Python | Filter a list based on the given list of strings

Given a List, the task is to filter elements from list based on another list of strings. These type of problems are quite common while scraping websites. 
Examples: 

Input:
List_string1 = ['key', 'keys', 'keyword', 'keychain', 'keynote']
List_string2 = ['home/key/1.pdf',
         'home/keys/2.pdf', 
         'home/keyword/3.pdf', 
         'home/keychain/4.pdf',
         'home/Desktop/5.pdf', 
         'home/keynote/6.pdf']
Output:
['home/Desktop/5.pdf']

Explanation: We filter only those element from 
list_string2 that do not have string in list_string1

Below are some ways to achieve the above task. 
Method #1: Using Iteration 

Python3

# Python code to filter element from list
# based on another list of string.

# List Initialization
Input = ['key', 'keys', 'keyword', 'keychain', 'keynote']
Input_string = ['home/key/1.pdf',
        'home/keys/2.pdf',
        'home/keyword/3.pdf',
        'home/keychain/4.pdf',
        'home/Desktop/5.pdf',
        'home/keynote/6.pdf']

Output = Input_string.copy()
temp = []

# Using iteration
for elem in Input_string:
    for n in Input:
        if n in elem:
            temp.append(elem)

for elem in temp:
    if elem in Output:
        Output.remove(elem)

# Printing
print("List of keywords are:", Input)
print("Given list:", Input_string)
print("filtered list is :", Output)

Output

List of keywords are: ['key', 'keys', 'keyword', 'keychain', 'keynote']
Given list: ['home/key/1.pdf', 'home/keys/2.pdf', 'home/keyword/3.pdf', 'home/keychain/4.pdf', 'home/Desktop/5.pdf', 'home/keynote/6.pdf']
filtered list is : ['home/Desktop/5.pdf']

Time Complexity: O(n*n), where n is the length of the list test_list 
Auxiliary Space: O(n) additional space of size n is created where n is the number of elements in the res list 

Method #2: Using list comprehension 

Python3

# Python code to filter element from list
# based on another list of string.

# List Initialization
Input = ['key', 'keys', 'keyword', 'keychain', 'keynote']
Input_string = ['home/key/1.pdf',
        'home/keys/2.pdf',
        'home/keyword/3.pdf',
        'home/keychain/4.pdf',
        'home/Desktop/5.pdf',
        'home/keynote/6.pdf']

# Using list comprehension
Output = [b for b in Input_string if
        all(a not in b for a in Input)]

# Printing
print("List of keywords are:", Input)
print("Given list:", Input_string)
print("filtered list is :", Output)

Output

List of keywords are: ['key', 'keys', 'keyword', 'keychain', 'keynote']
Given list: ['home/key/1.pdf', 'home/keys/2.pdf', 'home/keyword/3.pdf', 'home/keychain/4.pdf', 'home/Desktop/5.pdf', 'home/keynote/6.pdf']
filtered list is : ['home/Desktop/5.pdf']

Time Complexity: O(n) where n is the number of elements in the list 
Auxiliary Space: O(n), where n is the number of elements in the new output list 

Approach #3: Using the filter function and a lambda function

This approach uses the built-in filter function and a lambda function to filter the elements from the list. The lambda function returns True if none of the elements in Input are present in the element being considered, and False otherwise. The filter function filters the list Input_string based on the output of the lambda function.

Python3

# Python code to filter element from list
# based on another list of string.

# List Initialization
Input = ['key', 'keys', 'keyword', 'keychain', 'keynote']
Input_string = ['home/key/1.pdf',
         'home/keys/2.pdf',
         'home/keyword/3.pdf',
         'home/keychain/4.pdf',
         'home/Desktop/5.pdf',
         'home/keynote/6.pdf']

# Using the filter function and a lambda function
Output = list(filter(lambda x: all(y not in x for y in Input), Input_string))

# Printing
print("List of keywords are:", Input)
print("Given list:", Input_string)
print("filtered list is :", Output)
#This code is contributed by Edula Vinay Kumar Reddy

Output

List of keywords are: ['key', 'keys', 'keyword', 'keychain', 'keynote']
Given list: ['home/key/1.pdf', 'home/keys/2.pdf', 'home/keyword/3.pdf', 'home/keychain/4.pdf', 'home/Desktop/5.pdf', 'home/keynote/6.pdf']
filtered list is : ['home/Desktop/5.pdf']

Time complexity: O(n*m), where n is the length of Input_string and m is the length of Input
Auxiliary Space: O(n)