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

zip(list1, list2)

Answer 2

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

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

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 4

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 5

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 6

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

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

Answer 7

c=[]
for i, j in zip(list1,list2):
    c.append(i)
    c.append(j)
c.append(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

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 11

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 12

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 13

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

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

Using itertools and assuming that lists don't contain None:

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