Interleave multiple lists of the same length in Python

Question

In Python, is there a good way to interleave two lists of the same length?

Say I'm given [1,2,3] and [10,20,30]. I'd like to transform those into [1,10,2,20,3,30].

Answer 1

Having posted the question, I've realised that I can simply do the following:

[val for pair in zip(l1, l2) for val in pair]

where l1 and l2 are the two lists.

---

If there are N lists to interleave, then

lists = [l1, l2, ...]
[val for tup in zip(*lists) for val in tup]

Answer 2

For Python>=2.3, there's extended slice syntax:

>>> a = [0, 2, 4, 6, 8]
>>> b = [1, 3, 5, 7, 9]
>>> c = a + b
>>> c[::2] = a
>>> c[1::2] = b
>>> c
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]

The line c = a + b is used as a simple way to create a new list of exactly the right length (at this stage, its contents are not important). The next two lines do the actual work of interleaving a and b: the first one assigns the elements of a to all the even-numbered indexes of c; the second one assigns the elements of b to all the odd-numbered indexes of c.

Answer 3

Given

a = [1, 2, 3]
b = [10, 20, 30]
c = [100, 200, 300, 999]

Code

Assuming lists of equal length, you can get an interleaved list with itertools.chain and zip:

import itertools


list(itertools.chain(*zip(a, b)))
# [1, 10, 2, 20, 3, 30]

---

Alternatives

itertools.zip_longest

More generally with unequal lists, use zip_longest (recommended):

[x for x in itertools.chain(*itertools.zip_longest(a, c)) if x is not None]
# [1, 100, 2, 200, 3, 300, 999]

Many lists can safely be interleaved:

[x for x in itertools.chain(*itertools.zip_longest(a, b, c)) if x is not None]
# [1, 10, 100, 2, 20, 200, 3, 30, 300, 999]

more_itertools⁺

A library that ships with the roundrobin itertools recipe, interleave and interleave_longest.

import more_itertools


list(more_itertools.roundrobin(a, b))
# [1, 10, 2, 20, 3, 30]

list(more_itertools.interleave(a, b))
# [1, 10, 2, 20, 3, 30]

list(more_itertools.interleave_longest(a, c))
# [1, 100, 2, 200, 3, 300, 999]

yield from

Finally, for something interesting in Python 3 (though not recommended):

list(filter(None, ((yield from x) for x in zip(a, b))))
# [1, 10, 2, 20, 3, 30]

list([(yield from x) for x in zip(a, b)])
# [1, 10, 2, 20, 3, 30]

_{⁺Install using pip install more_itertools}

Answer 4

I needed a way to do this with lists of different sizes which the accepted answer doesn't address.

My solution uses a generator and its usage looks a bit nicer because of it:

def interleave(l1, l2):
    iter1 = iter(l1)
    iter2 = iter(l2)
    while True:
        try:
            if iter1 is not None:
                yield next(iter1)
        except StopIteration:
            iter1 = None
        try:
            if iter2 is not None:
                yield next(iter2)
        except StopIteration:
            iter2 = None
        if iter1 is None and iter2 is None:
            raise StopIteration()

And its usage:

>>> a = [1, 2, 3, 4, 5]
>>> b = ['a', 'b', 'c', 'd', 'e', 'f', 'g']
>>> list(interleave(a, b))
[1, 'a', 2, 'b', 3, 'c', 4, 'd', 5, 'e', 'f', 'g']
>>> list(interleave(b, a))
['a', 1, 'b', 2, 'c', 3, 'd', 4, 'e', 5, 'f', 'g']

Answer 5

Alternative:

>>> l1=[1,2,3]
>>> l2=[10,20,30]
>>> [y for x in map(None,l1,l2) for y in x if y is not None]
[1, 10, 2, 20, 3, 30]

This works because map works on lists in parallel. It works the same under 2.2. By itself, with None as the called functions, map produces a list of tuples:

>>> map(None,l1,l2,'abcd')
[(1, 10, 'a'), (2, 20, 'b'), (3, 30, 'c'), (None, None, 'd')]

Then just flatten the list of tuples.

The advantage, of course, is map will work for any number of lists and will work even if they are different lengths:

>>> l1=[1,2,3]
>>> l2=[10,20,30]
>>> l3=[101,102,103,104]
>>> [y for x in map(None,l1,l2,l3) for y in x if y in not None]
[1, 10, 101, 2, 20, 102, 3, 30, 103, 104]

Answer 6

I like aix's solution best. here is another way I think should work in 2.2:

>>> x=range(3)
>>> x
[0, 1, 2]
>>> y=range(7,10)
>>> y
[7, 8, 9]
>>> sum(zip(x,y),[])
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: can only concatenate list (not "tuple") to list
>>> sum(map(list,zip(x,y)),[])
[0, 7, 1, 8, 2, 9]

and one more way:

>>> a=[x,y]
>>> [a[i][j] for j in range(3) for i in (0,1)]
[0, 7, 1, 8, 2, 9]

and:

>>> sum((list(i) for i in zip(x,y)),[])
[0, 7, 1, 8, 2, 9]

Answer 7

[el for el in itertools.chain(*itertools.izip_longest([1,2,3], [4,5])) if el is not None]

As long as you don't have None that you want to keep

Answer 8

To answer the question's title of "Interleave multiple lists of the same length in Python", we can generalize the 2-list answer of @ekhumoro. This explicitly requires that the lists are the same length, unlike the (elegant) solution by @NPE

import itertools

def interleave(lists):
    """Interleave a list of lists.

    :param lists: List of lists; each inner length must be the same length.
    :returns: interleaved single list
    :rtype: list

    """
    if len(set(len(_) for _ in lists)) > 1:
        raise ValueError("Lists are not all the same length!")
    joint = list(itertools.chain(*lists))
    for l_idx, li in enumerate(lists):
        joint[l_idx::len(lists)] = li
    return joint

Examples:

>>> interleave([[0,2,4], [1, 3, 5]])
[0, 1, 2, 3, 4, 5]
>>> interleave([[0,2,4], [1, 3, 5], [10, 11, 12]])
[0, 1, 10, 2, 3, 11, 4, 5, 12]
>>> interleave([[0,2,4], [1, 3, 5], [10, 11, 12], [13, 14, 15]])
[0, 1, 10, 13, 2, 3, 11, 14, 4, 5, 12, 15]
>>> interleave([[0,2,4], [1, 3, 5], [10, 11, 12], [13, 14]])
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "<stdin>", line 10, in interleave
ValueError: Lists are not all the same length!
>>> interleave([[0,2,4]])
[0, 2, 4]

Answer 9

Too late to the party, and there is plenty of good answers but I would also like to provide a simple solution using extend() method:

list1 = [1, 2, 3]
list2 = [10, 20, 30]

new_list = []
for i in range(len(list1)):
    new_list.extend([list1[i], list2[i]])
print(new_list)

Output:

[1, 10, 2, 20, 3, 30]