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
Related
I have hundreds of bib references in a file, and they have the following syntax:
#article{tabata1999precise,
title={Precise synthesis of monosubstituted polyacetylenes using Rh complex catalysts.
Control of solid structure and $\pi$-conjugation length},
author={Tabata, Masayoshi and Sone, Takeyuchi and Sadahiro, Yoshikazu},
journal={Macromolecular chemistry and physics},
volume={200},
number={2},
pages={265--282},
year={1999},
publisher={Wiley Online Library}
}
I would like to title case (aka Proper Case) the journal name in Notepad++ using regular expression. For example, from Macromolecular chemistry and physics to Macromolecular Chemistry and Physics.
I am able to find all instances using:
(?<=journal\=\{).*?(?=\})
but I am unable to change the case via Edit > Convert Case to. Apparently it doesn't work on find all and I have to go one by one.
Next, I tried recording and running a macro but Notepad++ just hangs indefinitely when I try to run it (option to run until the end of the file).
So my question is: does anyone know the replace regex syntax I could use to change the case? Ideally, I would also like to use "|" exclusions for particular words such as " of ", " an ", " the ", etc. I tried to play with some of the examples provided here, but I was not able to integrate it into my look-aheads.
Thank you in advance, I'd appreciate any help.
This works for any number of words:
Ctrl+H
Find what: (?:journal={|\G)\K(?:(\w{4,})|(\w+))(\h*)
Replace with: \u$1\E$2$3
CHECK Wrap around
CHECK Regular expression
Replace all
Explanation:
(?: # non capture group
journal={ # literally
| # OR
\G # restart from last match position
) # end group
\K # forget all we have seen until this position
(?: # non capture group
(\w{4,}) # group 1, a word with 4 or more characters
| # OR
(\w+) # group 2, a word of any length
) # end group
(\h*) # group 3, 0 or more horizontal spaces
Replacement:
\u # uppercased the first letter of the following
$1 # content of group 1
\E # stop the uppercased
$2 # content of group 2
$3 # content of group 3
Screenshot (before):
Screenshot (after):
if the format is always in the form:
journal={Macromolecular chemistry and physics},
i.e. journal followed by 3 words then use the following:
Find: journal={(\w+)\s*(\w+)\s*(\w+)\s*(\w+)
Replace with: journal={\u\1 \u\2 \l\3 \u\4
You can modify that if you have more words to replace by adding more \u\x, where x is the position of the word.
Hope it helps to give you an idea to move forward for a better solution.
\u translates the next letter to uppercase (used for all other words)
\l translates the next letter to lowercase (used for the word "and")
\1 replaces the 1st captured () search group
\2 replaces the 2nd captured () search group
\3 replaces the 3rd captured () search group
I am very new to the world of regular expressions. I am trying to use Notepad++ using Regex for the following:
Input file is something like this and there are multiple such files:
Code:
abc
17
015
0 7
4.3
5/1
***END***
abc
6
71
8/3
9 0
***END***
abc
10.1
11
9
***END***
I need to be able to edit the text in all of these files so that all the files look like this:
Code:
abc
1,2,3,4,5
***END***
abc
6,7,8,9
***END***
abc
10,11,12
***END***
Also:
In some files the number of * around the word END varies, is there a way to generalize the number of * so I don't have to worry about it?
There is some additional data before abcs which does not need to be transposed, how do I keep that data as it is along with transposing the data between abc and ***END***.
Kindly help me. Your help is much appreciated!
Try the following find and replace, in regex mode:
Find: ^(\d+)\R(?!\*{1,}END\*{1,})
Replace: $1,
Demo
Here is an explanation of the regex pattern:
^ from the start of the line
(\d+) match AND capture a number
\R followed by a platform independent newline, which
(?!\*{1,}END\*{1,}) is NOT followed by ***END***
Note carefully the negative lookahead at the end of the pattern, which makes sure that we don't do the replacement on the final number in each section. Without this, the last number would bring the END marker onto the same line.
This will eplace only between "abc" and "***END***" with any number of asterisk.
