Python – All occurrences of Substring from the list of strings

Given a list of strings and a list of substring. The task is to extract all the occurrences of a substring from the list of strings.

Examples:

Input : test_list = [“gfg is best”, “gfg is good for CS”, “gfg is recommended for CS”]

subs_list = [“gfg”, “CS”] 

Output : [‘gfg is good for CS’, ‘gfg is recommended for CS’] 

Explanation : Result strings have both “gfg” and “CS”.

Input : test_list = [“gfg is best”, “gfg is recommended for CS”]

subs_list = [“gfg”] 

Output : [“gfg is best”, “gfg is recommended for CS”] 

Explanation : Result strings have “gfg”. 

Method #1 : Using loop + in operator

The combination of the above functions can be used to solve this problem. In this, we run a loop to extract all strings and also all substring in the list. The in operator is used to check for substring existence.

Python3

# Python3 code to demonstrate working of
# Strings with all Substring Matches
# Using loop + in operator

# initializing list
test_list = ["gfg is best", "gfg is good for CS",
             "gfg is recommended for CS"]

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

# initializing Substring List
subs_list = ["gfg", "CS"]

res = []
for sub in test_list:
    flag = 0
    for ele in subs_list:

        # checking for non existence of
        # any string
        if ele not in sub:
            flag = 1
            break
    if flag == 0:
        res.append(sub)

# printing result
print("The extracted values : " + str(res))

Output:

The original list is : [‘gfg is best’, ‘gfg is good for CS’, ‘gfg is recommended for CS’] The extracted values : [‘gfg is good for CS’, ‘gfg is recommended for CS’]

Time Complexity: O(n²)

Auxiliary Space: O(n)

Method #2 : Using all() + list comprehension

This is a one-liner approach with the help of which we can perform this task. In this, we check for all values existence using all(), and list comprehension is used to iteration of all the containers.

Python3

# Python3 code to demonstrate working of
# Strings with all Substring Matches
# Using all() + list comprehension

# initializing list
test_list = ["gfg is best", "gfg is good for CS",
             "gfg is recommended for CS"]

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

# initializing Substring List
subs_list = ["gfg", "CS"]

# using all() to check for all values
res = [sub for sub in test_list
       if all((ele in sub) for ele in subs_list)]

# printing result
print("The extracted values : " + str(res))

Output:

The original list is : [‘gfg is best’, ‘gfg is good for CS’, ‘gfg is recommended for CS’] The extracted values : [‘gfg is good for CS’, ‘gfg is recommended for CS’]

Time Complexity: O(n²)

Auxiliary Space: O(n)

Method #2: Using Counter() function

Python3

# Python3 code to demonstrate working of
# Strings with all Substring Matches
# Using Counter() function
from collections import Counter
# initializing list
test_list = ["gfg is best", "gfg is good for CS",
             "gfg is recommended for CS"]

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

# initializing Substring List
subs_list = ["gfg", "CS"]

res = []
for sub in test_list:
    flag = 0
    freq = Counter(sub.split())
    for ele in subs_list:

        # checking for non existence of
        # any string
        if ele not in freq.keys():
            flag = 1
            break
    if flag == 0:
        res.append(sub)

# printing result
print("The extracted values : " + str(res))

Output

The original list is : ['gfg is best', 'gfg is good for CS', 'gfg is recommended for CS']
The extracted values : ['gfg is good for CS', 'gfg is recommended for CS']

Time Complexity: O(n)

Auxiliary Space: O(n)