Many times we have this particular use-case in which we need to repeat alternate element of list K times. The problems of making a double clone has been discussed but this problem extends to allow a flexible variable to define the number of times the element has to be repeated. Let’s discuss certain ways in which this can be performed.
Method #1 : Using list comprehension This particular task requires generally 2 loops and list comprehension can perform this particular task in one line and hence reduce the lines of codes and improving code readability.
# Python3 code to demonstrate# Alternate Element Repetition# using list comprehension# initializing list of liststest_list = [4, 5, 6]
# printing original listprint("The original list : " + str(test_list))
# declaring magnitude of repetitionK = 3
# using list comprehension# Alternate Element Repetitionres = [ele for idx, ele in enumerate(test_list)
for i in range(K) if idx % 2 == 0]
# printing resultprint("The list after alternate repeating elements : " + str(res))
|
The original list : [4, 5, 6] The list after alternate repeating elements : [4, 4, 4, 6, 6, 6]
Time Complexity: O(n)
Auxiliary Space: O(n)
Method #2: Using itertools.chain.from_iterable() + itertools.repeat() This particular problem can also be solved using python inbuilt functions of itertools library. The repeat function, as name suggests does the task of repetition and grouping into a list is done by the from_iterable function.
# Python3 code to demonstrate# Alternate Element Repetition# using itertools.chain.from_iterable() + itertools.repeat()import itertools
# initializing list of liststest_list = [4, 5, 6]
# printing original listprint("The original list : " + str(test_list))
# declaring magnitude of repetitionK = 3
# using itertools.chain.from_iterable() + itertools.repeat()# Alternate Element Repetitionres = list(itertools.chain.from_iterable(itertools.repeat(ele, K)
for idx, ele in enumerate(test_list) if idx % 2 == 0))
# printing resultprint("The list after alternate repetition elements : " + str(res))
|
The original list : [4, 5, 6] The list after alternate repetition elements : [4, 4, 4, 6, 6, 6]
Time complexity: O(n) as it requires iterating through the original list of size n and repeating each element K times, resulting in a total of n*K operations.
Auxiliary space: O(n*K) as it creates a new list to store the result of the repeated elements.
Method #3 : Using extend() method and * operator
# Python3 code to demonstrate# Alternate Element Repetition# initializing list of liststest_list = [4, 5, 6]
# printing original listprint("The original list : " + str(test_list))
# declaring magnitude of repetitionK = 3
res=[]
for i in range(0,len(test_list)):
if(i%2==0):
res.extend([test_list[i]]*K)
# printing resultprint("The list after alternate repeating elements : " + str(res))
|
The original list : [4, 5, 6] The list after alternate repeating elements : [4, 4, 4, 6, 6, 6]
Time complexity: O(n), where n is the length of the input list. The loop runs n times.
Auxiliary space: O(K), where K is the magnitude of repetition.
Method #4: Using slicing and extend() method
Step-by-step approach:
- Initialize a list of lists named test_list with some values.
- Use list slicing to create a new list named alternate_list that contains every other element of the test_list.
- The slice test_list[::2] means start from the beginning of the list and take every second element. Store the result in alternate_list.
- Print the original list using the print() function. The statement is print(“The original list : ” + str(test_list)).
- Declare the magnitude of repetition K. In this case, K = 3.
- Create an empty list named res to store the final result.
- Use a loop to iterate over each element of the alternate_list. For each element, create a new list with the repeated element K times using the extend() method.
- Append the new list to the res list
- Print the final list after repeating alternate elements K times using the print() function.
Below is the implementation of the above approach:
# Python3 code to demonstrate# Alternate Element Repetition# initializing list of liststest_list = [4, 5, 6]
alternate_list = test_list[::2]
# printing original listprint("The original list : " + str(test_list))
# declaring magnitude of repetitionK = 3
res = []
for i in range(len(alternate_list)):
res.extend([alternate_list[i]]*K)
# printing resultprint("The list after alternate repeating elements : " + str(res))
|
The original list : [4, 5, 6] The list after alternate repeating elements : [4, 4, 4, 6, 6, 6]
The time complexity of this code is O(N), where N is the length of the original list test_list.
The auxiliary space used by this code is O(N), where N is the length of the original list test_list.
Method #5 : Here’s another approach using the numpy library:
Note: Install numpy module using command “pip install numpy”
# Python3 code to demonstrate# Alternate Element Repetitionimport numpy as np
# printing original listtest_list = [4, 5, 6]
K = 3
res = np.repeat(np.array(test_list[::2]), K).tolist()
# printing resultprint("The list after alternate repeating elements :", res)
#This code is contributed by Edula Vinay Kumar Reddy |
Output:
The list after alternate repeating elements : [4, 4, 4, 6, 6, 6]
Time Complexity: O(n), where n is the length of the input list. The time complexity is linear with the length of the list as it is just repeating and converting the elements in the list.
Auxiliary Space: O(n), where n is the length of the output list after repeating the elements. The space complexity is linear with the length of the output list as it requires memory to store the repeated elements.
Method 6: Using map()+ lambda function
Uses a map() function with a lambda function to repeat the selected elements K times, and the extend() function to append the repeated elements to a new list.
- Initialize the list test_list with some values.
- Print the original list using the print() function and string concatenation.
- Declare the magnitude of repetition as K.
- Initialize an empty list res to store the alternate repeated elements.
- Iterate through the elements in test_list using the enumerate() function.
- Check if the index of the current element is even using the modulo operator %.
- If the index is even, use a map() function with a lambda function to repeat the selected element K times.
- Use the extend() function to append the repeated elements to the res list.
- Print the resulting list using the print() function and string concatenation.
Below is the implementation of the above approach:
# Initializing list of liststest_list = [4, 5, 6]
# Printing original listprint("The original list: " + str(test_list))
# Declaring magnitude of repetitionK = 3
# Using map and lambda function for alternate element repetitionres = []
for i, val in enumerate(test_list):
if i % 2 == 0:
res.extend(map(lambda x: val, range(K)))
# Printing resultprint("The list after alternate repeating elements: " + str(res))
|
The original list: [4, 5, 6] The list after alternate repeating elements: [4, 4, 4, 6, 6, 6]
Time complexity: O(NK) where N is the length of the input list.
Auxiliary space: O(NK)
Method #7 : Using extend()+operator.mul() methods
Approach
- Initiate a for loop with index from 0 to length of list
- Find alternate element index by checking if the index is divisible by 2
- Access the alternate element by index and repeat element by using operator.mul() and append it to output list
- Display output list
# Python3 code to demonstrate# Alternate Element Repetition# initializing list of liststest_list = [4, 5, 6]
# printing original listprint("The original list : " + str(test_list))
# declaring magnitude of repetitionK = 3
res=[]
import operator
for i in range(0,len(test_list)):
if(i%2==0):
res.extend(operator.mul([test_list[i]],K))
# printing resultprint("The list after alternate repeating elements : " + str(res))
|
The original list : [4, 5, 6] The list after alternate repeating elements : [4, 4, 4, 6, 6, 6]
Time complexity: O(NK) where N is the length of the input list.
Auxiliary space: O(NK)