This article focuses on one of the operations of getting the unique list from a list that contains a possible duplicated. Remove duplicates from list operation has a large number of applications and hence, its knowledge is good to have.
Method 1: Using *set[]
This is the fastest and smallest method to achieve a particular task. It first removes the duplicates and returns a dictionary which has to be converted to list.
Python3
l = [1, 2, 4, 2, 1, 4, 5]
print["Original List: ", l]
res = [*set[l]]
print["List after removing duplicate elements: ", res]
Output:
Original list: [1, 2, 4, 2, 1, 4, 5] List after removing duplicate elements: [1, 2, 4, 5]
Method 2: Using list comprehension
This method has working similarly to the above method, but this is just a one-liner shorthand of a longer method done with the help of list comprehension.
Python3
test_list = [1, 3, 5, 6, 3, 5, 6, 1]
print [& quot
The original list is : & quot
+ str[test_list]]
res = []
[res.append[x] for x in
test_list if x not in res]
print [& quot
The list after removing duplicates : & quot
+ str[res]]
Output :
The original list is : [1, 3, 5, 6, 3, 5, 6, 1] The list after removing duplicates : [1, 3, 5, 6]
Method 3: Using set[]
This is the most popular way by which the duplicates are removed from the list. But the main and notable drawback of this approach is that the ordering of the element is lost in this particular method.
Python3
test_list = [1, 5, 3, 6, 3, 5, 6, 1]
print [& quot
The original list is : & quot
+ str[test_list]]
test_list = list[set[test_list]]
print [& quot
The list after removing duplicates : & quot
+ str[test_list]]
Output :
The original list is : [1, 5, 3, 6, 3, 5, 6, 1] The list after removing duplicates : [1, 3, 5, 6]
Method 4: Using list comprehension + enumerate[]
list comprehension coupled with enumerate function can also achieve this task. It basically looks for already occurred elements and skips adding them. It preserves the list ordering.
Python3
test_list =
[1, 5, 3, 6, 3, 5, 6, 1]
print [& quot
The original list is : & quot
+ str[test_list]]
res = [i for n, i in enumerate[test_list] if i not in test_list[:n]]
print
[& quot
The list after removing duplicates : & quot
+ str[res]]
Output :
The original list is : [1, 5, 3, 6, 3, 5, 6, 1] The list after removing duplicates : [1, 5, 3, 6]
Method 5: Using collections.OrderedDict.fromkeys[]
This is the fastest method to achieve a particular task. It first removes the duplicates and returns a dictionary which has to be converted to list. This works well in the case of strings also.
Python3
from collections import
OrderedDict
test_list = [1, 5, 3, 6, 3, 5, 6, 1]
print [& quot
The original list is : & quot
+ str[test_list]]
res = list[OrderedDict.fromkeys[test_list]]
print [& quot
The list after removing duplicates : & quot
+ str[res]]
Output:
The original list is : [1, 5, 3, 6, 3, 5, 6, 1] The list after removing duplicates : [1, 5, 3, 6]
Method 6: Using in, not in operators
Python3
test_list = [1, 5, 3, 6, 3, 5, 6, 1]
print ["The original list is : "+ str[test_list]]
res = []
for i in test_list:
if i not in res:
res.append[i]
print ["The list after removing duplicates : "+
str[res]]
Output
The original list is : [1, 5, 3, 6, 3, 5, 6, 1] The list after removing duplicates : [1, 5, 3, 6]