This article focuses on one of the operation of getting the unique list from a list that contains a possible duplicates and performing its product. This operations has large no. of applications and hence it’s knowledge is good to have.
Method 1 : Naive method + loop In naive method, we simply traverse the list and append the first occurrence of the element in new list and ignore all the other occurrences of that particular element. The task of product is performed using loop.
Python3
# Python 3 code to demonstrate # Unique values Multiplication# using naive methods + loop# getting Product def prod(val) : res = 1 for ele in val: res *= ele return res # initializing listtest_list = [1, 3, 5, 6, 3, 5, 6, 1]print ("The original list is : " + str(test_list))# using naive method + loop# Unique values Multiplication# from list res = []for i in test_list: if i not in res: res.append(i)res = prod(res)# printing list after productprint ("The unique elements product : " + str(res)) |
The original list is : [1, 3, 5, 6, 3, 5, 6, 1] The unique elements product : 90
Time Complexity: O(n*n), where n is the number of elements in the list “test_list”.
Auxiliary Space: O(n), where n is the number of elements in the list “test_list”.
Method 2 : Using set() + loop This is the most popular way by which the duplicated are removed from the list. After that the product of list can be performed using loop.
Python3
# Python 3 code to demonstrate # Unique values Multiplication# using set() + loop# getting Product def prod(val) : res = 1 for ele in val: res *= ele return res # initializing listtest_list = [1, 5, 3, 6, 3, 5, 6, 1]print ("The original list is : " + str(test_list))# Unique values Multiplication# using set() + loop res = prod(list(set(test_list)))# Unique values Multiplication# using set() + loop# printing resultprint ("The unique elements product : " + str(res)) |
The original list is : [1, 3, 5, 6, 3, 5, 6, 1] The unique elements product : 90
Time Complexity: O(n), where n is the number of elements in the list “test_list”.
Auxiliary Space: O(1), constant extra space required
Method #3:Using Counter() function
Python3
# Python 3 code to demonstrate# Unique values Multiplicationfrom collections import Counter# getting Productdef prod(val): res = 1 for ele in val: res *= ele return res# initializing listtest_list = [1, 5, 3, 6, 3, 5, 6, 1]print("The original list is : " + str(test_list))freq = Counter(test_list)# Unique values MultiplicationuniqueValues = freq.keys()res = prod(uniqueValues)# Unique values Multiplication# printing resultprint("The unique elements product : " + str(res)) |
The original list is : [1, 5, 3, 6, 3, 5, 6, 1] The unique elements product : 90
Time Complexity: O(N)
Auxiliary Space: O(N)
Method #4:Using Operator.countOf() method
Python3
# Python 3 code to demonstrate # Unique values Multiplicationimport operator as op# getting Product def prod(val) : res = 1 for ele in val: res *= ele return res # initializing listtest_list = [1, 3, 5, 6, 3, 5, 6, 1]print ("The original list is : " + str(test_list))# using naive method + loop# Unique values Multiplication# from list res = []for i in test_list: if op.countOf(res,i)==0: res.append(i)res = prod(res)# printing list after productprint ("The unique elements product : " + str(res)) |
The original list is : [1, 3, 5, 6, 3, 5, 6, 1] The unique elements product : 90
Auxiliary Space: O(N)
Time Complexity:O(N)
Method 5 : Using numpy
Note: Install numpy module using command “pip install numpy”
Python3
#Using numpy approach#Python 3 code to demonstrate#Unique values Multiplicationimport numpy as np#initializing listtest_list = [1, 3, 5, 6, 3, 5, 6, 1]print("The original list is : " + str(test_list))#using numpy approach#Unique values Multiplicationunique_list = np.unique(test_list)result = np.prod(unique_list)#printing list after productprint("The unique elements product : " + str(result)) |
Output:
The original list is : [1, 3, 5, 6, 3, 5, 6, 1]
The unique elements product : 90
Time Complexity: O(N)
Auxiliary Space: O(N)
Method 6 : Using List comprehension:
Python3
from functools import reducedef prod(val) : if not val: return 1 return reduce(lambda x, y: x * y, val)# initializing listtest_list = [1, 3, 5, 6, 3, 5, 6, 1]print ("The original list is : " + str(test_list))# using set to get unique values# and then using list comprehension to calculate productunique_list = list(set(test_list))result = prod(unique_list)# printing resultprint ("The unique elements product : " + str(result))#This code is contributed by Jyothi pinjala |
The original list is : [1, 3, 5, 6, 3, 5, 6, 1] The unique elements product : 90
Time Complexity: O(N)
Auxiliary Space: O(N)
Method 7: Using itertools groupby()
Python3
import itertools# initializing listtest_list = [1, 3, 5, 6, 3, 5, 6, 1]print ("The original list is : " + str(test_list))# using groupby function from itertools libraryunique_values = [key for key, group in itertools.groupby(sorted(test_list))]# getting the product of the unique valuesresult = 1for value in unique_values: result *= value# printing the resultprint("The unique values multiplication is:", result)#This code is contributed by Vinay Pinjala. |
The original list is : [1, 3, 5, 6, 3, 5, 6, 1] The unique values multiplication is: 90
Time Complexity: O(N)
Auxiliary Space: O(N)
