This article focuses on one of the operation of getting the unique list from a list that contains a possible duplicated and finding its product. This operations has large no. of applications and hence it’s knowledge is good to have.
Method 1 : Naive method
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 performing product is done using loop.
# Python 3 code to demonstrate # Duplication Removal List Product # using naive methods # getting Product def prod(val) : res = 1 for ele in val: res * = ele return res # initializing list test_list = [ 1 , 3 , 5 , 6 , 3 , 5 , 6 , 1 ] print ( "The original list is : " + str (test_list)) # using naive method # Duplication Removal List Product res = [] for i in test_list: if i not in res: res.append(i) res = prod(res) # printing list after removal print ( "Duplication removal list product : " + str (res)) |
The original list is : [1, 3, 5, 6, 3, 5, 6, 1] Duplication removal list product : 90
Method 2 : Using list comprehension
This method has working similar to the above method, but this is just a one-liner shorthand of longer method done with the help of list comprehension.
# Python 3 code to demonstrate # Duplication Removal List Product # using list comprehension # getting Product def prod(val) : res = 1 for ele in val: res * = ele return res # initializing list test_list = [ 1 , 3 , 5 , 6 , 3 , 5 , 6 , 1 ] print ( "The original list is : " + str (test_list)) # using list comprehension # Duplication Removal List Product res = [] [res.append(x) for x in test_list if x not in res] res = prod(res) # printing list after removal print ( "Duplication removal list product : " + str (res)) |
The original list is : [1, 3, 5, 6, 3, 5, 6, 1] Duplication removal list product : 90
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