I have the following string
abc|ghy|33d
The regex below matches it fine
^([\d\w]{3}[|]{1})+[\d\w]{3}$
The string changes but the characters separated by the pipe are always in 3's ... so we can have
krr|455
we can also have
ddc
Here's where the problem happens: The regex explained above doesn't match the string if there is only one set of letters ... i.e. "dcc"
Let's do this step by step.
Your regex :
^([\d\w]{3}[|]{1})+[\d\w]{3}$
We can already see some changes. [|]{1} is equivalent to \|.
Then, we see that you match the first part (aaa|) at least once (the + operator matches once at least). Also, \w matches numbers.
The * operator matches 0 or more. So :
^(?:\w{3}\|)*\w{3}$
works.
See here.
Explanation
^ Matches beggining of string
(?:something)* matches something zero time or more. the group is non-capturing as you won't need to
\w{3} matches 3 alphanumeric characters
\| matches |
$ matches end of string.
^[\d\w]{3}(?:[|][\d\w]{3}){0,2}$
You simply quantify the variable part.See demo.
https://regex101.com/r/tS1hW2/18
You can modify your regex as below:
^([\d\w]{3})(\|[\d\w]{3})*$
here first match 3 alphaNumeric and then alphaNum with | as prefix.
Demo
Your description is a little awkward, but I'm guessing you want to be able to match
abc
abc|def
abc|def|ghi
You can do that with
/^\w{3}(?:\|\w{3}){0,2}$/
Visualization
Explanation
^ — match beginning of string
\w{3} — match any 3 of [A-Za-z0-9_]
(? ... )? — non-capturing group, 0 or 1 matches
\| — literal | character
$ — end of string
If the goal is to match any amount of 3-letter segments, you can use
/^(?:\w{3}(?:\||$))+$/
Related
How can I get only the middle part of a combined name with PCRE regex?
name: 211103_TV_storyname_TYPE
result: storyname
I have used this single line: .(\d)+.(_TV_) to remove the first part: 211103_TV_
Another idea is to use (_TYPE)$ but the problem is that I don´t have in all variations of names a space to declare a second word to use the ^ for the first word and $ for the second.
The variation of the combined name is fix for _TYPE and the TV.
The numbers are changing according to the date. And the storyname is variable.
Any ideas?
Thanks
With your shown samples, please try following regex, this creates one capturing group which contains matched values in it.
.*?_TV_([^_]*)(?=_TYPE)
OR(adding a small variation of above solution with fourth bird's nice suggestion), following is without lazy match .*? unlike above:
_TV_([^_]*)(?=_TYPE)
Here is the Online demo for above regex
Explanation: Adding detailed explanation for above.
.*?_ ##Using Lazy match to match till 1st occurrence of _ here.
TV_ ##Matching TV_ here.
([^_]*) ##Creating 1st capturing group which has everything before next occurrence of _ here.
(?=_TYPE) ##Making sure previous values are followed by _TYPE here.
You could match as least as possible chars after _TV_ until you match _TYPE
\d_TV_\K.*?(?=_TYPE)
\d_TV_ Match a digit and _TV_
\K Forget what is matched until now
.*? Match as least as possible characters
(?=_TYPE) Assert _TYPE to the right
Regex demo
Another option without a non greedy quantifier, and leaving out the digit at the start:
_TV_\K[^_]*+(?>_(?!TYPE)[^_]*)*(?=_TYPE)
_TV_ Match literally
\K[^_]*+ Forget what is matched until now and optionally match any char except _
(?>_(?!TYPE)[^_]*)* Only allow matching _ when not directly followed by TYPE
(?=_TYPE) Assert _TYPE to the right
Regex demo
Edit
If you want to replace the 2 parts, you can use an alternation and replace with an empty string.
If it should be at the start and the end of the string, you can prepend ^ and append $ to the pattern.
