I want to find words starting with stop. and extract the string that follows the word. Each string should be in a new line.
Also the results file should not have any duplicates.
Example file:
example regex stop.variant1
stop stop.variant_2 examplestop
stopstopvariant
stop.variant_#_3
Result:
variant1
variant_2
variant_#_3
Ctrl+H
Find what: .*?(\bstop\.variant\S*)
Replace with: $1\n
CHECK Match case
CHECK Wrap around
CHECK Regular expression
CHECK . matches newline*
Replace all
Explanation:
.*? # 0 or more any character
( # group 1
\b # word boundary
stop\.variant # literally
\S* # 0 or more non spaces
) # end group
Screen capture:
Related
I have three lines of tab-separated values:
SELL 2022-06-28 12:42:27 39.42 0.29 11.43180000 0.00003582
BUY 2022-06-28 12:27:22 39.30 0.10 3.93000000 0.00001233
_____2022-06-28 12:27:22 39.30 0.19 7.46700000 0.00002342
The first two have 'SELL' or 'BUY' as first value but the third one has not, hence a Tab mark where I wrote ______:
I would like to capture the following using Regex:
My expression ^(BUY|SELL).+?\r\n\t does not work as it gets me this:
I do know why outputs this - adding an lazy-maker '?' obviously won't help. I don't get lookarounds to work either, if they are the right means at all. I need something like 'Match \r\n\t only or \r\n(?:^\t) at the end of each line'.
The final goal is to make the three lines look at this at the end, so I will need to replace the match with capturing groups:
Can anyone point me to the right direction?
Ctrl+H
Find what: ^(BUY|SELL).+\R\K\t
Replace with: $1\t
CHECK Match case
CHECK Wrap around
CHECK Regular expression
UNCHECK . matches newline
Replace all
Explanation:
^ # beginning of line
(BUY|SELL) # group 1, BUY or SELL
.+ # 1 or more any character but newline
\R # any kind of linebreak
\K # forget all we have seen until this position
\t # a tabulation
Replacement:
$1 # content of group 1
\t # a tabulation
Screenshot (before):
Screenshot (after):
You can use the following regex ((BUY|SELL)[^\n]+\n)\s+ and replace with \1\2.
Regex Match Explanation:
((BUY|SELL)[^\n]+\n): Group 1
(BUY|SELL): Group 2
BUY: sequence of characters "BUY" followed by a space
|: or
SELL: sequence of characters "SELL" followed by a space
[^\n]+: any character other than newline
\n: newline character
\s+: any space characters
Regex Replace Explanation:
\1: Reference to Group 1
\2: Reference to Group 2
Check the demo here. Tested on Notepad++ in a private environment too.
Note: Make sure to check the "Regular expression" checkbox.
Regex
I have a text file with the following lines:
asm-java-2.0.0-lib
cib-slides-3.1.0
lib-hibernate-common-4.0.0-beta
astp
act4lib-4.0.0
I want to remove everything from, including the '-' before the numbers begin so the results look like:
2.0.0-lib
3.1.0
4.0.0-beta
act4lib
Does anyone know the correct regex for this? So far I have come up with -\D.*(a-z)* but its got too many errors.
