Python insert list items into every 3rd slot of larger list-CodePudding

So I have two lists in python,


letters = ['1', '2', '4', '5', '7', '8']
list = ['3', '6', '9']

and I would like to merge these two lists into one. That looks like so:

['1', '2', '3', '4', '5', '6', '7', '8', '9']

Here is the code I am running


letters = ['1','2','4','5','7','8']
list = ['3','6','9']

new_list = []
n = 2

length = len(list)
i = 0

while i < length:
    for start_index in range(0, len(letters), n):
        new_list.extend(letters[start_index:start_index n])
        print(list[i])
        new_list.append(list[i])
        i  = 1

new_list.pop()
print(new_list)

which results in:

['1', '2', '3', '4', '5', '6', '7', '8']

whilst I am expecting

['1', '2', '3', '4', '5', '6', '7', '8', '9']

also the order is not alpha-numeric, the goal is to iterate through the smaller list and place each item into every 3rd slot of the larger list.

CodePudding user response：

Assuming that the numbers given are just placeholders, and not meant to actually be sorted:

One way to interleave items would be using iterators manually:

letters = ['1','2','4','5','7','8']
lst = ['3','6','9']

a = iter(letters)
b = iter(lst)
new_list = []
try:
    while True:
        if len(new_list) % 3 == 2:
            new_list.append(next(b))
        else:
            new_list.append(next(a))
except StopIteration:
    pass

Or for an entirely different, more terse, solution, you could use itertools.chain alongside zip():

import itertools

letters = ['1','2','4','5','7','8']
lst = ['3','6','9']

new_list = list(itertools.chain(*zip(letters[::2], letters[1::2], lst)))
# letters[::2]   ->  1, 4, 7
# letters[1::2]  ->  2, 5, 8
# lst            ->  3, 6, 9
#
# zip(...) -> (1, 2, 3), (4, 5, 6), (7, 8, 9)
# chain(...) -> 1, 2, 3, 4, 5, 6, 7, 8, 9

CodePudding user response：

I have edited your own code and used for loop instead of while:

letters = ['1','2','4','5','7','8']
myList = ['3','6','9']

new_list = []
length = len(myList)   len(letters)
j = 0
for i in range(1,length 1):
  if i % 3 == 0:
    new_list.append(myList[i//3 - 1])
    j -= 1
  else:
    new_list.append(letters[i   j - 1])
print(new_list)

Output

['1', '2', '3', '4', '5', '6', '7', '8', '9']

Let's have another example:

letters = ['a','b','c','d','e','f']
myList = ['1','2','3']

new_list = []
length = len(myList)   len(letters)
j = 0
for i in range(1,length 1):
  if i % 3 == 0:
    new_list.append(myList[i//3 - 1])
    j -= 1
  else:
    new_list.append(letters[i   j - 1])
print(new_list)

Output

['a', 'b', '1', 'c', 'd', '2', 'e', 'f', '3']

Note that this only works when the number of elements in the myList variable will be multiple of 3.

CodePudding user response：

In Python 3.10, you can use itertools.pairwise:

import itertools

letters = ['1','2','4','5','7','8']
new = ['3', '6', '9']

result = []
for pair, single in zip(itertools.pairwise(letters), new):
    result.extend(pair)
    result.append(single)

Or more generally (and for a pairwise replacement for before 3.10):

def by_n(iterable, n):
    return zip(*[iter(iterable)]*n)

result = []
for triple, single in zip(by_n(letters, 3), new):
    result.extend(triple)
    result.append(single)