What I am trying to do
I have a string that looks like this:
foobar
* Level1-1
* Level1-2
** Level2-1
** Level2-2
*** Level3-1
*** Level3-2
foo
foo
foo
bar
* Level1-1
foo
bar
foo
bar
* Level1-1
** Level2-1
foo
bar
I would like to use Regex to capture the lines starting with an Asterisk together, so given the string above I get the following three results captured together:
Result 1
* Level1-1
* Level1-2
** Level2-1
** Level2-2
*** Level3-1
*** Level3-2
Result 2
* Level1-1
Result 3
* Level1-1
** Level2-1
What I tried
I tried to use this regex with a multiline flag (/m):
^(?<Content>\*(.|\n|\r|\n\r)+)(?=[\n\r]+[^\*]+)
The regex as is understand/intended it:
^ = Line/String Start
(?<Content>\*(.|\n|\r|\n\r)+) = The Capture Group that multiline-matches all lines that start with an Asterisk
(?=[\n\r]+[^\*]+) = Positive Lookahead to match any line that does not start with/contain an Asterisk, thus ending the match.
I expected this regex to match what I need, but it actually matches the whole string apart from the first 2 lines and the last line of my string.
I know that I could easily match the single lines with the following regex (^\*.*), but I need the subsequent lines containing Asterisks to go into a single group instead of one group for each line.
The Question
I am not sure what I am doing wrong, especially with the positive lookahead part, and I would be very grateful for any advice on how I can achieve my goal.
I initially started trying to use multiline mode, but then gave up and fell back to a plain regex with no special modes:
(?<=^|\n)\*+.*(?:\n\*+.*)*
Demo
Explanation:
(?<=^|\n) assert that match begins with start of the line
\*+ match one or more stars
.* match rest of the line
(?:
\n match a newline
\*+ match one or more stars
.* match the rest of the line
)* zero or more times
Your regex ^(?<Content>\*(.|\n|\r|\n\r)+)(?=[\n\r]+[^\*]+) with the multiline flag matches too much because it is greedy.
You can make the pattern non greedy to get the desired matches:
^(?<Content>\*(?:.|\n|\r|\n\r)+?)(?=[\n\r]+[^\*]+)
See a regex101 demo.
Which is shorted written as this:
^(?<Content>\*[\s\S]+?)(?=\r?\n[^*])
But note that both patterns need to match a newline followed by at least a single char other than an asterix due to this part [^\*]
If there does not have to be a newline, you could write it like this:
^(?<Content>\*.*(?:\r?\n\*.*)*)
See another regex101 demo.
Related
I would like to mask the email passed in the maskEmail function. I'm currently facing a problem wherein the asterisk * is not repeating when i'm replacing group 2 and and 4 of my pattern.
Here is my code:
fun maskEmail(email: String): String {
return email.replace(Regex("(\\w)(\\w*)\\.(\\w)(\\w*)(#.*\\..*)$"), "$1*.$3*$5")
}
Here is the input:
tom.cat#email.com
cutie.pie#email.com
captain.america#email.com
Here is the current output of that code:
t*.c*#email.com
c*.p*#email.com
c*.a*#email.com
Expected output:
t**.c**#email.com
c****.p**#email.com
c******.a******#email.com
Edit:
I know this could be done easily with for loop but I would need this to be done in regex. Thank you in advance.
For your problem, you need to match each character in the email address that not is the first character in a word and occurs before the #. You can do that with a negative lookbehind for a word break and a positive lookahead for the # symbol:
(?<!\b)\w(?=.*?#)
The matched characters can then be replaced with *.
Note we use a lazy quantifier (?) on the .* to improve efficiency.
Demo on regex101
Note also as pointed out by #CarySwoveland, you can replace (?<!\b) with \B i.e.
\B\w(?=.*?#)
Demo on regex101
As pointed out by #Thefourthbird, this can be improved further efficiency wise by replacing the .*? with a [^\r\n#]* i.e.
\B\w(?=[^\r\n#]*#)
Demo on regex101
Or, if you're only matching single strings, just [^#]*:
\B\w(?=[^#]*#)
Demo on regex101
I suggest keeping any char at the start of string and a combination of a dot + any char, and replace any other chars with * that are followed with any amount of characters other than # before a #:
((?:\.|^).)?.(?=.*#)
Replace with $1*. See the regex demo. This will handle emails that happen to contain chars other than just word (letter/digit/underscore) and . chars.
