I have some data like
small_animal/Mouse
BigAnimal:Elephant
Not an animal.
What I want to get is:
Mouse
Elephant
Not an animal.
Thus, I need a regular expression that searches for / or : as follows: If one of these is found, take the text behind that character. If neither / nor : exists, take the whole string.
I tried a lot. For example this will work for mouse and elephant, but not for the third line:
(?<=:)[^:]*|(?<=/)[^/]*
And this will always give the full string ...
(?<=:)[^:]*|(?<=/)[^/]*|^.*$
My head is burning^^ Maybe, somebody can help? :) Thanks a lot!
EDIT:
#The fourth bird offered a nice solution for single characters. But what if I want to search for strings like
animal::Dog
Another123Cat
Not an animal.
How can I split on :: or 123?
You might use
^(?:[^:/]*[:/])?\K.+
^ Start of string
(?:[^:/]*[:/])? Optionally match any char except : or / till matching either : or /
\K Forget what is matched so far
.+ Match 1+ times any char
regex demo
If you don't want to cross a newline, you can extend the character class with [^:/\r\n]*
Another option could be using an alternation
^[^:/]*[:/]\K.+|.+
Regex demo
Or perhaps making use of a SKIP FAIL approach by matching what you want to omit
^[^:/]*[:/](*SKIP)(*F)|.+
Regex demo
If you want to use multiple characters, you might also use
^(?:(?:(?!123|::|[:/]).)*+(?:123|::|[:/]))?\K.+
Regex demo
Related
I'm having an issue with Regex.
I'm trying to match T0000001 (2, 3 and so on).
However, some of the lines it searches has what I can describe as positioners. These are shown as a question mark, followed by 2 digits, such as ?21.
These positioners describe a new position if the document were to be printed off the website.
Example:
T123?214567
T?211234567
I need to disregard ?21 and match T1234567.
From what I can see, this is not possible.
I have looked everywhere and tried numerous attempts.
All we have to work off is the linked image. The creators cant even confirm the flavour of Regex it is - they believe its Python but I'm unsure.
Regex Image
Update
Unfortunately none of the codes below have worked so far. I thought to test each code in live (Rather than via regex thinking may work different but unfortunately still didn't work)
There is no replace feature, and as mentioned before I'm not sure if it is Python. Appreciate your help.
Do two regex operations
First do the regex replace to replace the positioners with an empty string.
(\?[0-9]{2})
Then do the regex match
T[0-9]{7}
If there's only one occurrence of the 'positioners' in each match, something like this should work: (T.*?)\?\d{2}(.*)
This can be tested here: https://regex101.com/r/XhQXkh/2
Basically, match two capture groups before and after the '?21' sequence. You'll need to concatenate these two matches.
At first, match the ?21 and repace it with a distinctive character, #, etc
\?21
Demo
and you may try this regex to find what you want
(T(?:\d{7}|[\#\d]{8}))\s
Demo,,, in which target string is captured to group 1 (or \1).
Finally, replace # with ?21 or something you like.
Python script may be like this
ss="""T123?214567
T?211234567
T1234567
T1234434?21
T5435433"""
rexpre= re.compile(r'\?21')
regx= re.compile(r'(T(?:\d{7}|[\#\d]{8}))\s')
for m in regx.findall(rexpre.sub('#',ss)):
print(m)
print()
for m in regx.findall(rexpre.sub('#',ss)):
print(re.sub('#',r'?21', m))
Output is
T123#4567
T#1234567
T1234567
T1234434#
T123?214567
T?211234567
T1234567
T1234434?21
If using a replace functionality is an option for you then this might be an approach to match T0000001 or T123?214567:
Capture a T followed by zero or more digits before the optional part in group 1 (T\d*)
Make the question mark followed by 2 digits part optional (?:\?\d{2})?
Capture one or more digits after in group 2 (\d+).
Then in the replacement you could use group1group2 \1\2.
Using word boundaries \b (Or use assertions for the start and the end of the line ^ $) this could look like:
\b(T\d*)(?:\?\d{2})?(\d+)\b
Example Python
Is the below what you want?
Use RegExReplace with multiline tag (m) and enable replace all occurrences!
Pattern = (T\d*)\?\d{2}(\d*)
replace = $1$2
Usage Example:
I am not really a RegEx expert and hence asking a simple question.
I have a few parameters that I need to use which are in a particular pattern
For example
$$DATA_START_TIME
$$DATA_END_TIME
$$MIN_POID_ID_DLAY
$$MAX_POID_ID_DLAY
$$MIN_POID_ID_RELTM
$$MAX_POID_ID_RELTM
And these will be replaced at runtime in a string with their values (a SQL statement).
For example I have a simple query
select * from asdf where asdf.starttime = $$DATA_START_TIME and asdf.endtime = $$DATA_END_TIME
Now when I try to use the RegEx pattern
\$\$[^\W+]\w+$
I do not get all the matches(I get only a the last match).
I am trying to test my usage here https://regex101.com/r/xR9dG0/2
If someone could correct my mistake, I would really appreciate it.
