I need to replace a pattern matched by regex with another pattern using regex, in C++.
Example -
We have the following characters: "a" and "b"
I want to replace like this -
Original text -
aabaaaaaaabaaabab
Replacement -
abbabbbbbbbabbbab
Replace logic -
"aab" must be replaced by "abb",
"aaab" must be replaced by "abbb",
"aaaab" must be replaced by "abbbb",
and so on...
I found the following regex for getting the matches -
aa+b
What regex replace pattern must be applied to get the desired replacement?
Thanks.
If using lookarounds be a possibility for you, here is one solution. Match on the following pattern:
(?<=a)a(?=a*b)
and then replace with just a single b. The pattern says to match:
(?<=a) assert that at least one 'a' precedes (i.e. ignore first 'a')
a a letter 'a'
(?=a*b) which is following by zero or more 'a' then a 'b'
Here is working demo.
Here is a solution that is not limited to the letters 'a' and 'b'.
Matches of the following regular expression are to be replaced by the content of capture group 2.
(?<=(.))\1(?=\1*(.))
Demo
The regular expression is comprised of the following elements.
(?<= # begin a positive lookbehind
(.) # match any character and save to capture group 1
) # end positive lookbehind
\1 # match the content of capture group 1
(?= # begin a positive lookahead
\1* # match the content of capture group 1 zero or more times
(.) # match any character and save to capture group 2
) # end positive lookahead
Related
I want to pull out a base string (Wax) or (noWax) from a longer string, along with potentially any data before and after if the string is Wax. I'm having trouble getting the last item in my list below (noWax) to match.
Can anyone flex their regex muscles? I'm fairly new to regex so advice on optimization is welcome as long as all matches below are found.
What I'm working with in Regex101:
/(?<Wax>Wax(?:Only|-?\d+))/mg
Original string
need to extract in a capturing group
Loc3_341001_WaxOnly_S212
WaxOnly
Loc4_34412-a_Wax4_S231
Wax4
Loc3a_231121-a_Wax-4-S451
Wax-4
Loc3_34112_noWax_S311
noWax
Here is one way to do so, using a conditional:
(?<Wax>(no)?Wax(?(2)|(?:Only|-?\d+)))
See the online demo.
(no)?: Optional capture group.
(? If.
(2): Test if capture group 2 exists ((no)). If it does, do nothing.
|: Or.
(?:Only|-?\d+)
I assume the following match is desired.
the match must include 'Wax'
'Wax' is to be preceded by '_' or by '_no'. If the latter 'no' is included in the match.
'Wax' may be followed by:
'Only' followed by '_', in which case 'Only' is part of the match, or
one or more digits, followed by '_', in which case the digits are part of the match, or
'-' followed by one or more digits, followed by '-', in which case
'-' followed by one or more digits is part of the match.
If these assumptions are correct the string can be matched against the following regular expression:
(?<=_)(?:(?:no)?Wax(?:(?:Only|\d+)?(?=_)|\-\d+(?=-)))
Demo
The regular expression can be broken down as follows.