Details
((?:\.|^).)? - an optional capturing group matching a dot or start of string position and then any char other than a line break char
. - any char other than a line break char...
(?=.*#) - if followed with any 0 or more chars other than line break chars as many as possible and then #.
Kotlin code (with a raw string literal used to define the regex pattern so as not to have to double escape the backslash):
fun maskEmail(email: String): String {
return email.replace(Regex("""((?:\.|^).)?.(?=.*#)"""), "$1*")
}
See a Kotlin test online:
val emails = arrayOf<String>("captain.am-e-r-ica#email.com","my-cutie.pie+here#email.com","tom.cat#email.com","cutie.pie#email.com","captain.america#email.com")
for(email in emails) {
val masked = maskEmail(email)
println("${email}: ${masked}")
}
Output:
captain.am-e-r-ica#email.com: c******.a*********#email.com
my-cutie.pie+here#email.com: m*******.p*******#email.com
tom.cat#email.com: t**.c**#email.com
cutie.pie#email.com: c****.p**#email.com
captain.america#email.com: c******.a******#email.com
I'm new to regex, and would appreciate some guidance/help.
Currently, I'm looking to write an expression, that derives a certain part of text from the 2nd line of the provided text.
Here is the text:
123 anywhere Avenue
Winnipeg, Manitoba R3E 0L7
Canada
Pharmacy Manager: person person
Pharmacy Licence Holder/Owner: 123456 Manitoba Ltd.
see correct formatting with code here
My goal is to derive the 'Manitoba' string from the second line, however I'd like to make it dynamic rather than writing an expression to always fetch Manitoba as a static. I used the below code to target the second line:
(.*)(?=(\n.*){3}$)
(It matches 3 lines up from the last line, thus targeting the desired line)
I noticed, that within the dataset, that the Province (Manitoba) is always in between two spaces.
Is there any addition I can make to the code, so that the expression only targets the second line, then matches the first string in-between spaces?
Perhaps using a lazy expression with a positive lookaround?
If I target all matches in between spaces, it would take both 'Manitoba' and 'R3E 0L7' which I dont want.
I want it to only match the first piece of text in between spaces on the second line.
Any help is much appreciated :-)
Thanks.
One option could be to match the first line, then capture the second word in the second lines in capturing group 1.
Then match the rest of the second line and assert what follows is 3 times a line.
^.*\r?\n\S+[^\S\r\n]+(\S+).*(?=(?:\r?\n.*){3}$)
In parts:
^ Start of string
.*\r?\n Match the whole lines and a newline
\S+ Match 1+ non whitespace char (the first "word")
[^\S\r\n]+ Match 1+ times a whitespace char except newlines
(\S+) Capture group 1 Match 1+ times a non whitespace char (the second "word')
.* Match the rest of the line
(?= Positive lookahead, assert what follows on the right is
(?:\r?\n.*){3}$ Match 3 times a newline followed by 0+ times any except a newline and assert the end of the string
) Close lookahead
Regex demo
You could also turn the lookahead in to a match instead
^.*\r?\n\S+[^\S\r\n]+(\S+).*(?:\r?\n.*){3}$
Regex demo
I am trying to work out a regex expression but struggle with conditionals. I have a list of 100s of URLs that look like this:
/name/something/details/55334
/name/page/1/2
/name/somethingdifferent/34523
/name/page/1
/name/something/553/1
Bottom line is that I want to remove everything when a number appears apart from a scenario where the last thing before the number is a word 'page'.
1. /name/something/details/
2. /name/page/1/2
3. /name/somethingdifferent/
4. /name/page/1
5. /name/something
I will be removing it with Google Analytics Content Grouping or potentially with DataStudio. I already removed /name/ so I have:
1. /something/details/55334
2. /page/1/2
3. /somethingdifferent/34523
4. /page/1
5. /something/553/1
but want to add another rule and remove the numbers so I get:
1. /something/details/
2. /page/1/2
3. /somethingdifferent/
4. /page/1
5. /something
have already tried:
\(?(?=(page\/[0-9]+))(\2)|(\/\d+)
following the syntax of:
(?(?=condition))(IF)|(ELSE)
but it highlights all numbers after text.
