Python program to mask a list using values from another list

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 list
test_list = [5, 2, 1, 9, 8, 0, 4]

# printing original list
print("The original list is : " + str(test_list))

# initializing search list
search_list = [1, 10, 8, 3, 9]

# list comprehension iteration and in operator
# checking composition
res = [1 if ele in search_list else 0 for ele in test_list]

# printing result
print("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 list
test_list = [5, 2, 1, 9, 8, 0, 4]

# printing original list
print("The original list is : " + str(test_list))

# initializing search list
search_list = [1, 10, 8, 3, 9]

# list comprehension iteration and in operator
# checking composition
# set() removes duplicates
res = [1 if ele in set(search_list) else 0 for ele in test_list]

# printing result
print("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 list
test_list = [5, 2, 1, 9, 8, 0, 4]

# printing original list
print("The original list is : " + str(test_list))

# initializing search list
search_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 result
print("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 #3 : Using map() and lambda function

Python

# Python3 code to demonstrate working of
# Boolean composition mask in list

# initializing list
test_list = [5, 2, 1, 9, 8, 0, 4]

# printing original list
print("The original list is : " + str(test_list))

# initializing search list
search_list = [1, 10, 8, 3, 9]

res= map(lambda x : 1 if x in search_list else 0 , test_list)

# printing result
print("The Boolean Masked list : " , list(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*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 list
test_list = [5, 2, 1, 9, 8, 0, 4]

# printing original list
print("The original list is : " + str(test_list))

# initializing search list
search_list = [1, 10, 8, 3, 9]

# Create a Counter from search_list
counter = collections.Counter(search_list)

# Iterate through test_list and check if it's present in counter as key
res = [1 if counter.get(ele) else 0 for ele in test_list]

# printing result
print("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), where N is the number of elements in test_list
# Auxiliary Space : O(M), where M is the number of elements in search_list