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Python Error Invalid Group Reference

For example: >>> expr = r'\s+([A-Z][a-z]+[A-Z][a-z]+)\s' >>> parser=re.compile(expr) >>> parser.sub(repl, mystr) 'this is a string to Test whetherWikiLinkwill work ProPerly' The backreference \1 refers to the group 1 within the Recalll is crowed sourced knowledge vault, where community can create, curate and access qualitative knowledge, In form of small and precise topics. Use a named capture group: p = r'(?P[\,|\.]\d{1})$' re.sub(p, r"\g0", v) >>> p = r'(?P[\,|\.]\d{1})$' >>> v = '235,5' >>> re.sub(p, r"\g0", v) '235,50' python - regex group reference error - Are illegal immigrants more likely to commit crimes? http://caribtechsxm.com/python-error/python-error-in-processing-external-entity-reference.php

But in your case it's never necessary; \[*? Administration User List Committer List Help Tracker Documentation Tracker Development Report Tracker Problem Issue1519638 classification Title: Unmatched Group issue - workaround Type: enhancement Stage: resolved Components: Library (Lib), Regular Expressions Versions: This tool has some ready adaptable .apy files which the … calling Python function from C/C++ 10 replies Hello, I'm writing a simple calculator in C. add a comment| 1 Answer 1 active oldest votes up vote 7 down vote You can use \g based group substitution: >>> import re >>> re.sub("(..)(..)", r"\g<1>0\g<2>", "toto") 'to0to' From docs:

But re.sub() has another trick up its sleeve - the replacement 'expression' can be a callable which is passed the match object and returns the string to replace it with. Do you mean [.,]? –georg Dec 9 '13 at 11:18 add a comment| 3 Answers 3 active oldest votes up vote 2 down vote accepted Just wrap the group reference in A better example is probably re.sub("foo(?:b(ar)|foo)","\\1","foofoo") because this can't be simply repaired by refactoring the regex. sre_constants.error: invalid group reference >>> p = r'(?P[\,|\.]\d{1}$)' >>> re.sub(p, r'\g0', '1.2') '1.20' Pick some name better than just "frac" (which I pulled out my ...

How much are taxes for a postdoc in the United States? History Date User Action Args 2014-10-1008:45:02serhiy.storchakasetstatus: open -> closedresolution: fixedmessages: + msg228969stage: patch review -> resolved 2014-10-1008:16:35python-devsetnosy: + python-devmessages: + msg228966 2014-10-1007:50:01serhiy.storchakasetassignee: serhiy.storchaka 2014-10-0820:32:20pitrousetassignee: effbot -> (no value) 2014-09-1810:54:53serhiy.storchakasetfiles: + re_sub_unmatched_group.patchtype: Was Sigmund Freud "deathly afraid" of the number 62? I understand why this happends and maybe it's just me doing things the wrong way but I just can't see what I should do differently.

python regex share|improve this question asked Mar 5 '13 at 14:54 tone7 877 1 Please note that while nneonneo's post gives you the actual answer to your problem, PurityLake's remark Steps To Reproduce: playbook: https://gist.github.com/horrorhead/691035576a0c79a42c33#file-reproduce_replace_error-yml text file: https://gist.github.com/horrorhead/07f18c290500c1269452#file-replace-txt Expected Results: I had expected this to replace the line: value=sometext with: value: sometext\8somemoretext Actual Results: The play fails with the following error: Thanks to @m.buettner for the great link. http://stackoverflow.com/questions/20468085/regex-group-reference-error Join them; it only takes a minute: Sign up regex group reference error up vote 2 down vote favorite p = r'([\,|\.]\d{1}$)' re.sub(p, r"\1", v) works, but I want to add

Font identification dificulties Was the Oceanic flight 815 pilot the only one attacked by the monster? Check this Out Similar queries Points of reference on Stack Overflow Edit 1 - The Crazy World of replacing in Invalid HTML Algorithms: Bubble Sort - Python Algorithms: Selection Sort - more stack exchange communities company blog Stack Exchange Inbox Reputation and Badges sign up log in tour help Tour Start here for a quick overview of the site Help Center Detailed Whatever part of the match you want to show up as \1, you need to put in parentheses.

Then an empty string is returned instead of a None and the sub method is executed normally instead of raising the “unmatched group” error. check here Does anybody have experience with speeding things up? If I prepend the others with a letter it also works. Rearrange colors in BarChart Nested apply function at a list Why would breathing pure oxygen be a bad idea?

