Python | Get matching substrings in string
Last Updated :
24 Mar, 2023
The testing of a single substring in a string has been discussed many times. But sometimes, we have a list of potential substrings and check which ones occur in a target string as a substring. Let’s discuss certain ways in which this task can be performed.
Method #1: Using list comprehension
Using list comprehension is the naive and brute force method to perform this particular task. In this method, we try to get the matching string using the “in” operator and store it in the new list.
Python3
test_str = "GfG is good website";
test_list = ["GfG", "site", "CS", "Geeks", "Tutorial"]
print("The original string is : " + test_str)
print("The original list is : " + str(test_list))
res = [sub for sub in test_list if sub in test_str]
print("The list of found substrings : " + str(res))
|
Output
The original string is : GfG is good website
The original list is : ['GfG', 'site', 'CS', 'Geeks', 'Tutorial']
The list of found substrings : ['GfG', 'site']
Method #2: Using filter() + lambda
This task can also be performed using the filter function which performs the task of filtering out the resultant strings that is checked for existence using the lambda function.
Python3
test_str = "GfG is good website";
test_list = ["GfG", "site", "CS", "Geeks", "Tutorial"]
print("The original string is : " + test_str)
print("The original list is : " + str(test_list))
res = list(filter(lambda x: x in test_str, test_list))
print("The list of found substrings : " + str(res))
|
Output
The original string is : GfG is good website
The original list is : ['GfG', 'site', 'CS', 'Geeks', 'Tutorial']
The list of found substrings : ['GfG', 'site']
Method #3 : Using find() method.
find() method returns the position of the string passed as an argument in the given string or else returns -1
Python3
test_str = "GfG is good website"
test_list = ["GfG", "site", "CS", "Geeks", "Tutorial" ]
print("The original string is : " + test_str)
print("The original list is : " + str(test_list))
res=[]
for i in test_list:
if(test_str.find(i)!=-1 and i not in res):
res.append(i)
print("The list of found substrings : " + str(res))
|
Output
The original string is : GfG is good website
The original list is : ['GfG', 'site', 'CS', 'Geeks', 'Tutorial']
The list of found substrings : ['GfG', 'site']
Method #4 : Using re
In this approach using the re (regular expression) module, we import the re module and use its search() function to check for the presence of a substring in the target string. The search() function returns a match object if the substring is found, and None if it is not found. We can use this behavior to filter the list of potential substrings and only keep the ones that are present in the target string.
Python3
import re
test_str = "GfG is good website"
test_list = ["GfG", "site", "CS", "Geeks", "Tutorial"]
print("The original string is : " + test_str)
print("The original list is : " + str(test_list))
res = [sub for sub in test_list if re.search(sub, test_str)]
print("The list of found substrings : " + str(res))
|
Output
The original string is : GfG is good website
The original list is : ['GfG', 'site', 'CS', 'Geeks', 'Tutorial']
The list of found substrings : ['GfG', 'site']
The time complexity of the approach using the re module is O(n * m), where n is the length of the target string and m is the number of substrings in the list of potential substrings.
This is because for each substring in the list, we are searching the entire target string to see if the substring is present.
The auxiliary space is O(n), as we are creating a new list of substrings that are found in the target string, and this list will have a length of at most n.
Method #5 : Using index() method.
In python we index ( ) returns matching substring index()” method returns the index of the first occurrence of the substring in the string “index()” method raises a ValueError exception, which is handled by the try-except block.
Python3
test_str = "GfG is good website"
test_list = ["GfG", "site", "CS", "Geeks", "Tutorial"]
print("The original string is:", test_str)
print("The original list is:", test_list)
res = []
for sub in test_list:
try:
test_str.index(sub)
res.append(sub)
except ValueError:
pass
print("The list of found substrings:", res)
|
Output
The original string is: GfG is good website
The original list is: ['GfG', 'site', 'CS', 'Geeks', 'Tutorial']
The list of found substrings: ['GfG', 'site']
The time complexity of this program is O(n * m)
The auxiliary space of this program is O(n)
Method #6 : Using operator.contains() method
- Initiate a for loop to traverse list of substrings
- Check whether each substring is present in original string using operator.contains()
- If present append that substring to output list
- Display output list
Python3
test_str = "GfG is good website"
test_list = ["GfG", "site", "CS", "Geeks", "Tutorial" ]
print("The original string is : " + test_str)
print("The original list is : " + str(test_list))
res=[]
import operator
for i in test_list:
if(operator.contains(test_str,i) and i not in res):
res.append(i)
print("The list of found substrings : " + str(res))
|
Output
The original string is : GfG is good website
The original list is : ['GfG', 'site', 'CS', 'Geeks', 'Tutorial']
The list of found substrings : ['GfG', 'site']
Time Complexity : O(M*N) M -length of string N-length of substring list
Auxiliary Space : O(N) N – number of substrings present in string
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