how to this regex in similar and faster way ?

Question

re.search(r'data check\s*:\s*false$(\d+)$', content).group(1) if re.search(r'data check\s*:\s*false$(\d+)$', content) else "NA" how to this regex in similar and faster way ? i have so many regex to check how to do it in efficient way ?

Amira Bedhiafi · Answer

You're calling re.search() twice, once to check if there's a match, and again to get .group(1) . Instead, store the match in a variable: match = re.search(r'data check\s*:\s*false$(\d+)$', content) result = match.group(1) if match else "NA" If you're running the same regex many times (in a loop or on multiple lines), compile it once: pattern = re.compile(r'data check\s*:\s*false$(\d+)$') match = pattern.search(content) result = match.group(1) if match else "NA" If you're matching multiple regexes on the same input and each has a similar pattern : patterns = { "data_check_false": re.compile(r'data check\s*:\s*false$(\d+)$'), "another_check": re.compile(r'another pattern\s*:\s*(\d+)'), } results = {} for name, pattern in patterns.items(): match = pattern.search(content) results[name] = match.group(1) if match else "NA"

sasikala s · Answer

match = re.search(r'data check\s*:\s*false$(\d+)$', content) result = match.group(1) if match else "NA" re.search(...) looks for the pattern data check : false( ) in the content string. If a match is found, group(1) extracts the number inside the parentheses. If no match is found, it returns "NA" .

how to this regex in similar and faster way ?

Answers (2)