I need help creating a regular expression.
Here are two sample strings:
/path/to/file.jpg
/path/to/file.type.jpg
Respectively, I'm trying to capture:
file.jpg
file.type.jpg
But I want to capture the three as separate strings.
file,jpg
file,type,jpg
Note that I'm not capturing the periods.
I thought something like this could work (excluding the new lines):
([a-z]+)\.
[([a-z]+)[\.]{1}]?
([a-z]{3})
Guidance would be appreciated.
I'm wondering if there is another modified I would need to use to have it capture it properly.
The above expression errors out, by the way :(
I suggest you to use pattern
\/([^.]+)\.?([^.]+|)\.([^.]+)$
and you will have 3 groups: file, type (which will be empty, if not present) and extension
You'd have to use:
/(\w+)(\.(\w+))?\.(\w+){3,4}\b
Then capturing groups 1, 3 and 4 would be your: file(1) type(3) and jpg/png whatever(4)
Groups taken apart:
(\w+) - matches word characters 1 or more (equivalent of saying: {1, }
(\.(\w+))? - matches the 3rd group and with a dot in front, and makes the whole group optional ( ? )
(\w+) - as gr 1
(\w{3,4})\b - matchees 3 or 4 word characters ( {3,4} ) and ensures that after those chracters there are no other characters (word end - \b - ! if supported !)
You can use: "\/(?:\w+\/)+(\w+)\.?(\w+)?\.(\w+)" as regex.
Edit: didnt read about not matching dots.
Live Demo
This regex should work:
/(\w+)\.(\w+)(?:\.(\w+))?$/
Live Demo
Related
edit
I've realized I made a mistake when explaining myself. Apologies for that.
Most of the artifacts come from this path:
D:\Folder1\Folder2\Folder3\Folder4\Folder5\
then breaks into Artifact folders and its sub-folders like this:
D:\Folder1\Folder2\Folder3\Folder4\Folder5\ArtifactFolder\Artifact\Artifact-1.0\data.xxx
D:\Folder1\Folder2\Folder3\Folder4\Folder5\ArtifactFolder\Artifact\Artifact-1.1\data.xxx
D:\Folder1\Folder2\Folder3\Folder4\Folder5\ArtifactFolder\Artifact\Artifact-1.2\data.xxx
I would appreciate help with following thing:
I have this list (around 5k rows) of paths to different artifacts and they have different versions, to give you an example:
D:\Folder1\Folder2\Folder3\Folder4\Folder5\ArtifactFolder\Artifact\Artifact-1.0\data.xxx
D:\Folder1\Folder2\Folder3\Folder4\Folder5\ArtifactFolder\Artifact\Artifact-1.1\data.xxx
D:\Folder1\Folder2\Folder3\Folder4\Folder5\ArtifactFolder\Artifact\Artifact-1.2\data.xxx
D:\Folder1\Folder2\Folder3\Folder4\Folder5\ArtifactFolder2\Artifact\Artifact-1.1\data.xxx
D:\Folder1\Folder2\Folder3\Folder4\Folder5\ArtifactFolder2\Artifact\Artifact-1.2\data.xxx
D:\Folder1\Folder2\Folder3\Folder4\Folder5\ArtifactFolder2\Artifact\Artifact-1.3\data.xxx
D:\Folder1\Folder2\Folder3\Folder4\Folder5\ArtifactFolder3\Artifact\Artifact-1.2\data.xxx
D:\Folder1\Folder2\Folder3\Folder4\Folder5\ArtifactFolder3\Artifact\Artifact-1.3\data.xxx
And my goal to achieve is this:
D:\Folder1\Folder2\Folder3\Folder4\Folder5\ArtifactFolder\Artifact\Artifact-1.0\data.xxx
D:\Folder1\Folder2\Folder3\Folder4\Folder5\ArtifactFolder2\Artifact\Artifact-1.1\data.xxx
D:\Folder1\Folder2\Folder3\Folder4\Folder5\ArtifactFolder3\Artifact\Artifact-1.2\data.xxx
Basically to scope it down to just 1 version.
I've tried using ^(.*)(\n\1)+$ and $1. but that obviously didn't work. So I was wondering if you have an idea how to approach this. Greatly appreciate help, thanks!
You can use
Find what: ^(.*\.)(\d+)\\[^\\\n]+(\n\1\d+\\[^\\\n]+)+$
Replace: $1$2\\
See the regex demo. Details:
^ - start of a line (it is the default ^ behavior in Visual Studio Code)
(.*\.) - Group 1: any one or more chars other than line break chars as many as possible and then a .
(\d+) - Group 2:
\\ - a \ char
[^\\\n]+ - one or more chars other than \ and a line break
(\n\1\d+\\[^\\\n]+)+ - Group 3 capturing one or more sequences of a line break and then the value captured into Group 1, one or more digits, a \ char and then one or more chars other than \ and a line break
$ - end of a line.
Here is another attempt, see regex101 demo.
The basic idea is to isolate someText-\d?. in capture group 2.
Then look for $2 in following lines. What precedes $2 or follows $2 in those following lines can vary.
Find: ^(.*\\(?=.*\\))(.*-\d+\.)(.*\\?.*)(\n.*\2.*)*
Replace: $1$2$3
So here is the most interesting part: ^(.*\\(?=.*\\))(.*-\d+\.)
