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?$
Related
I've got a practical application for a vim regex where I'd like to remove numbers from the end of file location links. For example, if the developer is sloppy and just adds files and doesn't reuse file locations, you'll end up with something awful like this:
PATH_TO_MY_FILES>
PATH_TO_MY_FILES1>
...
PATH_TO_MY_FILES22>
PATH_TO_MY_FILES_ELSEWHERE>
PATH_TO_MY_FILES_ELSEWHERE1>
...
So all I want to do is to S&R and replace PATH_TO_MY_FILES*\d+ with PATH_TO_MY_FILES* using regex. Obviously I am not doing it quite right, so I was hoping someone here could not spoon feed the answer necessarily, but throw a regex buzzword my way to get me on track.
Here's what I have tried:
:%s\(PATH_TO_MY_FILES\w*\)\(\d+\)>:gc
But this doesn't work, i.e. if I just do a vim search on that, it doesn't find anything. However, if I use this:
:%s\(PATH_TO_MY_FILES\w*\)\(\d\)>:gc
It will match the string, but the grouping is off, as expected. For example, the string PATH_TO_MY_FILES22 will be grouped as (PATH_TO_MY_FILES2)(2), presumably because the \d only matches the 2, and the \w match includes the first 2.
Question 1: Why doesn't \d+ work?
If I go ahead and use the second string (which is wrong), Vim appears to find a match (even though the grouping is wrong), but then does the replacement incorrectly.
For example, given that we know the \d will only match the last number in the string, I would expect PATH_TO_MY_FILES22> to get replaced with PATH_TO_MY_FILES2>. However, instead it replaces it with this:
PATH_TO_MY_FILES2PATH_TO_MY_FILES22>gt
So basically, it looks like it finds PATH_TO_MY_FILES22>, but then replaces only the & with group 1, which is PATH_TO_MY_FILES2.
I tried another regex at Regexr.com to see how it would interpret my grouping, and it looked correct, but maybe a hack around my lack of regex understanding:
(PATH_TO_\D*)(\d*)>
This correctly broke my target string into the PATH part and the entire number, so I was happy. But then when I used this in Vim, it found the match, but still replaced only the &.
Question 2: Why is Vim only replacing the &?
Answer 1:
You need to escape the + or it will be taken literally. For example \d\+ works correctly.
Answer 2:
An unescaped & in the replacement portion of a substitution means "the entire matched text". You need to escape it if you want a literal ampersand.
I use Notepad++,
i need to search and replace entire word that contain a specific keyword.
Ex: someting HELP.blablabla.blabla someting
i would like to search entire text for words that contain the keyword "HELP" untill the first space OR the first comma.
In this case: HELP.blablabla.blabla
thanks a lot
Go to the search panel, check the regex checkbox on the bottom and try: (HELP)([^ ,]*)
Note: There are a space character after the ^
This regex means: Search for the entire word HELP (HELP) followed by anything that it isn't an space or an comma [^ ,] the ^ inside the brackets is a denial
Edit:
You can use just HELP[^ ,]* the parenthesis is just to create capturing groups if you need to use the specific groups to replace later. As pointed by #alphabravo
You say search and replace an entire word but if it were that simple then I wonder why a regular search and replace isn't sufficient. So I'm reading between the lines and assuming you want to match on full lines of text.
I think I've used npp enough to get the syntax right. I don't remember any eccentricities that would apply. Is the comma/space optional?
^[^, ]*HELP[^, ]*[, ]
I'm kinda thinking this one might be good enough:
^[^, ]*HELP
I am facing a problem whereby I am given a string that contains a path to a file and the file's name and I only want to extract the path (without the file's name)
For example, I will receive something like
C:\Users\OopsD\Projects\test.acdbd
and from that string I want to extract only
C:\Users\OopsD\Projects
I was trying to create a RegEx to match a backslash followed by a word, followed by a dot followed by another word - this is to match the
\test.acdbd
part and replace it with empty string so that the final result is
C:\Users\OopsD\Projects
Can anyone, familiar with RegEx, help me on this one? Also, I will be using regular expressions quite a lot in the future. Is there a (free) program I can download to create regular expressions?
Are you really sure you need to be using Regex for such as simple task? How about this:
Dim file As New IO.FileInfo(" C:\Users\OopsD\Projects\test.acdbd")
MsgBox(file.Directory.FullName)
Regarding the free program on Regex, I would definitely recommend http://www.gskinner.com/RegExr/ - using it all the time. But you always have to consider alternatives, before going the Regex way.
