Remove common elements from two list in Python
Last Updated :
28 Feb, 2023
Given two lists, the task is to write a Python program to remove all the common elements of two lists.
Examples:
Input : list1 = [1, 2, 3, 4, 5]
list2 = [4, 5, 6, 7, 8,]
Output : list1 = [1, 2, 3]
list2 = [6, 7, 8]
Explanation: Lists after removing common elements of both the lists i.e, 4 and 5.
Input : list1 = [1, 2, 3]
list2 = [1, 2, 3]
Output : list1 = []
list2 = []
Explanation: They have all the elements in common in
between them.
Method 1: Using Remove() Method
The remove() method removes the first matching element (which is passed as an argument) from the list.
Python3
def remove_common(a, b):
for i in a[:]:
if i in b:
a.remove(i)
b.remove(i)
print("list1 : ", a)
print("list2 : ", b)
if __name__ == "__main__":
a = [1, 2, 3, 4, 5]
b = [4, 5, 6, 7, 8]
remove_common(a, b)
|
Output
list1 : [1, 2, 3]
list2 : [6, 7, 8]
Time Complexity: O(n)
Auxiliary Space: O(1)
List comprehension gives a shorter syntax when you want to create a new list based on the elements of the existing list.
Python3
def remove_common(a, b):
a, b = [i for i in a if i not in b], [j for j in b if j not in a]
print("list1 : ", a)
print("list2 : ", b)
if __name__ == "__main__":
a = [1, 2, 3, 4, 5]
b = [4, 5, 6, 7, 8]
remove_common(a, b)
|
Time Complexity: O(n)
Space Complexity: O(n)
The difference operator – gets items in the first set but not in the second.
Python3
def remove_common(a, b):
a, b = list(set(a) - set(b)), list(set(b) - set(a))
print("list1 : ", a)
print("list2 : ", b)
if __name__ == "__main__":
a = [1, 2, 3, 4, 5]
b = [4, 5, 6, 7, 8]
remove_common(a, b)
|
Output
list1 : [1, 2, 3]
list2 : [8, 6, 7]
The difference() method in python returns a set that contains the difference between two sets i.e, the returned set contains items that exist only in the first set and excludes elements present in both sets.
Python3
def remove_common(a, b):
a, b = list(set(a).difference(b)), list(set(b).difference(a))
print("list1 : ", a)
print("list2 : ", b)
if __name__ == "__main__":
a = [1, 2, 3, 4, 5]
b = [4, 5, 6, 7, 8]
remove_common(a, b)
|
Output
list1 : [1, 2, 3]
list2 : [6, 7, 8]
Method #5: Using Counter() Function
Python3
from collections import Counter
def remove_common(a, b):
freq1 = Counter(a)
freq2 = Counter(b)
for key in freq1:
if key in freq2:
a.remove(key)
b.remove(key)
print("list1 : ", a)
print("list2 : ", b)
if __name__ == "__main__":
a = [1, 2, 3, 4, 5]
b = [4, 5, 6, 7, 8]
remove_common(a, b)
|
Output
list1 : [1, 2, 3]
list2 : [6, 7, 8]
This method uses the filterfalse() function along with the __contains__() method of sets to create new lists that contain only elements that are not present in the other list
Python3
from itertools import filterfalse
def remove_common(a, b):
a ,b= list(filterfalse(set(b).__contains__, a)), list(filterfalse(set(a).__contains__, b))
print("list1 : ", a)
print("list2 : ", b)
if __name__ == "__main__":
a = [1, 2, 3, 4, 5]
b = [4, 5, 6, 7, 8]
remove_common(a, b)
|
Output
list1 : [1, 2, 3]
list2 : [6, 7, 8]
Time complexity: O(N2).
Auxiliary space: O(1)
Another approach: Using set intersection
Python3
def remove_common(a, b):
common = set(a) & set(b)
a = [i for i in a if i not in common]
b = [i for i in b if i not in common]
print("list1 : ", a)
print("list2 : ", b)
a = [1, 2, 3, 4, 5]
b = [4, 5, 6, 7, 8]
remove_common(a, b)
|
Output
list1 : [1, 2, 3]
list2 : [6, 7, 8]
Time Complexity: O(nlogn), where n is the number of elements in the list. The set intersection operation takes O(nlogn) time.
Auxiliary Space: O(n), where n is the number of elements in the list. The set requires O(n) space to store the elements.
Explanation:
We convert both the lists into sets and use the set intersection operation to find the common elements.
Then we use list comprehension to remove the common elements from both the lists and return the updated lists.
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