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Regular expression: how to search for a sequence of one number alternated with 0s?

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Serbring
Serbring 2021 年 8 月 11 日
コメント済み: Serbring 2021 年 8 月 17 日
Hi all,
I need to craft a regular expression. What I need is to search any pattern where sequences of one number (i.e., 1 or 2) is alternated with a sequence of 0s. So, I search for patterns like the following:
'10111'
'22200002'
'300333'
How can I do it?
Thank you.
Best regards.
MM
  1 件のコメント
Walter Roberson
Walter Roberson 2021 年 8 月 12 日
Does the sequence always end with the same number? For example would 300330000 be valid? Would 300334 be valid on the grounds that it starts with alternating sequence, and then anything can be after that?
Is "no" alternations also valid? Such as 222 by itself, which is "a sequence of twos, followed by zero repetitions of (0's and 2's) ?
Is the pattern only ever at the beginning of the string, or could it be inside the string as well? If it is inside then that answers the first two questions.

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採用された回答

DGM
DGM 2021 年 8 月 12 日
Well, I'm certainly not the one to ask about regex. I thought I had this nailed down, but had to resort to this to get it to work in MATLAB. I bet it could be simpler.
vec = '5ad3515505546151g5454651333300342511324sgfb1565440444532152331450005534563asdf445341536404334400044453';
C = regexp(vec,'(\d)\1*0+\1+','match')
C = 1×6 cell array
{'55055'} {'3333003'} {'440444'} {'500055'} {'404'} {'44000444'}
(\d) matches and captures any numeric digit
\1* matches zero or more instances of the captured digit
0+ matches one or more zeros
\1+ matches one or more instances of the captured digit
  1 件のコメント
Serbring
Serbring 2021 年 8 月 17 日
Thank you so much. Is it possible to introduce tokens in lookahead/lookbehind expressions? I tried with the following, but it does not work. I have read in the web, that certains regular expression engines are not able to deal with tokens in was not able to do it.
C = regexp(S,'(?<=(\d)\1*)0+(?=(\1)+)','match')

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その他の回答 (1 件)

Stephen
Stephen 2021 年 8 月 12 日
編集済み: Stephen 2021 年 8 月 12 日
S = '5ad3515505546151g545460000051333300342511324sgfb15654404440044532152331450005534563asdf4453415364043344004044453';
C = regexp(S,'(\d)\1*(0+\1+)+','match')
C = 1×7 cell array
{'55055'} {'0000'} {'3333003'} {'4404440044'} {'500055'} {'404'} {'440040444'}
Note that \d also matches the zero character. If your definition of "one number" excludes zero, then use this instead:
C = regexp(S,'([1-9])\1*(0+\1+)+','match')
C = 1×6 cell array
{'55055'} {'3333003'} {'4404440044'} {'500055'} {'404'} {'440040444'}
  1 件のコメント
DGM
DGM 2021 年 8 月 12 日
See, excluding zero is the kind of obvious thing I expected I'd miss.
I didn't even think about continuing the alternation.

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