(?<=_) # positive lookbehind asserts previous character is '_'
(?: # begin non-capture group
(?:no)? # optionally match 'no'
Wax # match literal
(?: # begin non-capture group
(?:Only|\d+)? # optionally match 'Only' or >=1 digits
(?=_) # positive lookahead asserts next character is '_'
| # or
\-\d+ # match '-' followed by >= 1 digits
(?=-) # positive lookahead asserts next character is '-'
) # end non-capture group
) # end non-capture group
I'm trying to get UNIX group names using a regex (can't use groups because I can only get the process uid, so I'm using id <process_id> to get groups)
input looks like this
uid=1001(kawsay) gid=1001(kawsay) groups=1001(kawsay),27(sudo),44(video),997(gpio)\n
I'd like to capture kawsay, sudo, video and gpio
The only pieces I've got are:
a positive lookbehind to start capturing after groups: /(?<=groups)/
capture the parenthesis content: /\((\w+)\)/
Using PCRE's \G you may use this regex:
(?:\bgroups=|(?<!^)\G)[^(]*\(([^)]+)\)
Your intended matches are available in capture group #1
RegEx Demo
RegEx Details:
(?:: Start non-capture group
\bgroups=: Match word groups followed by a =
|: OR
(?<!^)\G: Start from end position of the previous match
): End non-capture group
[^(]*: Match 0 or more of any character that is not (
\(: Match opening (
([^)]+): Use capture group #1 to match 1+ of any non-) characters
\): Match closing )
You can use
(?:\G(?!\A)\),|\bgroups=)\d+\(\K\w+
See the regex demo. Details:
(?:\G(?!\A)\),|\bgroups=) - either of
\G(?!\A)\), - end of the previous match (\G operator matches either start of string or end of the previous match, so the (?!\A) is necessary to exclude the start of string location) and then ), substring
| - or
\bgroups= - a whole word groups (\b is a word boundary) and then a = char
\d+\( - one or more digits and a (
\K - match reset operator that makes the regex engine "forget" the text matched so far
\w+ - one or more word chars.
Here are two more ways to extract the strings of interest. Both return matches and do not employ capture groups. My preference is for second one.
str = "uid=1001(kawsay) gid=1001(kawsay) groups=1001(kawsay),27(sudo),44(video),997(gpio)\n"
Match substrings between parentheses that are not followed later in the string with "groups="
Match the regular expression
rgx = /(?<=\()(?!.*\bgroups=).*?(?=\))/
str.scan(rgx)
#=> ["kawsay", "sudo", "video", "gpio"]
Demo
See String#scan.
This expression can be broken down as follows.
(?<=\() # positive lookbehind asserts previous character is '('
(?! # begin negative lookahead
.* # match zero or more characters
\bgroups= # match 'groups=' preceded by a word boundary
) # end negative lookahead
.* # match zero or more characters lazily
(?=\)) # positive lookahead asserts next character is ')'
This may not be as efficient as expressions that employ \G (because of the need to determine if 'groups=' appears in the string after each left parenthesis), but that may not matter.
Extract the portion of the string following "groups=" and then match substrings between parentheses
First, obtain the portion of the string that follows "groups=":
rgx1 = /(?<=\bgroups=).*/
s = str[rgx1]
#=> "1001(kawsay),27(sudo),44(video),997(gpio)\n"
See String#[].
Then match the regular expression
rgx2 = /(?<=\()[^\)\r\n]+/
against s:
s.scan(rgx2)
#=> ["kawsay", "sudo", "video", "gpio"]
The regular expression rgx1 can be broken down as follows:
(?<=\bgroups=) # Positive lookbehind asserts that the current
# position in the string is preceded by`'groups'`,
# which is preceded by a word boundary
.* # match zero of more characters other than line
# terminators (to end of line)
rgx2 can be broken down as follows:
(?<=\() # Use a positive lookbehind to assert that the
# following character is preceded by '('
[^\)\r\n]+ # Match one or more characters other than
# ')', '\r' and '\n'
Note:
The operations can of course be chained: str[/(?<=\bgroups=).*/].scan(/(?<=\()[^\)\r\n]+/); and
rgx2 could alternatively be written /(?<=\().+?(?=\)), where ? makes the match of one or more characters lazy and (?=\)) is a positive lookahead that asserts that the match is followed by a right parenthesis.
This would probably be the fastest solution of those offered and certainly the easiest to test.
Extract all the string between 2 patterns:
Input:
test.output0 testx.output1 output3 testds.output2(\t)
Output:
output0 output1 ouput3 output2
Note: (" ") is the tab character.
You may try:
\.\w+$
Explanation of the above regex:
\. - Matches . literally. If you do not want . to be included in your pattern; please use (?<=\.) or simply remove ..
\w+ - Matches word character [A-Za-z0-9_] 1 or more time.
$ - Represents end of the line.
You can find the demo of the regex in here.