Ctrl+H
Find what: ^.*?(?=\d|$)
Replace with: LEAVE EMPTY
check Wrap around
check Regular expression
UNCHECK . matches newline
Replace all
Explanation:
^ # beginning of line
.*? # 0 or more any character but newline, not greedy
(?= # start lookahead, zero-length assertion that makes sure we have after
\d # a digit
| # OR
$ # end of line
) # end lookahead
Result for given example:
2.0.0-lib
3.1.0
4.0.0-beta
Another solution that deals with act4lib-4.0.0:
Ctrl+H
Find what: ^(?:.*-(?=\d)|\D+)
Replace with: LEAVE EMPTY
check Wrap around
check Regular expression
UNCHECK . matches newline
Replace all
Explanation:
^ # beginning of line
(?: # start non capture group
.* # 0 or more any character but newline
- # a dash
(?=\d) # lookahead, zero-length assertion that makes sure we have a digit after
| # OR
\D+ # 1 or more non digit
) # end group
Replacement:
\t # a tabulation, you may replace with what you want
Given:
asm-java-2.0.0-lib
cib-slides-3.1.0
lib-hibernate-common-4.0.0-beta
astp
act4lib-4.0.0
Result for given example:
2.0.0-lib
3.1.0
4.0.0-beta
4.0.0
Use
^\D+\-
If you want to completely remove lines without numbers then use this
^\D+(\-|$)
In case the packages contain numbers in their names like act4lib-4.0.0 then a longer variant is needed
^[\w-]+(\-(?=\d+\.\d+)|$)
It can be shortened to ^.+?(\-(?=\d+\.)|$) but I just want to be sure so I also check the minor version number
The ^ will match from the start of line
I have a large list of urls that has a unique numeric string in each, the string falls between a / and a ? I would like to remove all other text from notepad++ that are not these strings. for example
www.website.com/dsw/fv3n24nv1e4121v/123456789012?fwe=32432fdwe23f3 would end up as only 123456789012
I have figured out that the following regex \b\d{12}\b will get me the 12 digits, now I just need to remove all of the information that falls each side. I have had a look and found some posts that suggest replace with \t$1 , $1\n
, $1 , and /1 however all of these do the exact oposite of what I want and just remove the 12 digit string.
You can use this regex and replace it with empty string,
^[^ ]*\/|\?[^ ]*$
Demo
Explanation:
^[^ ]*\/ --> Matches anything expect space from start of string till it finds a /
\?[^ ]*$ --> Similarly, this matches anything except space starting from ? till end of input.
Ctrl+H
Find what: ^.*/([^?]+).*$
Replace with: $1
check Wrap around
check Regular expression
UNCHECK . matches newline
Replace all
Explanation:
^ # beginning of line
.* # 0 or more any character but newline
/ # a slash
([^?\r\n]+) # group 1, 1 or more any character that is not ? or line break
.* # 0 or more any character but newline
$ # end of line
Result for given example:
123456789012
I have a list of words, for example:
Good -> Bad
Sky -> Blue
Gray -> Black
etc...
What is the best why to do find&replace in notepad++?
I tried this:
FIND: (Good)|(Sky)|(Gray)
Replace: (?1Bad)(?2Blue)(?3Black)
but it doesn't work :(
any idea? or suggestions ?
There is however a workaround if you add this newline at the end of your text (it must be the last line, so don't press enter at the end):
#Good:Bad#Sky:Blue#Gray:Black#
and if you use this pattern:
(Good|Blue|Black)(?=(?:.*\R)++#(?>[^#]+#)*?\1:([^#]+))|\R.++(?!\R)
with this replacement:
$2
pattern details:
(Good|Blue|Black) # this part capture the word in group 1
(?= # then we reach the last line in a lookakead
(?:.*\R)++ # match all the lines until the last line
#(?>[^#]+#)*? # advance until the good value is found
\1 # the good value (backreference to the capture group 1)
: ([^#]+) # capture the replacement in group 2
) # close the lookbehind
| # OR
\R.++(?!\R) # match the last line (to remove it)
Note: to make the pattern more efficient, you can put it in a non capturing group and add a lookahead at the begining with all the first possible characters to quickly discard useless positions in the string:
(?=[GB\r\n])(?:\b(Good|Blue|Black)\b(?=(?:.*\R)++#(?>[^#]+#)*?\1:([^#]+))|\R.++(?!\R))
Trying to learn regular expressions. As a practice, I'm trying to find every word that appears exactly one time in my document -- in linguistics this is a hapax legemenon (http://en.wikipedia.org/wiki/Hapax_legomenon)
So I thought the following expression give me the desired result:
\w{1}
But this doesn't work. The \w returns a character not a whole word. Also it does not appear to be giving me characters that appear only once (it actually returns 25873 matches -- which I assume are all alphanumeric characters). Can someone give me an example of how to find "hapax legemenon" with a regular expression?
