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Python – Rear Kth elements

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  • Last Updated : 10 Jul, 2020

Given a list, the task is to extract all Kth elements from rear end.

Input : test_list = [3, 4, 5, 2, 1], K = 2
Output : [1, 5, 3]
Explanation : Every 2nd elements are extracted from rear starting from 1.

Input : test_list = [3, 4, 5], K = 1
Output : [5, 4, 3]
Explanation : Every elements are extracted from rear starting from 1.

Method #1 : Using loop
The is brute method to solve this problem. In this, we reverse and then perform iteration to get each Kth multiple element.




# Python3 code to demonstrate working of 
# Rear Kth elements
# Using loop
  
# initializing list
test_list = [3, 5, 7, 9, 10, 2, 8, 6
  
# printing original list
print("The original list is : " + str(test_list))
  
# initializing K 
K = 3
  
# Rear Kth elements
res = []
test_list.reverse()
for idx, ele in enumerate(test_list):
      
    # Extracting elements divisible by K
    if idx % K == 0:
        res.append(ele)
  
# printing result 
print("Rear Kth elements : " + str(res))
Output :
The original list is : [3, 5, 7, 9, 10, 2, 8, 6]
Rear Kth elements : [6, 10, 5]

 

Method #2 : Using list slicing
This is shorthand version solution to this problem. We use power of list slicing skip to skip Kth element, and also reverse it.




# Python3 code to demonstrate working of 
# Rear Kth elements
# Using list slicing
  
# initializing list
test_list = [3, 5, 7, 9, 10, 2, 8, 6
  
# printing original list
print("The original list is : " + str(test_list))
  
# initializing K 
K = 3
  
# Rear Kth elements
# Starting slicing from Rear (-1) and extracting all Kth elements
res = test_list[-1::-K]
  
# printing result 
print("Rear Kth elements : " + str(res))
Output :
The original list is : [3, 5, 7, 9, 10, 2, 8, 6]
Rear Kth elements : [6, 10, 5]

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