Pythonic way to combine two lists in an alternating fashion?

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?

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']

There's a recipe for this in the itertools documentation (note: for Python 3):

from itertools import cycle, islice

def roundrobin(*iterables):
    "roundrobin('ABC', 'D', 'EF') --> A D E B F C"
    # Recipe credited to George Sakkis
    num_active = len(iterables)
    nexts = cycle(iter(it).__next__ for it in iterables)
    while num_active:
        try:
            for next in nexts:
                yield next()
        except StopIteration:
            # Remove the iterator we just exhausted from the cycle.
            num_active -= 1
            nexts = cycle(islice(nexts, num_active))

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.

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']

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')

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'] 

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]]

Here's a one liner that does it:

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

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))]

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]

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']

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

itertools.zip_longest returns an iterator of tuple pairs with any missing elements in one list replaced with fillvalue=None (passing fillvalue=object lets you use None as a value). If you flatten these pairs, then filter fillvalue in a list comprehension, this gives:

>>> from itertools import zip_longest
>>> def merge(a, b):
...     return [
...         x for y in zip_longest(a, b, fillvalue=object)
...         for x in y if x is not object
...     ]
...
>>> merge("abc", "defgh")
['a', 'd', 'b', 'e', 'c', 'f', 'g', 'h']
>>> merge([0, 1, 2], [4])
[0, 4, 1, 2]
>>> merge([0, 1, 2], [4, 5, 6, 7, 8])
[0, 4, 1, 5, 2, 6, 7, 8]

Generalized to arbitrary iterables:

>>> def merge(*its):
...     return [
...         x for y in zip_longest(*its, fillvalue=object)
...         for x in y if x is not object
...     ]
...
>>> merge("abc", "lmn1234", "xyz9", [None])
['a', 'l', 'x', None, 'b', 'm', 'y', 'c', 'n', 'z', '1', '9', '2', '3', '4']
>>> merge(*["abc", "x"]) # unpack an iterable
['a', 'x', 'b', 'c']

Finally, you may want to return a generator rather than a list comprehension:

>>> def merge(*its):
...     return (
...         x for y in zip_longest(*its, fillvalue=object)
...         for x in y if x is not object
...     )
...
>>> merge([1], [], [2, 3, 4])
<generator object merge.<locals>.<genexpr> at 0x000001996B466740>
>>> next(merge([1], [], [2, 3, 4]))
1
>>> list(merge([1], [], [2, 3, 4]))
[1, 2, 3, 4]

If you're OK with other packages, you can try more_itertools.roundrobin:

>>> list(roundrobin('ABC', 'D', 'EF'))
['A', 'D', 'E', 'B', 'F', 'C']

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)]

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

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])

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.

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]

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),[])

using for loop also we can achive this easily:

list1 = ['f', 'o', 'o']
list2 = ['hello', 'world']
list3 = []

for i in range(len(list1)):
    #print(list3)
    list3.append(list1[i])
    if i < len(list2):
        list3.append(list2[i])
        
print(list3)

output :

['f', 'hello', 'o', 'world', 'o']

Further by using list comprehension this can be reduced. But for understanding this loop can be used.

I'd do the simple:

chain.from_iterable( izip( list1, list2 ) )

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

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
]

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

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