I'm trying to do a substitution where I need to find spaces followed by a number or a letter, with or without accents, to replace them with an underscore. I currently have this (note the space in the beginning):
\b(?=[a-zA-Z0-9àéèêëîïôöûü])
With the string test string école test, the replacement looks like this:
test_string école_test
I guess you got the problem, but just in case, the expected result is this:
test_string_école_test
The strangest thing is that if I just search for [a-zA-Z0-9àéèêëîïôöûü], it matches every single one of the letters, so my RegEx seems just fine...
Is this a bug or am I missing something?
Drop the \b - it's not essential to your query (you're already matching the space), and unicode support is patchy in regular expressions. The boundary detection is ASCII only in Sublime Text 2.
Related
I have a very large file containing thousands of sentences. In all of them, the first word of each sentence begins with lowercase, but I need them to begin with uppercase.
I looked through the site trying to find a regex to do this but I was unable to. I learned a lot about regex in the process, which is always a plus for my job, but I was unable to find specifically what I am looking for.
I tried to find a way of compiling the code from several answers, including the following:
Convert first lowercase to uppercase and uppercase to lowercase (regex?)
how to change first two uppercase character to lowercase character on each line in vim
Regex, two uppercase characters in a string
Convert a char to upper case using regular expressions (EditPad Pro)
But for different reasons none of them served my purpose.
I am working with a translation-specific application which accepts regex.
Do you think this is possible at all? It would save me hours of tedious work.
You can use this regex to search for the first letters of sentences:
(?<=[\.!?]\s)([a-z])
It matches a lowercase letter [a-z], following the end of a previous sentence (which might end with one of the following: [\.!?]) and a space character \s.
Then make a substitution with \U$1.
It doesn't work only for the very first sentence. I intentionally kept the regex simple, because it's easy to capitalize the very first letter manually.
Working example: https://regex101.com/r/hqwK26/1
UPD: If your software doesn't support \U, you might want to copy your text to Notepad++ and make a replacement there. The \U is fully supported, just checked.
UPD2: According to the comments, the task is slightly different, and just the first letters of each line should be capitalized.
There is a simple regex for that: ^([a-z]), with the same substitution pattern.
Here is a working example: https://regex101.com/r/hqwK26/2
Taking Ildar's answer and combining both of his patterns should work with no compromises.
(?<=[\.!?]\s)([a-z])|^([a-z])
This is basically saying, if first pattern OR second pattern. But because you're now technically extracting 2 groups instead of one, you'll have to refer to group 2 as $2. Which should be fine because only one of the patterns should be matched.
So your substitution pattern would then be as follows...
\U$1$2
Here's a working example, again based on Ildar's answer...
https://regex101.com/r/hqwK26/13
I have this text like below:
아니다
bukan
싫다
tidak suka
훌륭하다
bagus
And I am trying to remove the English line(English Alphabets) and attach it to the end of upper line(Korean Alphabets) like this:
아니다bukan
싫다tidak suka
훌륭하다bagus
Now, Finally find almost close regular expression, which is this:
[가-힣]\R
However, It makes the text file like this:
아니bukan
싫tidak suka
훌륭하bagus
The problem is removing the one word of Korean too.
How can I solve this problem?
C++ std::regex does not support Unicode property classes like \p{Hangul}, but you may use the equivalent character class, [\u1100-\u11FF\u302E\u302F\u3131-\u318E\u3200-\u321E\u3260-\u327E\uA960-\uA97C\uAC00-\uD7A3\uD7B0-\uD7C6\uD7CB-\uD7FB\uFFA0-\uFFBE\uFFC2-\uFFC7\uFFCA-\uFFCF\uFFD2-\uFFD7\uFFDA-\uFFDC], see this reference.
Besides, \R is not supported either. You may probably just use \r?\n to match Windows/Linux style line endings, or (?:\r\n?|\n) to also support MacOS line endings.
Next, if you match and consume a Korean char, when replacing, you need to capture it into a capturing group and use a backreference to the group in the replacement pattern.
So, you may use
([\u1100-\u11FF\u302E\u302F\u3131-\u318E\u3200-\u321E\u3260-\u327E\uA960-\uA97C\uAC00-\uD7A3\uD7B0-\uD7C6\uD7CB-\uD7FB\uFFA0-\uFFBE\uFFC2-\uFFC7\uFFCA-\uFFCF\uFFD2-\uFFD7\uFFDA-\uFFDC])(?:\r\n?|\n)
Replace with $1 to put back the Korean char into the resulting string.
