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Home/ Questions/Q 904715
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Editorial Team
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Editorial Team
Asked: May 15, 20262026-05-15T16:05:55+00:00 2026-05-15T16:05:55+00:00

Given a pattern p and a string s, suppose p is in lower case.

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Given a pattern p and a string s, suppose p is in lower case. Which of the following two is more efficient?

r = re.compile(r'p', RE.IGNORECASE)
r.match(s)

… or …

r = re.compile(r'p')
r.match(s.lower())
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  1. Editorial Team
    Editorial Team
    2026-05-15T16:05:56+00:00Added an answer on May 15, 2026 at 4:05 pm

    It’s really going to depend on the language and engine. s.lower() and re.IGNORECASE are generally only slow because they’re trying to deal with localization or Unicode strings (see this question). If the regex package you’re using deals with that, and the s.lower() method doesn’t, then the s.lower() method is a clear win. And vice-versa.

    In general, I’d expect the s.lower() method to be faster (it tends to be more optimized than regex matching). But in the example as given …

    r = re.compile(r'[Pp]')
    r.match(s)
    

    … is going to be faster than either of them.

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