\b\d{6}_TV_|_TYPE\b
\b\d{6}_TV_ A word boundary, match 6 digits and _TV_
| Or
_TYPE\b Match _TYPE followed by a word boundary
Regex demo
Here i put some additional Screenshots to the post. With the Documentation that appears on the help button. And you see the forms and what i see.
Documentation
The regular expressions we use are based on PCRE - Perl Compatible Regular Expressions. Full specification can be found here: http://www.pcere.org and http://perldoc.perl.org/perlre.html
Summary of some useful terms:
Metacharacters
\ Quote the next metacharacter
^ Match the beginning of the line
. Match any character (except newline)
$ Match the end of the line (or before newline at the end)
| Alternation
() Grouping
[] Character class
Quantifiers
* Match 0 or more times
+ Match 1 or more times
? Match 1 or 0 times
{n} Match exactly n times
{n,} Match at least n times
{n,m} Match at least n but not more than m times
Charcter Classes
\w Match a "word" character (alphanumeric plus mao}
\W Match a non-"word" character
\s Match a whitespace character
\S Match a non-whitespace character
\d Match a digit character
\D Match a non-digit character
Capture buffers
The bracketing construct (...) creates capture buffers. To refer to
Within the same pattern, use \1 for the first, \2 for the second, and so on. Outside the match use "$" instead of "". The \ notation works in certain circumstances outside the match. See the warning below about \1 vs $1 for details.
Referring back to another part of the match is called a backreference.
Examples
Replace story with certain prefix letters M N or E to have the prefix "AA":
`srcPattern "(M|N|E ) ([A-Za-z0-9\s]*)"`
`trgPattern "AA$2" `
`"N StoryWord1 StoryWord2" -> "AA StoryWord1 StoryWord2"`
`"E StoryWord1 StoryWord2" -> "AA StoryWord1 StoryWord2"`
`"M StoryWord1 StoryWord2" -> "AA StoryWord1 StoryWord2"`
"NoMatchWord StoryWord1 StoryWord2" -> "NoMatchWord StoryWord1 StoryWord2" (no match found, name remains the same)
So I need to match the following:
1.2.
3.4.5.
5.6.7.10
((\d+)\.(\d+)\.((\d+)\.)*) will do fine for the very first line, but the problem is: there could be many lines: could be one or more than one.
\n will only appear if there are more than one lines.
In string version, I get it like this: "1.2.\n3.4.5.\n1.2."
So my issue is: if there is only one line, \n needs not to be at the end, but if there are more than one lines, \n needs be there at the end for each line except the very last.
Here is the pattern I suggest:
^\d+(?:\.\d+)*\.?(?:\n\d+(?:\.\d+)*\.?)*$
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,}$/gm;
/* Slice is used to drop the dot at the end, otherwise resulting in
* an empty string on split.
*
* "1.2.3.".split(".") //=> ["1", "2", "3", ""]
* "1.2.3.".slice(0, -1) //=> "1.2.3"
* "1.2.3".split(".") //=> ["1", "2", "3"]
*/
console.log(
text.match(regex)
.map(match => match.slice(0, -1).split("."))
);
For more info about regex flags/modifiers have a look at: Regular Expression Reference: Mode Modifiers
For example, this is the regular expression
([a]{2,3})
This is the string
aaaa // 1 match "(aaa)a" but I want "(aa)(aa)"
aaaaa // 2 match "(aaa)(aa)"
aaaaaa // 2 match "(aaa)(aaa)"
However, if I change the regular expression
([a]{2,3}?)
Then the results are
aaaa // 2 match "(aa)(aa)"
aaaaa // 2 match "(aa)(aa)a" but I want "(aaa)(aa)"
aaaaaa // 3 match "(aa)(aa)(aa)" but I want "(aaa)(aaa)"
My question is that is it possible to use as few groups as possible to match as long string as possible?
How about something like this:
(a{3}(?!a(?:[^a]|$))|a{2})
This looks for either the character a three times (not followed by a single a and a different character) or the character a two times.
