Python | Append Odd element twice

Given a list of number, the task is to create a new list from the initial list with the condition to append every odd element twice.

Below are some ways to achieve the above task.

Method #1: Using list comprehension

# Python code to create a new list from initial list 
# with condition to append every odd element twice. 
  
# List initialization 
Input = [1, 2, 3, 8, 9, 11] 
  
# Using list comprehension  
Output = [elem for x in Input for elem in (x, )*(x % 2 + 1)] 
  
# printing  
print("Initial list is:'", Input) 
print("New list is:", Output) 

Output:

Initial list is:' [1, 2, 3, 8, 9, 11]
New list is: [1, 1, 2, 3, 3, 8, 9, 9, 11, 11]

Method #2: Using itertools

# Python code to create a new list from initial list 
# with condition to append every odd element twice. 
   
# Importing 
from itertools import chain 
   
# List initialization 
Input = [1, 2, 3, 8, 9, 11] 
   
# Using list comprehension  and chain 
Output = list(chain.from_iterable([i]  
              if i % 2 == 0 else [i]*2 for i in Input)) 
   
# printing  
print("Initial list is:'", Input) 
print("New list is:", Output) 

Output:

Initial list is:' [1, 2, 3, 8, 9, 11]
New list is: [1, 1, 2, 3, 3, 8, 9, 9, 11, 11]

Method #3: Using Numpy array

# Python code to create a new list from initial list 
# with condition to append every odd element twice. 
  
# Importing 
import numpy as np 
  
# List initialization 
Input = [1, 2, 3, 8, 9, 11] 
Output = [] 
  
# Using Numpy repeat 
for x in Input: 
    (Output.extend(np.repeat(x, 2, axis = 0)) 
      if x % 2 == 1 else Output.append(x)) 
  
# printing  
print("Initial list is:'", Input) 
print("New list is:", Output) 

Output:

Initial list is:' [1, 2, 3, 8, 9, 11]
New list is: [1, 1, 2, 3, 3, 8, 9, 9, 11, 11]

My Personal Notes

Suggest a Topic

Recommended Posts: