Python | Multiple indices Replace in String

Last Updated : 16 May, 2023

Sometimes, while working with Python Stings, we can have a problem, in which we need to perform the replace of characters based on several indices of String. This kind of problem can have applications in many domains. Lets discuss certain ways in which this task can be performed.

Method #1 : Using loop + join() This is brute force way in which this task can be performed. In this, we iterate for each character and substitute with replace character if that is one.

Python3

# Python3 code to demonstrate working of 
# Multiple indices Replace in String 
# Using loop + join() 
 
# initializing string 
test_str = 'geeksforgeeks is best'
 
# printing original string 
print("The original string is : " + test_str) 
 
# initializing list 
test_list = [2, 4, 7, 10] 
 
# initializing repl char 
repl_char = '*'
 
# Multiple indices Replace in String 
# Using loop + join() 
temp = list(test_str) 
for idx in test_list: 
    temp[idx] = repl_char 
res = ''.join(temp) 
 
# printing result 
print("The String after performing replace : " + str(res)) 

Output :

The original string is : geeksforgeeks is best
The String after performing replace : ge*k*fo*ge*ks is best

Method #2 : Using list comprehension + join() The combination of above functions can also be used to perform this task. In this, we perform similar task as above, just in one liner format using list comprehension.

Python3

# Python3 code to demonstrate working of 
# Multiple indices Replace in String 
# Using list comprehension + join() 
 
# initializing string 
test_str = 'geeksforgeeks is best'
 
# printing original string 
print("The original string is : " + test_str) 
 
# initializing list 
test_list = [2, 4, 7, 10] 
 
# initializing repl char 
repl_char = '*'
 
# Multiple indices Replace in String 
# Using list comprehension + join() 
temp = list(test_str) 
res = [repl_char if idx in test_list else ele for idx, ele in enumerate(temp)] 
res = ''.join(res) 
 
# printing result 
print("The String after performing replace : " + str(res)) 

Output :

The original string is : geeksforgeeks is best
The String after performing replace : ge*k*fo*ge*ks is best

The Time and Space Complexity for all the methods are the same:

Time Complexity: O(n)

Space Complexity: O(n)

Method #3: Using map() + join() + enumerate() methods and lambda to replace characters in string

Step by step Algorithm:

Initialize input string, list of indices to replace, and replacement character.
Use the enumerate() function to get a tuple of index and character for each character in the input string.
Use the map() function to iterate over each tuple and apply the lambda function.
In the lambda function, check if the index of the character is in the list of indices to replace. If yes, replace the character with the replacement character. If no, keep the character as is.
Join the resulting characters using the join() function to get the final string.
Return the final string.

Python3

test_str = 'geeksforgeeks is best'
test_list = [2, 4, 7, 10]
repl_char = '*'
res = ''.join(map(lambda x: repl_char if x[0] in test_list else x[1], enumerate(test_str)))
print("The String after performing replace : " + str(res))

Output

The String after performing replace : ge*k*fo*ge*ks is best

Time complexity: O(n), where n is the length of the input string.
Auxiliary Space: O(n), where n is the length of the input string. This is because we are creating a new string object to store the final result.

Method #4: Using replace() method inside a loop

Step-by-step explanation:

Initialize the string test_str, the list test_list, and the replacement character repl_char.
Print the original string.
Loop through the elements of test_list.
For each element i in test_list, we replace the character at the i-th position in test_str with repl_char. We achieve this by slicing the string into two parts: the part before the i-th character, and the part after the i-th character then concatenate the repl_char between these two parts.
We print the resulting string.

Python3

# initializing string 
test_str = 'geeksforgeeks is best'
 
# printing original string 
print("The original string is : " + test_str) 
 
# initializing list 
test_list = [2, 4, 7, 10] 
 
# initializing repl char 
repl_char = '*'
 
# Multiple indices Replace in String 
# Using replace() inside a loop
for i in test_list:
    test_str = test_str[:i] + repl_char + test_str[i+1:]
 
# printing result 
print("The String after performing replace : " + str(test_str))

Output

The original string is : geeksforgeeks is best
The String after performing replace : ge*k*fo*ge*ks is best

Time complexity: O(n*m), where n is the length of the string and m is the length of the list.
Auxiliary space: O(n), where n is the length of the string. This is because we are modifying the original string in-place, and not creating any new data structures.