if zipped[1] != None else [*lst, zipped[0]], zip_longest(list1, list2),[])

Answer 20

I'd do the simple:

chain.from_iterable( izip( list1, list2 ) )

It'll come up with an iterator without creating any additional storage needs.

Answer 21

I'm too old to be down with list comprehensions, so:

import operator
list3 = reduce(operator.add, zip(list1, list2))