This will get your Artifact-1. or Artifact-17. or someText-2. into capture group 2. Because using a positive lookahead (?=.*\\) the following group 2 (.*-\d+\.) will be in the last directory only. And then (.*\\?.*) gathers the rest of that line into group 3.
Finally (\n.*\2.*)* checks to see if there is a backreference to group 2, \2, in any following lines. [Technically, that backreference could be anywhere in a line, even the beginning, that can be fixed if necessary - let me know if you need that for your data. See safer regex101 demo if 'someText-/d.' could appear anywhere and should be ignored if not last directory and use that find.]
You can not use a single capture group for the whole line using ^(.*), as you want to repeat only the part before the last dot using a backreference and that will not work capturing the whole line.
Therefore you have to capture the digits in the first match in a separate capture group to keep it in the replacement.
If you want to match all following lines with the same text before the last dot, you can use a repeating group:
^\s*(.*\.)(\d+\\[^\\\r\n]*)(?:\r?\n\s*\1\d*\\[^\\\r\n]*)+
The pattern matches:
^ Start of string
\s* Match optional whitespace chars
(.*\.) Capture group 1, match till the last dot
(\d+\\[^\\\r\n]*) Capture group 2, match 1+ digits, \ and optional chars other than \ or a newline
(?: Non capture group
\r?\n\s*\1 Match a newline and a backreference to group 1
\d+\\[^\\\r\n]* Same pattern as in the first part
)+ Close the non capture group and repeat 1+ times
See a regex demo.
In the replacement use the 2 capture groups $1$2
The replacement will look like
D:\Folder1\Folder2\Folder3\Folder4\Folder5\ArtifactFolder\Artifact\Artifact-1.0\data.xxx
D:\Folder1\Folder2\Folder3\Folder4\Folder5\ArtifactFolder2\Artifact\Artifact-1.1\data.xxx
D:\Folder1\Folder2\Folder3\Folder4\Folder5\ArtifactFolder3\Artifact\Artifact-1.2\data.xxx
I am trying to use replace in Sublime using regular expressions but I'm stuck. I tried various combinations but don't seem to be getting there.
This is the input and my desired output:
Input: N_BBP_c_46137_n
Output : BBP
I tried combinations of:
[^BBP]+\b
\*BBP*+\g
But none of the above (and many others) don't seem to work.
To turn N_BBP_c_46137_n into BBP and according to the comment just want that entire long name such as N_BBP_ to be replaced by only BBP* you might also use a capture group to keep BBP.
\bN_(BBP)_\S*
\bN_ Match N preceded by a word boundary
(BBP) Capture group 1, match BBP (or use [A-Z]+ to match 1+ uppercase chars)
_\S* Match _ followed by 0+ times a non whitespace char
In the replacement use the first capturing group $1
Regex demo
You may use
(N_)[^_]*(_c_\d+_n)
Replace with ${1}some new value$2.
Details
(N_) - Group 1 ($1 or ${1} if the next char is a digit): N_
[^_]* - any 0 or more chars other than _
-(_c_\d+_n) - Group 2 ($2): _c_, 1 or more digits and then _n.
See the regex demo.
I'm trying to extract match patterns like:
AA/8G+8G+8G+8G/WITHOUT *
AA/8*G+8*G+8*G+8*G/WITH *
AA/8G+8*G+8*G+8*G/MIXED, THIS IS NOT SUPPORTED YET
using the following regular expression:
https://regex101.com/r/zemJ8H/1
but it matches only 8G+8G+8G+
because the pattern is identified as 8G**+**
Is there any way to include the last 8G (without +) in the group?
This expression takes care of all three cases, including the mixed one:
(?<=/)([0-9]{1,2})[*]*([GM])[+]?(\1[*]*\2[+]?)+
Demo
The idea is to separate out the capturing of digits (new \1) from letters (the \2) and use both captures in the repeating group (\1[\*]*\2[\+]?)+ at the end of the expression.
AA/(8\*?G\+?)+/
https://regex101.com/r/zemJ8H/2
What about something like this? AA followed by a plan followed by any number of sequences containing an 8, optionally a *, a G , and optionally a following +, finally followed by a +.
I'm tying to write some regex matching a start and end of a string.
start:
https://www.example.com.au/
end:
-end
Example input/match:
Input IsMatch
https://www.example.com.au/hithere-end Y
https://www.example.com.au/hi-there-end Y
https://www.example.com.au/hithere-endx N
https://www.example.com.au/end N
This is what i have so far:
^https?://(www\.)?example\.com\.au/[A-z](\-end)$
Any help?
Thanks.
Try this pattern:
^https?:\/\/(?:www\.)?example\.com\.au\/(.+)-end$
Changes from your pattern:
/ are escaped (with \, 3 times).
The first group changed to a non-capturing one (?:).
[A-z] matches a single capital letter. Changed to (.+)
(a capturing group).
Removed parentheses from the last group (you don't want to capture it), hence \ is also not needed.
The "middle part" you want to capture is in group 1.
Check this:
^(https?://(www\.)?example\.com\.au/)[A-z]*(-end)$
Should work.
Try this C# code
Somestring.StartsWith("https://www.example.com.au/")
Somestring.EndsWith("-end")
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.