See the regex demo online.
The regex for the set of all Korean characters in unicode is this:
\p{Hangul}
There is more information here: https://www.regular-expressions.info/unicode.html
Maybe you also need a + after your group of characters?
Use the [\p{Hangul}]+\R regular expression instead of what you're using now.
I’m working with a text file with 200.000+ lines in Notepad++. Each line has only one word. I need to strip out and remove all words which only contains one letter (e.g.: I) and words which contains only two letters (e.g.: as).
I thought I could just pas in regular regex like this [a-zA-Z]{1,2} but I does not recognize anything (I’m trying to Mark them).
I’ve done manual search and I know that there do exists words of that length so therefor can it only be my regex code that’s wrong. Anyone knows how to do this in Notepad++ ???
Cheers,
- Mestika
If you want to remove only the words but leave the lines empty, this works:
^[a-zA-Z]{1,2}$
Replace this with an empty string. ^ and $ are anchors for the beginning and the end of a line (because Notepad++'s regexes work in multi-line mode).
If you want to remove the lines completely, search for this:
^[a-zA-Z]{1,2}\r\n
And replace with an empty string. However, this won't work before Notepad++ 6, so make sure yours is up-to-date.
Note that you will have to replace \r\n with the specific line-endings of your file!
As Tim Pietzker suggested, a platform independent solution that also removes empty lines would be:
^[a-zA-Z]{1,2}[\r\n]+
A platform-independent solution that does not remove empty lines but only those with one or two letters would be:
^[a-zA-Z]{1,2}(\r\n?|\n)
I don't use Notepad++ but my guess is it could be because you have too many matches - try including word boundaries (your exp will match every set of 2 letters)
\b[a-zA-Z]{1,2}\b
The regex you specified should find 1-or-2 characters (even in Notepad++'s Find-dialog), but not in the way you'd think. You want to have the regex make sure it starts at the beginning of the line and ends at the end with ^ and $, respecitevely:
^[a-zA-Z]{1,2}$
Notepad++ version 6.0 introduced the PCRE engine, so if this doesn't work in your current version try updating to the most recent.
You seem to use the version of Notepad++ that doesn't support explicit quantifiers: that's why there's no match at all (as { and } are treated as literals, not special symbols).
The solution is to use their somewhat more lengthy replacement:
\w\w?
... but that's only part of the story, as this regex will match any symbol, and not just short words. To do that, you need something like this:
^\w\w?$
I want to find all strings containing at least 1 Cyrillic character (basically /.*[А-я].*/) but with exception of comments.
Comment is a string or part of a string which starts with 2 or more / characters.
Currently I get this regex which do some part of the trick:
^(?=^.*?[А-я]+).*?((?=[\/]{2,})|(^(?:(?![\/]{2,}).)*$))
But I'd like to get less bloated and faster expression.
And as additional question: could anyone explain why this one is working? I combined it by trial-and-error but I'm not sure I completely understood how it works, because when I try to change it in any part - it stops working.
The following regex will match any cyrllic character that is not preceded by a double forward slash
(?<!/{2}.*)[А-я]
It specifies that it should not be preceded by a double slash by using a negative lookbehind.
You haven't specified what flavour of regex your using, but be aware some flavours don't support lookarounds. For example PCRE (javascript) doesn't. You are using 3 of them in your regex, so i presume its ok.
I'm using vim to do some pattern matching on a text file. I've enabled search highlighting so that I know exactly what is getting matched on each search and am getting confused.
Consider searching for [a-z]* on the following text:123456789abcdefghijklmnopqrstuvwxyxz987654321ABCDEFGHIJKLMNOPQRSTUVWQXZ
I expected this search to match zero or more consecutive characters that are in the range [a-z]. Instead, I get a match on the entire line.
Should this be the expected behaviour?
Thanks,
Andrew
It's matching the empty strings that occur after every character. It has no way of highlighting empty ranges, so it looks like everything is highlighted.
Try searching for [a-z]\+ instead.
Empty string matches [a-z]*... therefore this thing is matching everywhere. Perhaps you want to cut down some of the cases by doing [a-z]+ (1 or more), or [a-z]{4,} (4 or more).
You're not getting a match on the entire line, you're getting a match on every character. Your pattern also matches nothing at all, which is matched by every single character.