Ctrl+H
Find what: (?:(?<=^abc)\R|\G(?!^)).+\K\R(?!\*+END\*+)
Replace with: ,
CHECK Match case
CHECK Wrap around
CHECK Regular expression
UNCHECK . matches newline*
Replace all
Explanation:
(?: # non capture group
(?<=^abc) # positive look behind, make sure we have "abc" at the beginning of line before
\R # any kind of linebreak
| # OR
\G # restart from last match position
(?!^) # negative look ahead, make sure we are not at the beginning of line
) # end group
.+ # 1 or more any character but newline
\K # forget all we have seen until this position
\R # any kind of linebreak
(?!\*+END\*+) # negative lookahead, make sure we haven't ***END*** after
Screen capture (before):
Screen capture (after):
I have a list of thousands of records within a .txt document.
some of them look like these records
201910031044 "00059" "11.31AG" "Senior Champion"
201910031044 "00060" "GBA146" "Junior Champion"
201910031044 "00999" "10.12G" "ProAM"
201910031044 "00362" "113.1LI" "Abcd"
Whenever a record similar to this occurs I'd like to get rid of the last words/numbers/etc in the last quotation marks (like "Senior Champion", "Junior Champion" etc. There are many possibilities here)
e.g. (before)
201910031044 "00059" "11.31AG" "Senior Champion"
after
201910031044 "00059" "11.31AG"
I tried the following regex but it wouldn't work.
Search: ^([0-9]{17,17} + "[0-9]{8,8}" + "[a-zA-Z0-9]").*$
Replace: \1 (replace string)
OK I forgot the . (dot) sign however even if I do not have a . (dot) sign it would not work. Not sure if it has anything to do when using the + sign used more than once.
I'd like to get rid of the last words/numbers/etc in the last quotation marks
This does the job:
Ctrl+H
Find what: ^.+\K\h+".*?"$
Replace with: LEAVE EMPTY
CHECK Wrap around
CHECK Regular expression
UNCHECK . matches newline*
Replace all
Explanation:
^ # beginning of line
.+ # 1 or more any character but newline
\K # forget all we have seen until this position
\h+ # 1 or more horizontal spaces
".*?" # something inside quotes
$ # end of line
Screen capture (before):
Screen capture (after):
The RegEx looks for the 4th double quote:
^(?:[^"]*\"){4}([^|]*)
You can see this demo: https://regex101.com/r/wJ9yS6/163
You will still need to parse the lines, so probably easier opening in excel or parsing using code as a CSV.
You have a problem with the count of your characters:
you specify that the line should start with exactly 17 digits ([0-9]{17,17}). However, there are only 12 digits in the data 201910031044.
you can specify exactly 12 digits by using {12} or if it could be 12-17, then {12,17}. I'll assume exactly 12 based on the current data.
similarly, for the second column you specify that it's exactly 8 digits surrounded by quotes ("[0-9]{8,8}") but it only has 5 digits surrounded by quotes.
again, you can specify exactly 5 with {5} or 5-8 with {5,8}. I will assume exactly 5.
finally, there is no quantifier for the final field, so the regex tries to match exactly one character that is a letter or a number surrounded by quotes "[a-zA-Z0-9]".
I'm not sure if there is any limit on the number of characters, so I would go with one or more using + as quantifier "[a-zA-Z0-9]+" - if you can have zero or more, then you can use *, or if it's any other count from m to n, then you can use {m,n} as before.
Not a character count problem but the final column can also have dots but the regex doesn't account for. You can just add . inside the square brackets and it will only match dot characters. It's usually used as a wildcard but it loses its special meaning inside a character class ([]), so you get "[a-zA-Z0-9.]+"
Putting it all together, you get
Search: ^([0-9]{12} + "[0-9]{5}" + "[a-zA-Z0-9.]+").*$
Replace: \1
Which will get rid of anything after the third field in Notepad++.
This can be shortened a bit by using \d instead of [0-9] for digits and \s+ for whitespace instead of +. As a benefit, \s will also match other whitespace like tabs, so you don't have to manually account for those. This leads to
Search: ^(\d{12}\s+"\d{5}"\s+"[a-zA-Z0-9.]+").*$
Replace: \1
If you want to get rid of the last words/numbers/etc in the last quotation marks you could capture in a group what is before that and match the last quotation marks and everything between it to remove it using a negated character class.
