Python – Extract Indices of substring matches
Last Updated :
05 May, 2023
Given a String List, and a substring, extract list of indices of Strings, in which that substring occurs.
Input : test_list = ["Gfg is good", "for Geeks", "I love Gfg", "Gfg is useful"], K = "Gfg"
Output : [0, 2, 3]
Explanation : "Gfg" is present in 0th, 2nd and 3rd element as substring.
Input : test_list = ["Gfg is good", "for Geeks", "I love Gfg", "Gfg is useful"], K = "good"
Output : [0]
Explanation : "good" is present in 0th substring.
Method #1: Using loop + enumerate()
This is the brute way in which this task can be done. In this, we iterate for all the elements along with their indices using enumerate(), and conditional statements are used to get the required result.
Python3
test_list = ["Gfg is good", "for Geeks", "I love Gfg", "Its useful"]
K = "Gfg"
print("The original list : " + str(test_list))
res = []
for idx, ele in enumerate(test_list):
if K in ele:
res.append(idx)
print("The indices list : " + str(res))
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Output
The original list : ['Gfg is good', 'for Geeks', 'I love Gfg', 'Its useful']
The indices list : [0, 2]
Time complexity : O(n*m)
Space Complexity : O(n)
Method #2 : Using list comprehension + enumerate()
This is yet another way in which this task can be solved. In this, we perform the similar task as above method using list comprehension and enumerate() is used to get compact solution.
Python3
test_list = ["Gfg is good", "for Geeks", "I love Gfg", "Its useful"]
K = "Gfg"
print("The original list : " + str(test_list))
res = [idx for idx, val in enumerate(test_list) if K in val]
print("The indices list : " + str(res))
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Output
The original list : ['Gfg is good', 'for Geeks', 'I love Gfg', 'Its useful']
The indices list : [0, 2]
The Time and Space Complexity for all the methods are the same:
Time Complexity: O(n)
Auxiliary Space: O(n)
Method #3 : Using find() method
Python3
test_list = ["Gfg is good", "for Geeks", "I love Gfg", "Its useful"]
K = "Gfg"
print("The original list : " + str(test_list))
res = []
for i in range(0,len(test_list)):
if test_list[i] .find(K)!=-1 :
res.append(i)
print("The indices list : " + str(res))
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Output
The original list : ['Gfg is good', 'for Geeks', 'I love Gfg', 'Its useful']
The indices list : [0, 2]
Time complexity : O(n*m)
Space Complexity : O(n)
Method #5 : Using operator.contains() method
Approach
- Initiate a for loop to traverse list of strings
- Check for K in each string using operator.contains() method
- If yes append the index of string in list to output list
- Display output list
Python3
test_list = ["Gfg is good", "for Geeks", "I love Gfg", "Its useful"]
K = "Gfg"
print("The original list : " + str(test_list))
res = []
import operator
for i in range(0,len(test_list)):
if(operator.contains(test_list[i],K)):
res.append(i)
print("The indices list : " + str(res))
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Output
The original list : ['Gfg is good', 'for Geeks', 'I love Gfg', 'Its useful']
The indices list : [0, 2]
Time Complexity : O(M*N) M – length of strings list N -length of each string
Auxiliary Space : O(N) N – number of strings with K as substring
Method #6: Using map() function + lambda function
Another approach to solve the given problem is by using the map() function along with the lambda function. The map() function applies a function to each element of a sequence and returns an iterator of the results.
step-by-step approach
Initialize the list and substring K.
Use the map() function to apply the lambda function to each element of the list and get a list of True/False values indicating if the substring K is present in the element or not.
Convert the list of True/False values into a list of 1’s and 0’s using the map() function and another lambda function.
Use the enumerate() function to get the indices of the list of 1’s and 0’s.
Extract the indices of the elements that have 1 in the list of 1’s and 0’s.
Print the result.
Python3
test_list = ["Gfg is good", "for Geeks", "I love Gfg", "Its useful"]
K = "Gfg"
print("The original list : " + str(test_list))
bool_list = list(map(lambda x: K in x, test_list))
int_list = list(map(lambda x: 1 if x else 0, bool_list))
enum_list = list(enumerate(int_list))
res = [i for i, j in enum_list if j == 1]
print("The indices list : " + str(res))
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Output
The original list : ['Gfg is good', 'for Geeks', 'I love Gfg', 'Its useful']
The indices list : [0, 2]
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 list.
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