Breakdown:
( # Start of the capturing group.
a{3} # Matches the character 'a' exactly three times.
(?! # Start of a negative Lookahead.
a # Matches the character 'a' literally.
(?: # Start of the non-capturing group.
[^a] # Matches any character except for 'a'.
| # Alternation (OR).
$ # Asserts position at the end of the line/string.
) # End of the non-capturing group.
) # End of the negative Lookahead.
| # Alternation (OR).
a{2} # Matches the character 'a' exactly two times.
) # End of the capturing group.
Here's a demo.
Note that if you don't need the capturing group, you can actually use the whole match instead by converting the capturing group into a non-capturing one:
(?:a{3}(?!a(?:[^a]|$))|a{2})
Which would look like this.
Try this Regex:
^(?:(a{3})*|(a{2,3})*)$
Click for Demo
Explanation:
^ - asserts the start of the line
(?:(a{3})*|(a{2,3})*) - a non-capturing group containing 2 sub-sequences separated by OR operator
(a{3})* - The first subsequence tries to match 3 occurrences of a. The * at the end allows this subsequence to match 0 or 3 or 6 or 9.... occurrences of a before the end of the line
| - OR
(a{2,3})* - matches 2 to 3 occurrences of a, as many as possible. The * at the end would repeat it 0+ times before the end of the line
-$ - asserts the end of the line
Try this short regex:
a{2,3}(?!a([^a]|$))
Demo
How it's made:
I started with this simple regex: a{2}a?. It looks for 2 consecutive a's that may be followed by another a. If the 2 a's are followed by another a, it matches all three a's.
This worked for most cases:
However, it failed in cases like:
So now, I knew I had to modify my regex in such a way that it would match the third a only if the third a is not followed by a([^a]|$). So now, my regex looked like a{2}a?(?!a([^a]|$)), and it worked for all cases. Then I just simplified it to a{2,3}(?!a([^a]|$)).
That's it.
EDIT
If you want the capturing behavior, then add parenthesis around the regex, like:
(a{2,3}(?!a([^a]|$)))
I have this scenario:
Ex1:
Valid:
12345678|abcdefghij|aaaaaaaa
Invalid:
12345678|abcdefghijk|aaaaaaaaa
Which means that between pipes the maximum length is 8. How can I make in the regex?
I put this
^(?:[^|]+{0,7}(?:\|[^|]+)?$ but it´s not working
Try the following pattern:
^.{1,8}(?:\|.{1,8})*$
The basic idea is to match between one and eight characters, followed by | and another 1 to 8 characters, that term repeated zero or more times. Explore the demo with any data you want to see how it works.
Sample data:
123
12345678
abcdefghi (no match)
12345678|abcdefgh|aaaaaaaa
12345678|abcdefghijk|aaaaaaaaa (no match)
Demo here:
Regex101
When you want to match delimited data, you should refrain from using plain unrestricted .. You need to match parts between |, so you should consider [^|] negated character class construct that matches any char but |.
Since you need to limit the number of the pattern occurrences of the negated character class, restrict it with a limiting quantifier {1,8} that matches 1 to 8 consecutive occurrences of the quantified subpattern.
Use
^[^|]{1,8}(?:\|[^|]{1,8})*$
See the regex demo.
Details
^ - start of a string
[^|]{1,8} - any 1 to 8 chars other than |
(?:\|[^|]{1,8})* - 0 or more consecutive sequences of:
\| - a literal pipe symbol
[^|]{1,8} - any 1 to 8 chars other than |
$ - end of string.
Then, the [^|] can be restricted further as per requirements. If you only need to validate a string that has ASCII letters, digits, (, ), +, ,, ., /, :, ?, whitespace and -, you need to use
^[A-Za-z0-9()+,.\/:?\s-]{1,8}(?:\|[A-Za-z0-9()+,.\/:?\s-]{1,8})*$
See another 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.