The task of replacing substrings in a list of strings involves iterating through each string and substituting specific words with their corresponding replacements. For example, given a list a = ['GeeksforGeeks', 'And', 'Computer Science'] and replacements b = [['Geeks', 'Gks'], ['And', '&'], ['Computer', 'Comp']], the updated list would be ['GksforGks', '&', 'Comp Science'] .

Using re

This method is the most efficient for replacing multiple substrings in a list using a single-pass regex operation. It compiles all replacement terms into a pattern, allowing for fast, optimized substitutions without multiple iterations.

import re

# list of strings
a = ['GeeksforGeeks', 'is', 'Best', 'For', 'Geeks', 'And', 'Computer Science']

# list of word replacements 
b = [['Geeks', 'Gks'], ['And', '&'], ['Computer', 'Comp']]

pattern = re.compile("|".join(re.escape(key) for key, _ in b))
replacement_map = dict(b) # Convert `b` to dictionary
a = [pattern.sub(lambda x: replacement_map[x.group()], ele) for ele in a]

print(a)

import re

​

# list of strings

a = ['GeeksforGeeks', 'is', 'Best', 'For', 'Geeks', 'And', 'Computer Science']

​

# list of word replacements 

b = [['Geeks', 'Gks'], ['And', '&'], ['Computer', 'Comp']]

​

pattern = re.compile("|".join(re.escape(key) for key, _ in b))

replacement_map = dict(b) # Convert `b` to dictionary

a = [pattern.sub(lambda x: replacement_map[x.group()], ele) for ele in a]

​

print(a)

['GksforGks', 'is', 'Best', 'For', 'Gks', '&', 'Comp Science']

Explanation: It compiles a regex pattern to match target substrings and uses a dictionary for quick lookups. It applies re.sub() with a lambda function for single-pass replacements, ensuring efficient and accurate modifications.

Using str.replace()

This approach iterates over the replacement dictionary and applies .replace() to each string. While more readable and simple, it loops multiple times and making it less efficient for large datasets.

# list of strings
a = ['GeeksforGeeks', 'is', 'Best', 'For', 'Geeks', 'And', 'Computer Science']

# list of word replacements
b = [['Geeks', 'Gks'], ['And', '&'], ['Computer', 'Comp']]

replace_map = dict(b) # convert `b` to dictionary

for key, val in replace_map.items():
    a = [ele.replace(key, val) for ele in a]

print(a)

# list of strings

a = ['GeeksforGeeks', 'is', 'Best', 'For', 'Geeks', 'And', 'Computer Science']

​

# list of word replacements

b = [['Geeks', 'Gks'], ['And', '&'], ['Computer', 'Comp']]

​

replace_map = dict(b) # convert `b` to dictionary

​

for key, val in replace_map.items():

    a = [ele.replace(key, val) for ele in a]

​

print(a)

['GksforGks', 'is', 'Best', 'For', 'Gks', '&', 'Comp Science']

Explanation: for loop iterates over each key-value pair, replacing occurrences in the string list using str.replace(), ensuring all substrings are updated efficiently.

Using nested loops

A straightforward method that manually iterates through each string and replaces the substrings one by one. While easy to implement, it is slower for larger lists due to multiple iterations.

# list of strings
a = ['GeeksforGeeks', 'is', 'Best', 'For', 'Geeks', 'And', 'Computer Science']

# list of word replacements
b = [['Geeks', 'Gks'], ['And', '&'], ['Computer', 'Comp']]

replace_map = dict(b) # convert `b` to dictionary

for key, val in replace_map.items():
    for i in range(len(a)):
        if key in a[i]:
            a[i] = a[i].replace(key, val)

print(a)

# list of strings

a = ['GeeksforGeeks', 'is', 'Best', 'For', 'Geeks', 'And', 'Computer Science']

​

# list of word replacements

b = [['Geeks', 'Gks'], ['And', '&'], ['Computer', 'Comp']]

​

replace_map = dict(b) # convert `b` to dictionary

​

for key, val in replace_map.items():

    for i in range(len(a)):

        if key in a[i]:

            a[i] = a[i].replace(key, val)

​

print(a)

['GksforGks', 'is', 'Best', 'For', 'Gks', '&', 'Comp Science']

Explanation: for loop iterates through each string in the list, checking for substring matches and replacing them using str.replace(), ensuring all occurrences are updated systematically.