If you're trying to do this as a learning exercise, you picked a very hard problem :)
First of all, here is the solution:
\b(\w+)\b(?<!\b\1\b.*\b\1\b)(?!.*\b\1\b)
Now, here is the explanation:
We want to match a word. This is \b\w+\b - a run of one or more (+) word characters (\w), with a 'word break' (\b) on either side. A word break happens between a word character and a non-word character, so this will match between (e.g.) a word character and a space, or at the beginning and the end of the string. We also capture the word into a backreference by using parentheses ((...)). This means we can refer to the match itself later on.
Next, we want to exclude the possibility that this word has already appeared in the string. This is done by using a negative lookbehind - (?<! ... ). A negative lookbehind doesn't match if its contents match the string up to this point. So we want to not match if the word we have matched has already appeared. We do this by using a backreference (\1) to the already captured word. The final match here is \b\1\b.*\b\1\b - two copies of the current match, separated by any amount of string (.*).
Finally, we don't want to match if there is another copy of this word anywhere in the rest of the string. We do this by using negative lookahead - (?! ... ). Negative lookaheads don't match if their contents match at this point in the string. We want to match the current word after any amount of string, so we use (.*\b\1\b).
Here is an example (using C#):
var s = "goat goat leopard bird leopard horse";
foreach (Match m in Regex.Matches(s, #"\b(\w+)\b(?<!\b\1\b.*\b\1\b)(?!.*\b\1\b)"))
Console.WriteLine(m.Value);
Output:
bird
horse
It can be done in a single regex if your regex engine supports infinite repetition inside lookbehind assertions (e. g. .NET):
Regex regexObj = new Regex(
#"( # Match and capture into backreference no. 1:
\b # (from the start of the word)
\p{L}+ # a succession of letters
\b # (to the end of a word).
) # End of capturing group.
(?<= # Now assert that the preceding text contains:
^ # (from the start of the string)
(?: # (Start of non-capturing group)
(?! # Assert that we can't match...
\b\1\b # the word we've just matched.
) # (End of lookahead assertion)
. # Then match any character.
)* # Repeat until...
\1 # we reach the word we've just matched.
) # End of lookbehind assertion.
# We now know that we have just matched the first instance of that word.
(?= # Now look ahead to assert that we can match the following:
(?: # (Start of non-capturing group)
(?! # Assert that we can't match again...
\b\1\b # the word we've just matched.
) # (End of lookahead assertion)
. # Then match any character.
)* # Repeat until...
$ # the end of the string.
) # End of lookahead assertion.",
RegexOptions.Singleline | RegexOptions.IgnorePatternWhitespace);
Match matchResults = regexObj.Match(subjectString);
while (matchResults.Success) {
// matched text: matchResults.Value
// match start: matchResults.Index
// match length: matchResults.Length
matchResults = matchResults.NextMatch();
}
If you are trying to match an English word, the best form is:
[a-zA-Z]+
The problem with \w is that it also includes _ and numeric digits 0-9.
If you need to include other characters, you can append them after the Z but before the ]. Or, you might need to normalize the input text first.
Now, if you want a count of all words, or just to see words that don't appear more than once, you can't do that with a single regex. You'll need to invest some time in programming more complex logic. It may very well need to be backed by a database or some sort of memory structure to keep track of the count. After you parse and count the whole text, you can search for words that have a count of 1.
(\w+){1} will match each word.
After that you could always perfrom the count on the matches....
Higher level solution:
Create an array of your matches:
preg_match_all("/([a-zA-Z]+)/", $text, $matches, PREG_PATTERN_ORDER);
Let PHP count your array elements:
$tmp_array = array_count_values($matches[1]);
Iterate over the tmp array and check the word count:
foreach ($tmp_array as $word => $count) {
echo $word . ' ' . $count;
}
Low level but does what you want:
Pass your text in an array using split:
$array = split('\s+', $text);
Iterate over that array:
foreach ($array as $word) { ... }
Check each word if it is a word:
if (!preg_match('/[^a-zA-Z]/', $word) continue;
Add the word to a temporary array as key:
if (!$tmp_array[$word]) $tmp_array[$word] = 0;
$tmp_array[$word]++;
After the loop. Iterate over the tmp array and check the word count:
foreach ($tmp_array as $word => $count) {
echo $word . ' ' . $count;
}