Python – Interleave two lists of different length

Given two lists of different lengths, the task is to write a Python program to get their elements alternatively and repeat the list elements of the smaller list till the larger list elements get exhausted.

Examples:

Input : test_list1 = ['a', 'b', 'c'], test_list2 = [5, 7, 3, 0, 1, 8, 4]
Output : ['a', 5, 'b', 7, 'c', 3, 'a', 0, 'b', 1, 'c', 8, 'a', 4]
Explanation : Alternate elements from 1st list are printed in 
              cyclic manner once it gets exhausted. Then after exhaustion, 
              again 1st list starts from 'a', with elements left in 2nd list.

Input : test_list1 = [3, 8, 7], test_list2 = [5, 7, 3, 0, 1, 8]
Output : [3, 5, 8, 7, 7, 3, 3, 0, 8, 1, 7, 8]
Explanation : Alternate elements from 1st list are printed in cyclic manner
              once it gets exhausted. 3, 5, 8, 7, 7, 3.. Then after exhaustion, 
              again 1st list starts from 3, with elements left in 2nd list

Method #1 : Using zip() + cycle() + list comprehension 

In this, the repetition of elements in the smaller list is handled using cycle(), and joining is done using zip(). List comprehension performs the task of interleaving simultaneously.

Python3

# Python3 code to demonstrate working of
# Repetitive Interleave 2 lists
# Using zip() + cycle() + list comprehension
from itertools import cycle

# initializing lists
test_list1 = list('abc')
test_list2 = [5, 7, 3, 0, 1, 8, 4]

# printing original lists
print("The original list 1 is : " + str(test_list1))
print("The original list 2 is : " + str(test_list2))

# zip() combining list, after Repetitiveness using cycle()
res = [ele for comb in zip(cycle(test_list1), test_list2) for ele in comb]

# printing result
print("The interleaved list : " + str(res))

Output

The original list 1 is : ['a', 'b', 'c']
The original list 2 is : [5, 7, 3, 0, 1, 8, 4]
The interleaved list : ['a', 5, 'b', 7, 'c', 3, 'a', 0, 'b', 1, 'c', 8, 'a', 4]

Time Complexity: O(n²)
Auxiliary Space: O(n)

Method #2 : Using chain() + zip() + cycle()

Most operations as above method, the only difference being interleaving task is performed using chain().

Python3

# Python3 code to demonstrate working of
# Repetitive Interleave 2 lists
# Using chain() + zip() + cycle()
from itertools import cycle, chain

# initializing lists
test_list1 = list('abc')
test_list2 = [5, 7, 3, 0, 1, 8, 4]

# printing original lists
print("The original list 1 is : " + str(test_list1))
print("The original list 2 is : " + str(test_list2))

# zip() combining list, after Repetitiveness using cycle()
# chain() gets interleaved done
res = list(chain(*zip(cycle(test_list1), test_list2)))

# printing result
print("The interleaved list : " + str(res))

Output

The original list 1 is : ['a', 'b', 'c']
The original list 2 is : [5, 7, 3, 0, 1, 8, 4]
The interleaved list : ['a', 5, 'b', 7, 'c', 3, 'a', 0, 'b', 1, 'c', 8, 'a', 4]

Time Complexity: O(n²)
Auxiliary Space: O(n)

Method 3: using the built-in function map(). 

Python3

from itertools import chain, cycle

# initializing lists
test_list1 = list('abc')
test_list2 = [5, 7, 3, 0, 1, 8, 4]

# Interleaving the lists
# using map() method
res = list(chain(*map(lambda x, y: [x, y], cycle(test_list1), test_list2)))

# printing result
print("The interleaved list : " + str(res))

Output

The interleaved list : ['a', 5, 'b', 7, 'c', 3, 'a', 0, 'b', 1, 'c', 8, 'a', 4]

Time Complexity: O(n^2)
Auxiliary Space: O(n)