Thanks for your help.
sampak
Try ^(\/page.*|[^0-9]*), works with your example.
A Version incl. name: ^(page[\/\d]*|[^\d\s])*
One option might be to match not a whitespace or digit while not matching /page.
Then match a forward slash and 1+ digits followed by any char 0+ times to omit that from the result.
^((?:(?!\/page)[^\d\s])*\/)\d.*
In parts
^ Start of string
( Capture group 1
(?: Non capturing group
(?!\/page) Negative lookahead, assert what is directly to the right is not
[^\d\s] Match any char except a digit or whitespace char
)* Close non capturing group and repeat 0+ times
\/ Match /
) Close group 1
\d.* Match a digit followed by any char except a newline 0+ times
In the replacement use the first capturing group
Regex demo
If you also want to remove /name you could use:
^\/name((?:(?!\/page)[^\d\s])*\/)\d.*
Regex demo
So I need to match the following:
1.2.
3.4.5.
5.6.7.10
((\d+)\.(\d+)\.((\d+)\.)*) will do fine for the very first line, but the problem is: there could be many lines: could be one or more than one.
\n will only appear if there are more than one lines.
In string version, I get it like this: "1.2.\n3.4.5.\n1.2."
So my issue is: if there is only one line, \n needs not to be at the end, but if there are more than one lines, \n needs be there at the end for each line except the very last.
Here is the pattern I suggest:
^\d+(?:\.\d+)*\.?(?:\n\d+(?:\.\d+)*\.?)*$
Demo
Here is a brief explanation of the pattern:
^ from the start of the string
\d+ match a number
(?:\.\d+)* followed by dot, and another number, zero or more times
\.? followed by an optional trailing dot
(?:\n followed by a newline
\d+(?:\.\d+)*\.?)* and another path sequence, zero or more times
$ end of the string
You might check if there is a newline at the end using a positive lookahead (?=.*\n):
(?=.*\n)(\d+)\.(\d+)\.((\d+)\.)*
See a regex demo
Edit
You could use an alternation to either match when on the next line there is the same pattern following, or match the pattern when not followed by a newline.
^(?:\d+\.\d+\.(?:\d+\.)*(?=.*\n\d+\.\d+\.)|\d+\.\d+\.(?:\d+\.)*(?!.*\n))
Regex demo
^ Start of string
(?: Non capturing group
\d+\.\d+\. Match 2 times a digit and a dot
(?:\d+\.)* Repeat 0+ times matching 1+ digits and a dot
(?=.*\n\d+\.\d+\.) Positive lookahead, assert what follows a a newline starting with the pattern
| Or
\d+\.\d+\. Match 2 times a digit and a dot
(?:\d+\.)* Repeat 0+ times matching 1+ digits and a dot
*(?!.*\n) Negative lookahead, assert what follows is not a newline
) Close non capturing group
(\d+\.*)+\n* will match the text you provided. If you need to make sure the final line also ends with a . then (\d+\.)+\n* will work.
Most programming languages offer the m flag. Which is the multiline modifier. Enabling this would let $ match at the end of lines and end of string.
The solution below only appends the $ to your current regex and sets the m flag. This may vary depending on your programming language.
var text = "1.2.\n3.4.5.\n1.2.\n12.34.56.78.123.\nthis 1.2. shouldn't hit",
regex = /((\d+)\.(\d+)\.((\d+)\.)*)$/gm,
match;
while (match = regex.exec(text)) {
console.log(match);
}
You could simplify the regex to /(\d+\.){2,}$/gm, then split the full match based on the dot character to get all the different numbers. I've given a JavaScript example below, but getting a substring and splitting a string are pretty basic operations in most languages.
var text = "1.2.\n3.4.5.\n1.2.\n12.34.56.78.123.\nthis 1.2. shouldn't hit",
regex = /(\d+\.){2,}$/gm;
/* Slice is used to drop the dot at the end, otherwise resulting in
* an empty string on split.
