Python – Extract Indices of substring matches

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

# Python3 code to demonstrate working of
# Extract Indices of substring matches
# Using loop + enumerate()

# initializing list
test_list = ["Gfg is good", "for Geeks", "I love Gfg", "Its useful"]

# initializing K
K = "Gfg"

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

# using loop to iterate through list
res = []
for idx, ele in enumerate(test_list):
  if K in ele:
    res.append(idx)

# printing result
print("The indices list : " + str(res))

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

# Python3 code to demonstrate working of
# Extract Indices of substring matches
# Using list comprehension + enumerate()

# initializing list
test_list = ["Gfg is good", "for Geeks", "I love Gfg", "Its useful"]

# initializing K
K = "Gfg"

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

# using list comprehension and enumerate to offer compact solution
res = [idx for idx, val in enumerate(test_list) if K in val]

# printing result
print("The indices list : " + str(res))

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

# Python3 code to demonstrate working of
# Extract Indices of substring matches

# initializing list
test_list = ["Gfg is good", "for Geeks", "I love Gfg", "Its useful"]

# initializing K
K = "Gfg"

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

# using loop to iterate through list
res = []
for i in range(0,len(test_list)):
    if test_list[i] .find(K)!=-1 :
        res.append(i)

# printing result
print("The indices list : " + str(res))

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 

1. Initiate a for loop to traverse list of strings
2. Check for K in each string using operator.contains() method
3. If yes append the index of string in list to output list
4. Display output list

Python3

# Python3 code to demonstrate working of
# Extract Indices of substring matches

# initializing list
test_list = ["Gfg is good", "for Geeks", "I love Gfg", "Its useful"]

# initializing K
K = "Gfg"

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

# using loop to iterate through list
res = []
import operator
for i in range(0,len(test_list)):
    if(operator.contains(test_list[i],K)):
        res.append(i)

# printing result
print("The indices list : " + str(res))

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

# Python3 code to demonstrate working of
# Extract Indices of substring matches

# initializing list
test_list = ["Gfg is good", "for Geeks", "I love Gfg", "Its useful"]

# initializing K
K = "Gfg"

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

# using map() function with lambda function to get the list of True/False values
bool_list = list(map(lambda x: K in x, test_list))

# using map() function with lambda function to convert bool_list into a list of 1's and 0's
int_list = list(map(lambda x: 1 if x else 0, bool_list))

# using enumerate() function to get the indices of the int_list
enum_list = list(enumerate(int_list))

# extracting the indices of the elements that have 1 in the int_list
res = [i for i, j in enum_list if j == 1]

# printing result
print("The indices list : " + str(res))

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.