The regex that you are looking for is as below:
[^/]+$
where,
^ (caret):Matches at the start of the string the regex pattern is applied to. Matches a position rather than a character. Most regex flavors have an option to make the caret match after line breaks (i.e. at the start of a line in a file) as well.
$ (dollar):Matches at the end of the string the regex pattern is applied to. Matches a position rather than a character. Most regex flavors have an option to make the dollar match before line breaks (i.e. at the end of a line in a file) as well. Also matches before the very last line break if the string ends with a line break.
+ (plus):Repeats the previous item once or more. Greedy, so as many items as possible will be matched before trying permutations with less matches of the preceding item, up to the point where the preceding item is matched only once.
More reference can be found out at this link.
Many Regex softwares and tools are out there. Some of them are:
www.gskinner.com/RegExr/
www.txt2re.com
Rubular- It is not just for Ruby.
I got a string like this:
PREFIX-('STRING WITH SPACES TO REPLACE')
and i need this:
PREFIX-('STRING_WITH_SPACES_TO_REPLACE')
I'm using Notepad++ for the Regex Search and Replace, but i'm shure every other Editor capable of regex replacements can do it to.
I'm using:
PREFIX-\('(.*)(\s)(.*)'\)
for search and
PREFIX-('\1_\3')
for replace
but that replaces only one space from the string.
The regex search feature in Notepad++ is very, very weak. The only way I can see to do this in NPP is to manually select the part of the text you want to work on, then do a standard find/replace with the In selection box checked.
Alternatively, you can run the document through an external script, or you can get a better editor. EditPad Pro has the best regex support I've ever seen in an editor. It's not free, but it's worth paying for. In EPP all I had to do was this:
search: ((?:PREFIX-\('|\G)[^\s']+)\s+
replace: $1_
EDIT: \G matches the position where the previous match ended, or the beginning of the input if there was no previous match. In other words, the first time you apply the regex, \G acts like \A. You can prevent that by adding a negative lookahead, like so:
((?:PREFIX-\('|(?!\A)\G)[^\s']+)\s+
If you want to prevent a match at the very beginning of the text no matter what it starts with, you can move the lookahead outside the group:
(?!\A)((?:PREFIX-\('|\G)[^\s']+)\s+
And, just in case you were wondering, a lookbehind will work just as well as a lookahead:
((?:PREFIX-\('|(?<!\A)\G)[^\s']+)\s+
You have to keep matching from the beggining of the string untill you can match no more.
find /(PREFIX-\('[^\s']*)\s([^']*'\))/
replace $1_$2
like: while (/(PREFIX-\('[^\s']*)\s([^']*'\))/$1_$2/) {}
How about using Replace all for about 20 times? Or until you're sure no string contains more spaces
Due to nature of regex, it's not possible to do this in one step by normal regular expression.
But if I be in your place, I do such replaces in several steps:
find such patterns and mark them with special character
(Like replacing STRING WITH SPACES TO REPLACE with #STRING WITH SPACES TO REPLACE#
Replace #([^#\s]*)\s to #\1_ server times.
Remove markers!
I studied a little the regex tool in Notepad++ because I didn't know their possibilities.
I conclude that they aren't powerful enough to do what you want.
Your are obliged to learn and use a programming language having a real regex capability. There are a number of them. Personnaly, I use Python. It would take 1 mn to do what you want with it
You'd have to run the replace several times for each space but this regex will work
/(?<=PREFIX-\(')([^\s]+)\s+/g
Replace with
\1_ or $1_
See it working at http://refiddle.com/10z
I am having issues doing a string replacement in Notepad++, and need some help.
My file:
LastName,(tab)FirstName[optional]MiddleName
Some times there is data that has a middle name, sometimes not.
Public,JohnQ.
Doe,John
Clinton,WilliamJefferson
would be:
Public(tab)John(tab)Q
Doe(tab)John
Clinton(tab)William(tab)Jefferson
I want to split it out into this:
LastName(tab)FirstName(tab)MiddleName
Thanks for adding the sample input. It helps immensely to have that around. Try this and see if it does what you want.
Find, making sure Match case is checked:
([A-Z][a-z]*),([A-Z][a-z]*)(.*)
Replace with:
\1(tab)\2(tab)\3
Of course, (tab) is actually a tab character that you have to place in the replacement string yourself.
An ugly regex like this works for me on the example you've provided:
(\w+),(\w+?)(([A-Z]\w*\.?)?)\n
replace with
\1\t\2\t\3\n
Note:
This only works if the middle name starts with a letter in the A-Z. You might be able to replace [A-Z] with [[:upper:]] if notepad++ supports it (I don't know).
I need that second bracket around the middle name part because I need to match at least an empty string when there is no middle name.