If what is between the values can be spaces or tabs, you could use [ \t]+ to match those (using \s could also match a newline)
Note that {17,17} and {8,8} may also be written as {17} and {8} which in this case should be {12} and {5}
^([0-9]{12}[ \t]+"[0-9]{5}"[ \t]+"[a-zA-Z0-9.]+")[ \t]{2,}"[^"\r\n]+"
In parts
^ Start of string
( Capture group 1
[0-9]{12}[ \t]+ Match 12 digits and 1+ spaces or tabs
"[0-9]{5}"[ \t]+ Match 5 digits between " and 1+ spaces or tabs
"[a-zA-Z0-9.]+" Match 1+ times any of the listed between "
) Close group
[ \t]{2,} Match 1+ times
"[^"\r\n]+"
In the replacement use group 1 $1
Regex demo
Before
After
I need to check occurrences where I have put one whitespace after a full-stop, and replace it by 2 spaces. I have the Regex for it, but Atom seems to call in invalid.
(?<=\.|\") {1,}(?=[a-zA-Z])
Conditions:
1 spaces after period.
If period in with a closing double quote, then 1 space after the quote.
The above regex works perfectly for my conditions however Atom is not able to validate it. I need to use it for existing files.
You may use
([."]) ([a-zA-Z])
and replace with $1 $2. See the regex demo and a regex graph:
Details
([."]) - Group 1 (its value is referred to with $1 backreference from the replacement pattern): . or "
- a space (use \s to match any whitespace)
([a-zA-Z]) - Group 2 ($2): an ASCII letter.
I have a text file with the following text:
andal-4.1.0.jar
besc_2.1.0-beta
prov-3.0.jar
add4lib-1.0.jar
com_lab_2.0.jar
astrix
lis-2_0_1.jar
Is there any way i can split the name and the version using regex. I want to use the results to make two columns 'Name' and 'Version' in excel.
So i want the results from regex to look like
andal 4.1.0.jar
besc 2.1.0-beta
prov 3.0.jar
add4lib 1.0.jar
com_lab 2.0.jar
astrix
lis 2_0_1.jar
So far I have used ^(?:.*-(?=\d)|\D+) to get the Version and -\d.*$ to get the Name separately. The problem with this is that when i do it for a large text file, the results from the two regex are not in the same order. So is there any way to get the results in the way I have mentioned above?
Ctrl+H
Find what: ^(.+?)[-_](\d.*)$
Replace with: $1\t$2
check Wrap around
check Regular expression
UNCHECK . matches newline
Replace all
Explanation:
^ # beginning of line
(.+?) # group 1, 1 or more any character but newline, not greedy
[-_] # a dash or underscore
(\d.*) # group 2, a digit then 0 or more any character but newline
$ # end of line
Replacement:
$1 # content of group 1
\t # a tabulation, you may replace with what you want
$2 # content of group 2
Result for given example:
andal 4.1.0.jar
besc 2.1.0-beta
prov 3.0.jar
add4lib 1.0.jar
com_lab 2.0.jar
astrix
lis 2_0_1.jar
Not quite sure what you meant for the problem in large file, and I believe the two regex you showed are doing opposite as what you said: first one should get you the name and second one should give you version.
Anyway, here is the assumption I have to guess what may make sense to you:
"Name" may follow by - or _, followed by version string.
"Version" string is something preceded by - or _, with some digit, followed by a dot or underscore, followed by some digit, and then any string.
If these assumption make sense, you may use
^(.+?)(?:[-_](\d+[._]\d+.*))?$
as your regex. Group 1 is will be the name, Group 2 will be the Version.
Demo in regex101: https://regex101.com/r/RnwMaw/3
Explanation of regex
^ start of line
(.+?) "Name" part, using reluctant match of
at least 1 character
(?: )? Optional group of "Version String", which
consists of:
[-_] - or _
( ) Followed by the "Version" , which is
\d+ at least 1 digit,
[._] then 1 dot or underscore,
\d+ then at least 1 digit,
.* then any string
$ end of line