How to match a string with pythons regex with optional character, but only if that optional character is preceded by another character

Question

I need to match a string that optionally ends with numbers, but only if the numbers aren't preceded by a 0.

so AAAA should match, AAA1 should, AA20 should, but AA02 should not.

I can figure out the optionality of it, but I'm not sure if python has a "preceded by" or "followed by" flag.

if s.isalnum() and re.match("^[A-Z]+[1-9][0-9]*$", s):
        return True

Another idea by use of a lookahead: ^[A-Z]+(?!0)\d*$ (Just to mention, prefer selected answer) — bobble bubble
– bobble bubble, Commented Nov 19, 2022 at 3:14

Andrej Kesely · Accepted Answer · 2022-11-19 02:31:38Z

1

Try:

^[A-Z]+(?:[1-9][0-9]*)?$

^[A-Z]+ - match letters from the beginning of string

(?:[1-9][0-9]*)? - optionally match a number that doesn't start from 0

$ - end of string

answered Nov 19, 2022 at 2:31

Andrej Kesely

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2 Comments

AA3209 should not be match according to the OP. Also, your pattern does not match A000, which, in the absence of more info, should also be a match.

@TimBiegeleisen - you seem to "understand" this much differently than the rest of us.