Given two lists, the task is to write a python program that marks 1 for elements present in the other list else mark 0.
Input : test_list = [5, 2, 1, 9, 8, 0, 4], search_list = [1, 10, 8, 3, 9]
Output : [0, 0, 1, 1, 1, 0, 0]
Explanation : 1, 9, 8 are present in test_list at position 2, 3, 4 and are masked by 1. Rest are masked by 0.Input : test_list = [5, 2, 1, 19, 8, 0, 4], search_list = [1, 10, 8, 3, 9]
Output : [0, 0, 1, 0, 1, 0, 0]
Explanation : 1, 8 are present in test_list at position 2, 4 and are masked by 1. Rest are masked by 0.
Method #1: Using list comprehension
In this, we iterate through search list and in operator used to check composition using list comprehension and assign 1 for presence and 0 for absence.
Python3
# Python3 code to demonstrate working of# Boolean composition mask in list# Using list comprehension# initializing listtest_list = [5, 2, 1, 9, 8, 0, 4]# printing original listprint("The original list is : " + str(test_list))# initializing search listsearch_list = [1, 10, 8, 3, 9]# list comprehension iteration and in operator# checking compositionres = [1 if ele in search_list else 0 for ele in test_list]# printing resultprint("The Boolean Masked list : " + str(res)) |
Output:
The original list is : [5, 2, 1, 9, 8, 0, 4] The Boolean Masked list : [0, 0, 1, 1, 1, 0, 0]
Time Complexity: O(n)
Auxiliary Space: O(n)
Method #2: Using set() + list comprehension
In this, duplicate elements are removed from the search list to reduce search space using set(). Rest all the operations are similar to the above method.
Python3
# Python3 code to demonstrate working of# Boolean composition mask in list# Using set() + list comprehension# initializing listtest_list = [5, 2, 1, 9, 8, 0, 4]# printing original listprint("The original list is : " + str(test_list))# initializing search listsearch_list = [1, 10, 8, 3, 9]# list comprehension iteration and in operator# checking composition# set() removes duplicatesres = [1 if ele in set(search_list) else 0 for ele in test_list]# printing resultprint("The Boolean Masked list : " + str(res)) |
Output:
The original list is : [5, 2, 1, 9, 8, 0, 4] The Boolean Masked list : [0, 0, 1, 1, 1, 0, 0]
Time Complexity: O(n)
Space Complexity: O(n)
Method #3 : Using count() method
Python3
# Python3 code to demonstrate working of# Boolean composition mask in list# initializing listtest_list = [5, 2, 1, 9, 8, 0, 4]# printing original listprint("The original list is : " + str(test_list))# initializing search listsearch_list = [1, 10, 8, 3, 9]res=[]for i in test_list: if(search_list.count(i)>=1): res.append(1) else: res.append(0)# printing resultprint("The Boolean Masked list : " + str(res)) |
The original list is : [5, 2, 1, 9, 8, 0, 4] The Boolean Masked list : [0, 0, 1, 1, 1, 0, 0]
Time Complexity: O(n)
Auxiliary Space: O(n)
Method #3 : Using map() and lambda function
Python
# Python3 code to demonstrate working of# Boolean composition mask in list# initializing listtest_list = [5, 2, 1, 9, 8, 0, 4]# printing original listprint("The original list is : " + str(test_list))# initializing search listsearch_list = [1, 10, 8, 3, 9]res= map(lambda x : 1 if x in search_list else 0 , test_list)# printing resultprint("The Boolean Masked list : " , list(res)) |
The original list is : [5, 2, 1, 9, 8, 0, 4]
('The Boolean Masked list : ', [0, 0, 1, 1, 1, 0, 0])
Time complexity: O(n*n), where n is the length of the test_list. The map() and lambda function takes O(n*n) time
Auxiliary Space: O(n), extra space of size n is required
# Method 4 : Using collections.Counter()
Python3
import collections# initializing listtest_list = [5, 2, 1, 9, 8, 0, 4] # printing original listprint("The original list is : " + str(test_list)) # initializing search listsearch_list = [1, 10, 8, 3, 9]# Create a Counter from search_listcounter = collections.Counter(search_list)# Iterate through test_list and check if it's present in counter as keyres = [1 if counter.get(ele) else 0 for ele in test_list] # printing resultprint("The Boolean Masked list : " + str(res)) |
The original list is : [5, 2, 1, 9, 8, 0, 4] The Boolean Masked list : [0, 0, 1, 1, 1, 0, 0]
# Time Complexity : O(N), where N is the number of elements in test_list
# Auxiliary Space : O(M), where M is the number of elements in search_list
