Python being the language of magicians can be used to perform many tedious and repetitive tasks in a easy and concise manner and having the knowledge to utilize this tool to the fullest is always useful. One such small application can be finding sum of cubes of list in just one line. Let’s discuss certain ways in which this can be performed.
Method #1 : Using reduce() + lambda The power of lambda functions to perform lengthy tasks in just one line, allows it combined with reduce which is used to accumulate the subproblem, to perform this task as well. Works with only Python 2.
Python
# Python code to demonstrate # Cube Summation in List # using reduce() + lambda # initializing list test_list = [ 3 , 5 , 7 , 9 , 11 ] # printing original list print ( "The original list is : " + str (test_list)) # using reduce() + lambda # Cube Summation in List res = reduce ( lambda i, j: i + j * j * j, [test_list[: 1 ][ 0 ] * * 3 ] + test_list[ 1 :]) # printing result print ( "The sum of cubes of list is : " + str (res)) |
The original list is : [3, 5, 7, 9, 11] The sum of cubes of list is : 2555
Time Complexity: O(n)
Auxiliary Space: O(1)
Method #2 : Using map() + sum() The similar solution can also be obtained using the map function to integrate and sum function to perform the summation of the cube number.
Python3
# Python3 code to demonstrate # Cube Summation in List # using sum() + map() # initializing list test_list = [ 3 , 5 , 7 , 9 , 11 ] # printing original list print ( "The original list is : " + str (test_list)) # using sum() + map() # Cube Summation in List res = sum ( map ( lambda i : i * i * i, test_list)) # printing result print ( "The sum of cubes of list is : " + str (res)) |
The original list is : [3, 5, 7, 9, 11] The sum of cubes of list is : 2555
Time complexity: O(n), where n is the length of the input list.
Auxiliary space: O(1), constant extra space is required
Method #3 : Using math.pow() and sum() methods
Python3
# Python code to demonstrate # Cube Summation in List # initializing list test_list = [ 3 , 5 , 7 , 9 , 11 ] # printing original list print ( "The original list is : " + str (test_list)) # Cube Summation in List x = [] for i in test_list: import math x.append(math. pow (i, 3 )) res = int ( sum (x)) # printing result print ( "The sum of cubes of list is : " + str (res)) |
The original list is : [3, 5, 7, 9, 11] The sum of cubes of list is : 2555
Time Complexity : O(N)
Auxiliary Space : O(N)
Method #4 : Using list comprehension:
Python3
# Initializing list test_list = [ 3 , 5 , 7 , 9 , 11 ] # Printing original list print ( "The original list is :" , test_list) # Using list comprehension res = sum ([num * * 3 for num in test_list]) # Printing result print ( "The sum of cubes of list is :" , res) #This code is contributed by Jyothi pinjala. |
The original list is : [3, 5, 7, 9, 11] The sum of cubes of list is : 2555
Time Complexity: O(N)
Auxiliary Space: O(1)
Method#5: Using for loop
Python3
def sum_of_cubes(lst): res = 0 for num in lst: res + = num * * 3 return res test_list = [ 3 , 5 , 7 , 9 , 11 ] print ( "The original list is:" , test_list) print ( "The sum of cubes of list is:" , sum_of_cubes(test_list)) #this code is contributed By Vinay pinjala. |
The original list is: [3, 5, 7, 9, 11] The sum of cubes of list is: 2555
Time Complexity: O(N)
Auxiliary Space: O(1)