Thanks. http://caribtechsxm.com/python-error/python-error-log-mac.php share|improve this answer edited Nov 29 '12 at 23:42 answered Nov 29 '12 at 23:34 abarnert 165k11198275 1 Thanks for the explanation. variableTest is a string - verified with type(variableTest) I'm totally lost as to why this is. Can anyone identify the city in this photo?

I'm struggling with Python's regular expressions. msg69558 - (view) Author: Brandon Mintern (BMintern) Date: 2008-07-11 16:52 Looking at your code example, that solution seems quite obvious now, and I wouldn't even call it a "workaround". Thanks again Bernard On 9/8/05, Kent Johnson wrote: > Bernard Lebel wrote: > > Hello, > > > > I have a string, and I use a regular this content Nested apply function at a list How can I Improve gameplay for new players, as a new player?

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The workaround does not apply to the situation where the captured text is on one side of an "or" grouping, rather than just being optional.

  1. Not the answer you're looking for?
  2. Group 0 is the entire match.
  3. The main problem is I need to parse string to evaluate for example 1+2*(3-1)/2.5.
  4. For example: >>> expr = r'\s+([A-Z][a-z]+[A-Z][a-z]+)\s' >>> parser=re.compile(expr) >>> parser.sub(repl, mystr) 'this is a string to Test whetherWikiLinkwill work ProPerly' The backreference \1 refers to the group 1 within the
  5. Suggestion error with Coveo search box more hot questions lang-py about us tour help blog chat data legal privacy policy work here advertising info mobile contact us feedback Technology Life /
  6. This behaviour, while by design, is unwanted because this type of matching usually requests that a blank match be returned (i.e.
  7. About 11 results You can use \g based group substitution: >>> import re >>> re.sub("(..)(..)", r"\g<1>0\g<2>", "toto") 'to0to' \g uses the corresponding group number; \g<2> is therefore equivalent to \2, but
  8. If those answers do not fully address your question, please ask a new question.
  9. asked 3 years ago viewed 2724 times active 3 years ago Blog Stack Overflow Podcast #92 - The Guerilla Guide to Interviewing Related 2738How do I check whether a file exists

When I run the oRe.sub() call, I get an error: Traceback (most recent call last): File "", line 1, in ? How to swap characters? 0 8 Years Ago data = re.sub(u"\u102F*\u102D", u"\u102D\\2\u102F", data) data = re.sub(u"\u1031*\u103B", u"\u103B\\2\u1031", data) data = re.sub(u"\u1001*\u102C", u"\u1001\\2\u102B", data) data = re.sub(u"\u1002*\u102C", u"\u1001\\2\u102B", data) data = re.sub(u"\u1004*\u102C", I'm trying to remove groups of text in parentheses that come at the end of a string, but if the content in a pair of parentheses is a number, I want So instead of this “(.+?)?” use this “(|.+?)” (without the double quotes).

The back ref indeed returns a None. I don't know what you were trying to accomplish, but if you wanted to replace the three brackets with one, you should have used + instead of *?, and you should more stack exchange communities company blog Stack Exchange Inbox Reputation and Badges sign up log in tour help Tour Start here for a quick overview of the site Help Center Detailed http://caribtechsxm.com/python-error/python-error-message-invalid-syntax.php It never makes sense to have a reluctant quantifier as the last thing in a regex.

Whatever part of the match you want to show up as \1, you need to put in parentheses. more stack exchange communities company blog Stack Exchange Inbox Reputation and Badges sign up log in tour help Tour Start here for a quick overview of the site Help Center Detailed I'm new to python and so far, I simply love it! Kind regards, Gerard.

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msg108670 - (view) Author: Ezio Melotti (ezio.melotti) * Date: 2010-06-26 01:09 It would be nice if you could port 'pieces' of #2636 to Python, in order to fix this and other For example, if sub were 123\1456, there's no way to tell whether that means 123, followed by group 1, followed by 456, or 123 followed by group 1456, or… Further reading View More at http://stackoverflow.com/questions/12994368/regexp-wrong-special-... more hot questions question feed lang-py about us tour help blog chat data legal privacy policy work here advertising info mobile contact us feedback Technology Life / Arts Culture / Recreation

If you had used raw strings as @Blender suggested (correctly--that was another error), your output would have been the original string: >>> re.sub(r"(\[*?)",r"\1","x[[[y") 'x[[[y' It's matching an empty string at every