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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'] .
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.
['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.
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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.
['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.
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.
['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.