python merge two lists (even/odd elements)

Question

Given two lists, I want to merge them so that all elements from the first list are even-indexed (preserving their order) and all elements from second list are odd-indexed (a

Answer 1

Use the roundrobin recipe from itertools:

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))
>>> list(roundrobin(x,y))
[0, 3, 1, 4, 2]

Answer 2

This is simple enough although not nearly as flexible as roundrobin:

def paired(it1, it2):
    it2 = iter(it2)
    for item in it1:
        yield item
        yield next(it2)

tested in 2.7.5:

>>> x = [0, 1, 2]
>>> y = [3, 4]
>>> print list(paired(x, y))
[0, 3, 1, 4, 2]

Note that it stops as soon as the list y runs out (because next(it2) raises StopIteration).

Answer 3

Here's something you can use. (Use list(izip_longest(...)) for Py2x)

>>> from itertools import chain
>>> from itertools import zip_longest
>>> list(filter(lambda x: x != '', chain.from_iterable(zip_longest(x, y, fillvalue = ''))))
[0, 3, 1, 4, 2]

This works for arbitrary length lists like follows -

>>> x = [0, 1, 2, 3, 4]
>>> y = [5, 6]
>>> list(filter(lambda x: x != '', chain.from_iterable(zip_longest(x, y, fillvalue = ''))))
[0, 5, 1, 6, 2, 3, 4]

Explanation on it's working -

1. zip_longest(...) with a fill value zips the lists and fills in the given fill value for iterables of unequal length. So, for your original example, it evaluates to something like [(0, 3), (1, 4), (2, '')]
2. We need to flatten the result because this method gives us a list of tuples. For that we use chain.from_iterable(...) giving us something like [0, 3, 1, 4, 2, ''].
3. We now use filter(...) to remove all occurences of '' and we get the required answer.

Answer 4

It can be done with slicing. Do count and slice in terminal:

>>> list1=['Apple','Mango','Orange']
>>> list2=['One','Two','Three']
>>> list = [None]*(len(list1)+len(list2))
>>> list[::2] = list1
>>> list[1::2] = list2
>>> list

Output:

 ['Apple', 'One', 'Mango', 'Two', 'Orange', 'Three']

Answer 5

You can simply do:

for i,v in enumerate(y):
    x.insert(2*i+1,v)

this takes the advantage that insert will use the last index when it is overpassed.

One example:

x = [0,1,2,3,4,5]
y = [100, 11,22,33,44,55,66,77]
print x
# [0, 100, 1, 11, 2, 22, 3, 33, 4, 44, 5, 55, 66, 77]

Answer 6

Try this:

x = [0,1,2,10,11]
y = [3,4]

n =  2*max([len(x),len(y)])
res = n *[None]
res[:2*len(x):2] = x
res[1:2*len(y):2] = y
res = [x for x in res if x!=None]
print res

It should work for unevenly long lists.