Python - Replace Different characters in String at Once

Python – Replace Different characters in String at Once

Last Updated : 02 Sep, 2020

Given a mapping of characters to be replaced with corresponding values, perform all replaces at one, in one liner.

Input : test_str = ‘geeksforgeeks is best’, map_dict = {‘e’:’1′, ‘b’:’6′}
Output : g11ksforg11ks is 61st
Explanation : All e are replaced by 1 and b by 6.

Input : test_str = ‘geeksforgeeks’, map_dict = {‘e’:’1′, ‘b’:’6′}
Output : g11ksforg11ks
Explanation : All e are replaced by 1 and b by 6.

Method #1 : Using join() + generator expression

In this, we perform the task of getting characters present in dictionary and map them to their values by dictionary access, all other characters are appended unchanged, the result is converted back to dictionary using join().

Python3

# Python3 code to demonstrate working of  
# Replace Different characters in String at Once 
# using join() + generator expression  
  
# initializing string 
test_str = 'geeksforgeeks is best'
  
# printing original String 
print("The original string is : " + str(test_str)) 
  
# initializing mapping dictionary 
map_dict = {'e':'1', 'b':'6', 'i':'4'} 
  
# generator expression to construct vals  
# join to get string  
res = ''.join(idx if idx not in map_dict else map_dict[idx] for idx in test_str) 
  
# printing result  
print("The converted string : " + str(res))  

Output

The original string is : geeksforgeeks is best
The converted string : g11ksforg11ks 4s 61st

Method #2 : Using regex + lambda

This is complex way to approach problem. In this, we construct appropriate regex using lambda functions and peform the required task of replacement.

Python3

# Python3 code to demonstrate working of  
# Replace Different characters in String at Once 
# using regex + lambda 
import re 
  
# initializing string 
test_str = 'geeksforgeeks is best'
  
# printing original String 
print("The original string is : " + str(test_str)) 
  
# initializing mapping dictionary 
map_dict = {'e':'1', 'b':'6', 'i':'4'} 
  
# using lambda and regex functions to achieve task  
res = re.compile("|".join(map_dict.keys())).sub(lambda ele: map_dict[re.escape(ele.group(0))], test_str) 
  
# printing result  
print("The converted string : " + str(res))  

Output

The original string is : geeksforgeeks is best
The converted string : g11ksforg11ks 4s 61st

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