Im trying to match a string to that containsthree consecutive characters at the beginning of the line and the same six consecutive characters at the end.
for example
CCC i love regex CCCCCC
the C's would be highlighted from search
I have found a way to find get the first 3 and the last six using these two regex codes but im struggling to combine them
^([0-9]|[aA-zZ])\1\1 and ([0-9]|[aA-zZ])\1\1\1\1\1$
appreciate any help
If you want just one regular expression to "highlight" only the 1st three characters and last six, maybe use:
(?:^([0-9A-Za-z])\1\1(?=.*\1{6}$)|([0-9A-Za-z])\2{5}(?<=^\2{3}.*)$)
See an online demo
(?: - Open non-capture group to allow for alternations;
^([0-9A-Za-z])\1\1(?=.*\1{6}$) - Start-line anchor with a 1st capture group followed by two backreferences to that same group. This is followed by a positive lookahead to assert that the very last 6 characters are the same;
| - Or;
([0-9A-Za-z])\2{5}(?<=^\2{3}.*)$ - The alternative is to match a 2nd capture group with 5 backreferences to the same followed by a positive lookbehind (zero-width) to check that the first three characters are the same.
Now, if you don't want to be too strict about "highlighting" the other parts, just use capture groups:
^(([0-9A-Za-z])\2\2).*(\2{6})$
See an online demo. Where you can now refer to both capture group 1 and 3.
Related
Regex is great, but I can't for the life of me figure out how I'd express the following constraint, without spelling out the whole permutation:
2 of any digit [0-9]
3 of any other digit [0-9] excluding the above
4 of any third digit [0-9] excluding the above
I've got this monster, which is clearly not a good way of doing this, as it grows exponentially with each additional set of digits:
^(001112222|001113333|001114444|001115555|001116666|0001117777|0001118888|0001119999|0002220000|...)$
OR
^(0{2}1{3}2{4}|0{2}1{3}3{4}|0{2}1{3}4{4}|0{2}1{3}5{4}|0{2}1{3}6{4}|0{2}1{3}7{4}|0{2}1{3}8{4}|...)$
Looks like the following will work:
^((\d)\2(?!.+\2)){2}\2(\d)\3{3}$
It may look a bit tricky, using recursive patterns, but it may look more intimidating then it really is. See the online demo.
^ - Start string anchor.
( - Open 1st capture group:
(\d) - A 2nd capture group that does capture a single digit ranging from 0-9.
\2 - A backreference to what is captured in this 2nd group.
(?!.+\2) - Negative lookahead to prevent 1+ characters followed by a backreference to the 2nd group's match.
){2} - Close the 1st capture group and match this two times.
\2 - A backreference to what is most recently captured in the 2nd capture group.
(\d) - A 3rd capture group holding a single digit.
\3{3} - Exactly three backreferences to the 3rd capture group's match.
$ - End string anchor.
EDIT:
Looking at your alternations it looks like you are also allowing numbers like "002220000" as long as the digits in each sequence are different to the previous sequence of digits. If that is the case you can simplify the above to:
^((\d)\2(?!.\2)){2}\2(\d)\3{3}$
With the main difference is the "+" modifier been taken out of the pattern to allow the use of the same number further on.
See the demo
Depending on whether your target environment/framework/language supports lookaheads, you could do something like:
^(\d)\1(?!\1)(\d)\2\2(?!\1|\2)(\d)\3\3\3$
First capture group ((\d)) allows us to enforce the "two identical consecutive digits" by referencing the capture value (\1) as the next match, after which the negative lookahead ensures the next sequence doesn't start with the previous digit - then we just repeat this pattern twice
Note: If you want to exclude only the digit used in the immediately preceding sequence, change (?!\1|\2) to just (?!\2)
Using PCRE, I want to capture only and all digits in a line which follows a line in which a certain string appears. Say the string is "STRING99". Example:
car string99 house 45b
22 dog 1 cat
women 6 man
In this case, the desired result is:
221
As asked a similar question some time ago, however, back then trying to capture the numbers in the SAME line where the string appears ( Regex (PCRE): Match all digits conditional upon presence of a string ). While the question is similar, I don't think the answer, if there is one at all, will be similar. The approach using the newline anchor ^ does not work in this case.
I am looking for a single regular expression without any other programming code. It would be easy to accomplish with two consecutive regex operations, but this not what I'm looking for.
Maybe you could try:
(?:\bstring99\b.*?\n|\G(?!^))[^\d\n]*\K\d
See the online demo
(?: - Open non-capture group:
\bstring99\b - Literally match "string99" between word-boundaries.
.*?\n - Lazy match up to (including) nearest newline character.
| - Or:
\G(?!^) - Asserts position at the end of the previous match but prevent it to be the start of the string for the first match using a negative lookahead.
) - Close non-capture group.
[^\d\n]* - Match 0+ non-digit/newline characters.
\K - Resets the starting point of the reported match.
\d - Match a digit.
Need some help in regexp matching pattern.
The text goes like here (it's subtitles for video)
...
223
00:20:47,920 --> 00:20:57,520
- Hello! This is good subtitle text.
- Yes! How are you, stackoverflow?
224
00:20:57,520 --> 00:21:11,120
Wow, seems amazing.
- We're good, thanks.
Like, you know, everyone is happy around here with their laptops.
225
00:21:11,120 --> 00:21:14,440
- Understood. Some dumb text
...
I need a set of groups:
startTime, endTime, text
For now my achievements are not very good. I can get startTime, endTime and some text, but not all the text, only the last sentence. I've attached a screenshot.
As you can see, group 3 is capturing text, but only last sentence.
Please, explain me what I'm doing wrong.
Thank you.
Accounting for the possibility there is no new-line character after the final text of your string; Would the following work for you:
(\d\d:\d\d:\d\d,\d\d\d)[ >-]*?((?1))\n(.*?(?=\n\n|\Z))
See the online demo
(\d\d:\d\d:\d\d,\d\d\d) - The same pattern as you used to capture starting time in 1st capture group.
[ >-]*? - 0+ (but lazy) character from the character class up to:
((?1)) - A 2nd capture group which matches the same pattern as 1st group.
\n - A newline-character.
(.*?(?=\n\n|\Z)) - A 3rd capture group that captures anything (including newline with the s-flag) up to a positive lookahead for either two newline characters or the end of the whole string.
Note, some (not all) engines allow for backreferencing a previous subpattern. I guess the app you are using does not. Therefor you can swap the (?1) with your own pattern to capture the 2nd group.
Another option is to use a pattern that would capture all lines in group 3 that do not start with 3 digits.
(\d\d:\d\d:\d\d,\d\d\d) --> (\d\d:\d\d:\d\d,\d\d\d)((?:\r?\n(?!\d\d\d\b).*)*)
Explanation
(\d\d:\d\d:\d\d,\d\d\d) Capture group 1 Match a time like pattern
--> Match literally
(\d\d:\d\d:\d\d,\d\d\d) Capture group 2 Same pattern as group 1
( Capture group 3
(?: Non capture group
\r?\n(?!\d\d\d\b).* Match a newline and assert using a negative lookahead that the line does not start with 3 digits followed by word boundary. If that is the case, match the whole line
)* Optionally repeat all lines
) Close group 3
Regex demo
A bitmore specific pattern could be matching all lines that do not start with 3 digits or a start/end time like pattern.
^(\d\d:\d\d:\d\d,\d\d\d)[^\S\r\n]+-->[^\S\r\n]+(\d\d:\d\d:\d\d,\d\d\d)((?:\r?\n(?!\d+$|\d\d:\d\d:\d\d,\d\d\d\b).*)*)
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
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.