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Asked: 2026-06-16T00:04:37+00:00

I have the string consisting times and I wanted to format the times to

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I have the string consisting times and I wanted to format the times to 12hrs format.

I tried,

pat1 = "(\d+):(\d+)?\s*pm"
pat1 = "(\d+):(\d+)?\s*am"
hooString = "673 HOO: Mon. 7:30 am - 6pm, Tue. 7:30am - 6 pm, Wed. 7:30 am - 6 pm, Thurs. 7:30 am - 6 pm, Fri. 8 am - 5 pm,"
hooString  = sub(compile(pat1, IGNORECASE), "\1:\2 AM", hooString)
hooString  = sub(compile(pat2, IGNORECASE), "\1:\2 PM", hooString)
print hooString

I got the output as,

673 HOO: Mon. 7:30 AM - 6pm, Tue. 7:30 AM - 6 pm, Wed. 7:30 AM - 6 pm, Thurs. 7:30 AM - 6 pm, Fri. 8 am - 5 pm,

But I required “8am or 8 am” also to be replaced to “8 AM”, so I slightly changed regex to 

pat1 = "(\d+):?(\d+)?\s*am"
pat2 = "(\d+):?(\d+)?\s*pm"

It throws me “unmatched group” exception. because of the 2nd group doesn’t exists in input string.

How could I get my expected result without exception for both the conditions “%H:%M %P” as well as “%H %p”

Also If I could get the result for “8 am” as “8:00 AM” that could be the best answer.

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1 Answer

  1. Editorial Team

    Why have two groups in the first place?

    pat1 = r"(\d+(?:[:]\d+)?)\s*am"
pat1 = r"(\d+(?:[:]\d+)?)\s*pm"

    Note the use of a raw string. Otherwise you might get problems with escaping.

    What did I do? I just stuff the whole time into one group, and made :\d+ optional (with a non-capturing group). Replacement would of course be \1 AM and \1 PM.

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