Merge Lists In Python By Placing Every Nth Item From One List And Others From Another?

I have two lists, list1 and list2. Here len(list2) << len(list1). Now I want to merge both of the lists such that every nth element of final list is from list2 and the others

Solution 1:

Making the larger list an iterator makes it easy to take multiple elements for each element of the smaller list:

list1 = ['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h']
list2 = ['x', 'y'] 
n = 3

iter1 = iter(list1)
res = []
for x in list2:
    res.extend([next(iter1) for _ in range(n - 1)])
    res.append(x)
res.extend(iter1)

>>> res
['a', 'b', 'x', 'c', 'd', 'y', 'e', 'f', 'g', 'h']

This avoids insert which can be expensive for large lists because each time the whole list needs to be re-created.

Solution 2:

To preserve the original list, you could try the following:

result = copy.deepcopy(list1)
index = n - 1
for elem in list2:
    result.insert(index, elem)
    index += n

result

['a', 'b', 'x', 'c', 'd', 'y', 'e', 'f', 'g', 'h']

Solution 3:

Using the itertools module and the supplementary more_itertools package, you can construct an iterable solution a couple different ways. First the imports:

import itertools as it, more_itertools as mt

This first one seems the cleanest, but it relies on more_itertools.chunked().

it.chain(*mt.roundrobin(mt.chunked(list1, n-1), list2))

This one uses only more_itertools.roundrobin(), whose implementation is taken from the itertools documentation, so if you don't have access to more_itertools you can just copy it yourself.

mt.roundrobin(*([iter(list1)]*(n-1) + [list2]))

Alternatively, this does nearly the same thing as the first sample without using any more_itertools-specific functions. Basically, grouper can replace chunked, but it will add Nones at the end in some cases, so I wrap it in it.takewhile to remove those. Naturally, if you are using this on lists which actually do contain None, it will stop once it reaches those elements, so be careful.

it.takewhile(lambda o: o is not None,
   it.chain(*mt.roundrobin(mt.grouper(n-1, list1), list2))
)

I tested these on Python 3.4, but I believe these code samples should also work in Python 2.7.

Solution 4:

What about the below solution? However I don't have a better one...

>>> list1 = ['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h']
>>> list2 = ['x', 'y']
>>> n = 2
>>> for i in range(len(list2)):
...     list1.insert(n, list2[i])
...     n += 3
...     
... 
>>> list1
['a', 'b', 'x', 'c', 'd', 'y', 'e', 'f', 'g', 'h']

n is 2 because the index of third element in a list is 2, since it starts at 0.

Solution 5:

list(list1[i-1-min((i-1)//n, len(list2))] if i % n or (i-1)//n >= len(list2) else list2[(i-1)//n] for i in range(1, len(list1)+len(list2)+1))

Definitely not pythonic, but I thought it might be fun to do it in a one-liner. More readable (really?) version:

list(
    list1[i-1-min((i-1)//n, len(list2))]
        if i % n or (i-1)//n >= len(list2)
        else
    list2[(i-1)//n]
        for i in range(1, len(list1)+len(list2)+1)
)

Basically, some tinkering around with indexes and determining which list and which index to take next element from.