*
* "1.2.3.".split(".") //=> ["1", "2", "3", ""]
* "1.2.3.".slice(0, -1) //=> "1.2.3"
* "1.2.3".split(".") //=> ["1", "2", "3"]
*/
console.log(
text.match(regex)
.map(match => match.slice(0, -1).split("."))
);
For more info about regex flags/modifiers have a look at: Regular Expression Reference: Mode Modifiers
First of all I apologize if this question is too naive or has been repeated earlier. I tried to find it in the forum but I'm posting it as a question because I failed to find an answer.
I have a data frame with column names as follows;
head(rownames(u))
[1] "A17-R-Null-C-3.AT2G41240" "A18-R-Null-C-3.AT2G41240" "B19-R-Null-C-3.AT2G41240"
[4] "B20-R-Null-C-3.AT2G41240" "A21-R-Transgenic-C-3.AT2G41240" "A22-R-Transgenic-C-3.AT2G41240"
What I want is to use regex in R to extract the string in between the first dash and the last period.
Anticipated results are,
[1] "R-Null-C-3" "R-Null-C-3" "R-Null-C-3"
[4] "R-Null-C-3" "R-Transgenic-C-3" "R-Transgenic-C-3"
I tried following with no luck...
gsub("^[^-]*-|.+\\.","\\2", rownames(u))
gsub("^.+-","", rownames(u))
sub("^[^-]*.|\\..","", rownames(u))
Would someone be able to help me with this problem?
Thanks a lot in advance.
Shani.
Here is a solution to be used with gsub:
v <- c("A17-R-Null-C-3.AT2G41240", "A18-R-Null-C-3.AT2G41240", "B19-R-Null-C-3.AT2G41240", "B20-R-Null-C-3.AT2G41240", "A21-R-Transgenic-C-3.AT2G41240", "A22-R-Transgenic-C-3.AT2G41240")
gsub("^[^-]*-([^.]+).*", "\\1", v)
See IDEONE demo
The regex matches:
^[^-]* - zero or more characters other than -
- - a hyphen
([^.]+) - Group 1 matching and capturing one or more characters other than a dot
.* - any characters (even including a newline since perl=T is not used), any number of occurrences up to the end of the string.
This can easily be achieved with the following regex:
-([^.]+)
# look for a dash
# then match everything that is not a dot
# and save it to the first group
See a demo on regex101.com. Outputs are:
R-Null-C-3
R-Null-C-3
R-Null-C-3
R-Null-C-3
R-Transgenic-C-3
R-Transgenic-C-3
Regex
-([^.]+)\\.
Description
- matches the character - literally
1st Capturing group ([^\\.]+)
[^\.]+ match a single character not present in the list below
Quantifier: + Between one and unlimited times, as many times as possible, giving back as needed [greedy]
. matches the character . literally
\\. matches the character . literally
Debuggex Demo
Output
MATCH 1
1. [4-14] `R-Null-C-3`
MATCH 2
1. [29-39] `R-Null-C-3`
MATCH 3
1. [54-64] `R-Null-C-3`
MATCH 4
1. [85-95] `R-Null-C-3`
MATCH 5
1. [110-126] `R-Transgenic-C-3`
MATCH 6
1. [141-157] `R-Transgenic-C-3`
This seems an appropriate case for lookarounds:
library(stringr)
str_extract(v, '(?<=-).*(?=\\.)')
where
(?<= ... ) is a positive lookbehind, i.e. it looks for a - immediately before the next captured group;
.* is any character . repeated 0 or more times *;
(?= ... ) is a positive lookahead, i.e. it looks for a period (escaped as \\.) following what is actually captured.
I used stringr::str_extract above because it's more direct in terms of what you're trying to do. It is possible to do the same thing with sub (or gsub), but the regex has to be uglier:
sub('.*?(?<=-)(.*)(?=\\.).*', '\\1', v, perl = TRUE)
.*? looks for any character . from 0 to as few as possible times *? (lazy evaluation);
the lookbehind (?<=-) is the same as above;
now the part we want .* is put in a captured group (...), which we'll need later;
the lookahead (?=\\.) is the same;
.* captures any character, repeated 0 to as many as possible times (here the end of the string).
The replacement is \\1, which refers to the first captured group from the pattern regex.