Result Snap:
EDIT 2 by OP:
According to your latest edit; this might be helpful.
.*?\.?(\w+)(?=\t)
Explanation:
.*? - Match everything other than new line lazily.
\.? - Matches . literally zero or one time.
(\w+) - Represents a capturing group matching the word-characters one or more times.
(?=\t) - Represents a positive look-ahead matching tab.
$1 - For the replacement part $1 represents the captured group and a white-space to separate the output as desired by you. Or if you want to restore tab then use the replacement $1\t.
Please find the demo of the above regex in here.
Result Snap 2:
Try matching on the following pattern:
Find: (?<![^.\s])\w+(?!\S)
Here is an explanation of the above pattern:
(?<![^.\s]) assert that what precedes is either dot, whitespace, or the start of the input
\w+ match a word
(?!\S) assert that what follows is either whitespace of the end of the input
Demo
There are 5 examples as below, and I am trying to find 3,4,5 while excluding 1,2.
ABC-abc
abc-ABC
ABC-ABC
ABC
vABC-ABC-ABCv
The current expression I use is:
(?!(\w*[A-Z]{2,}-[a-z]+\w*|\w*[a-z]+-[A-Z]{2,}\w*))(\w*-?[A-Z]{2,}-?\w*)
I utilize (\w*-?[A-Z]{2,}-?\w*) to get all possibility of all examples first.
I then use (?!...|...) to put two exclusion conditions.
The first exclusion condition is \w*[A-Z]{2,}-[a-z]+\w* and the second is \w*[a-z]+-[A-Z]{2,}\w*.
This expression works to exclude 1.ABC-abc but not abc-ABC.
I searched a lot and found some people say this way is not something regex is "good" at. Is there any solution or improvement I can do to get rid of abc-ABC.
Appreciate any help or opinion.
As I understand strings are to be rejected if they contain a hyphen that is preceded by a lower-case letter and followed by an upper-case letter, or vice-versa; else they are to be accepted. If so, the following regular expression could be used.
^(?!.*(?:[a-z]-[A-Z]|[A-Z]-[a-z]))
Demo
The regex engine performs the following operations.
^ # match beginning of line
(?! # begin a negative lookahead
.* # match 0+ characters
(?: # begin a non-capture group
[a-z]-[A-Z] # match a lc letter, '-', uc letter
| # or
[A-Z]-[a-z] # match an uc letter, '-', lc letter
) # end non-capture group
) # end negative lookahead
My regex knowledge is pretty limited, but I'm trying to write/find an expression that will capture the following string types in a document:
DO match:
ADY123
AD12ADY
1HGER_2
145-DE-FR2
Bicycle1
2Bicycle
128D
128878P
DON'T match:
BICYCLE
183-329-193
3123123
Is such an expression possible? Basically, it should find any string containing letters AND digits, regardless of whether the string contains a dash or underscore. I can find the first two using the following two regex:
/([A-Z][0-9])\w+/g
/([0-9][A-Z)\w+/g
But searching for possible dashes and hyphens makes it more complicated...
Thanks for any help you can provide! :)
MORE INFO:
I've made slight progress with: ([A-Z|a-z][0-9]+-*_*\w+) but it doesn't capture strings with more than one hyphen.
I had a document with a lot of text strings and number strings, which I don't want to capture. What I do want is any product code, which could be any length string with or without hyphens and underscores but will always include at least one digit and at least one letter.
You can use the following expression with the case-insensitive mode:
\b((?:[a-z]+\S*\d+|\d\S*[a-z]+)[a-z\d_-]*)\b
Explanation:
\b # Assert position at a word boundary
( # Beginning of capturing group 1
(?: # Beginning of the non-capturing group
[a-z]+\S*\d+ # Match letters followed by numbers
| # OR
\d+\S*[a-z]+ # Match numbers followed by letters
) # End of the group
[a-z\d_-]* # Match letter, digit, '_', or '-' 0 or more times
) # End of capturing group 1
\b # Assert position at a word boundary
Regex101 Demo