通过放置一个列表中的每个第n个项目和另一个列表中的其他项目来合并Python中的列表？

Question

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 from list1 .

For example:

list1 = ['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h']

list2 = ['x', 'y']

n = 3

Now the final list should be:

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

What is the most Pythonic way to achieve this?

I want to add all elements of list2 to the final list, final list should include all elements from list1 and list2 .

Answer 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.

Answer 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']

Answer 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 None s 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.

Answer 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.

Answer 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.

Answer 6

Yet another way, calculating the slice steps:

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

res = []
m = n - 1
start, end = 0, m
for x in list2:
    res.extend(list1[start:end])
    res.append(x)
    start, end = end, end + m
res.extend(list1[start:])

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

Answer 7

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

for index, item in enumerate(list2):
    new[n * (index + 1) - 1: n * (index + 1) - 1] = item

print(new)

Answer 8

I admire @David Z's use of more_itertools . Updates to the tools can simplify the solution:

import more_itertools as mit

n = 3
groups = mit.windowed(list1, n-1, step=n-1)
list(mit.flatten(mit.interleave_longest(groups, list2)))
# ['a', 'b', 'x', 'c', 'd', 'y', 'e', 'f', 'g', 'h']

Summary: list2 is being interleaved into groups from list1 and finally flattened into one list.

Notes

1. groups : n-1 size sliding windows, eg [('a', 'b'), ('c', 'd'), ('e', 'f'), ('g', 'h')]
2. interleave_longest is presently equivalent to roundrobin
3. None is the default fillvalue. Optionally remove with filter(None, ...)

Answer 9

Maybe here is another solution, slice the list1 the correct index then add the element of list2 into list1 .

>>> list1 = ['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h']
>>> list2 = ['x', 'y'] 
>>> n = 3
>>> for i in range(len(list2)):
...     list1 = list1[:n*(i+1) - 1] + list(list2[i]) + list1[n*(i+1)-1:]
... 
>>> list1
['a', 'b', 'x', 'c', 'd', 'y', 'e', 'f', 'g', 'h']