Thanks!
This will do the job:
\$\$\w+/g
See Demo
Just Some clarifications why your regex is doing what is doing:
\$\$[^\W+]\w+$
Unescaped $ char means end of string, so, your pattern is matching something that must be on the end of the string, that's why its getting only the last match.
This group [^\W+] doesn't really makes sense, groups starting with [^..] means negate the chars inside here, and \W is the negation of words, and + inside the group means literally the char +, so you are saying match everything that is Not a Not word and that is not a + sign, i guess that was not what you wanted.
To match the next word just \w+ will do it. And the global modifier /g ensures that you will not stop on the first match.
This should work - Based on what you said you wanted to match this should work . Also it won't match $$lower_case_strings if that's what you wanted. If not, add the "i" flag also.
\${2}[A-Z_]+/g
Ok, I know that it is a question often asked, but I did not manage to get what I wanted.
I am looking for a regular expression in order to find a pattern that does not contain a particular substring.
I want to find an url that does not contains the b parameter.
http://www.website.com/a=789&c=146 > MATCH
http://www.website.com/a=789&b=412&c=146 > NOT MATCH
Currently, I have the following Regex:
\bhttp:\/\/www\.website\.com\/((?!b=[0-9]+).)*\b
But I am wrong with the \b, the regex match the beginning of th string and stop when it find b=, instead of not matching.
See: http://regex101.com/r/fN3zU5/3
Can someone help me please?
Just use a lookahead to check anything following the URL must be a space or line end.
\bhttp:\/\/www\.website\.com\/(?:(?!b=[0-9]+).)*?\b(?= |$)
DEMO
use this:
^http:\/\/www\.website\.com\/((?!b=[0-9]+)).*$
\b only matches word endings.
^ matches start and end of string
and you dont even need to do it that complicated, If you dont want the url with the b parameter use this:
^http:\/\/www\.website\.com\/(?!b).*$
demo here : http://regex101.com/r/fN3zU5/5
import re
pattern=re.compile(r"(?!.*?b=.*).*")
print pattern.match(x)
This will look ahead if there is a "b=" present.A negative lookahead means it will not match that string.
You had a look at this possibility:
http://regex101.com/r/fN3zU5/6
^http:\/\/www\.website\.com\/[ac\=\d&]*$
only allow &,=,a,c and digits
complete url in group and there should not be a "b=" parameter
if you have more options and you dont want to list them all:
you dont allow a 'b' to be part of your parameters
^http:\/\/www\.website\.com\/[^b]*$
http://regex101.com/r/fN3zU5/7
^http:\/\/www\.website\.com\/(?!.*?b=.*?).*$ works too here "b=" is permitted at any position of the parameter string so you could even have the "b" string as a value of a parameter.
See
http://regex101.com/r/fN3zU5/8
This is what you want. ^http:\/\/www\.website\.com\/(([^b]=[0-9]+).)*$
Its a simple pattern not flexible but it works :
http:\/\/www\.website\.com\/+a=+\w+&+c=+\w+
I'm having trouble with lookaround in regex.
Here the problem : I have a big file I want to edit, I want to change a function by another keeping the first parameter but removing the second one.
Let say we have :
func1(paramIWantToKeep, paramIDontWant)
or
func1(func3(paramIWantToKeep), paramIDontWant)
I want to change with :
func2(paramIWantToKeep) in both case.
so I try using positive lookahead
func1\((?=.+), paramIDontWant\)
Now, I just try not to select the first parameter (then I'll manage to do the same with the parenthesis).
But it doesn't work, it appears that my regex, after ignoring the positive look ahead (.+) look for (, paramIDontWant\)) at the same position it was before the look ahead (so the opening parenthesis)
So my question is, how to continue a regex after a matching group, here after (.+).
Thanks.
PS: Sorry for the english and/or the bad construction of my question.
Edit : I use Sublime Text
The first thing you need to understand is that a regex will always match a consecutive string. There will never be gaps.
Therefore, if you want to replace 123abc456 with abc, you can't simply match 123456 and remove it.
Instead, you can use a capturing group. This will allow you to remember a section of the regex for later.
For example, to replace 123abc456 with abc, you could replace this regex:
\d+([a-z]+)\d+
with this string:
$1
What that does is actually replaces the match with the contents of the first capturing group. In this case, the capturing group was ([a-z]+), which matches abc. Thus, the entire match is replaced with just abc.
An example you may find more useful:
Given:
func1(foo, bar)
replacing this regex:
\w+\((\w+),\s*\w+\)
with this string:
func2($1)
results in:
func2(foo)
import re
t = "func1(paramKeep,paramLose)"
t1 = "func1(paramKeep,((paramLose(dog,cat))))"
t2 = "func1(func3(paramKeep),paramDont)"
t3 = "func1(func3(paramKeep),paramDont,((i)),don't,want,these)"
reg = r'(\w+\(.*?(?=,))(,.*)(\))'
keep,lose,end = re.match(reg,t).groups()
print(keep+end)
keep,lose,end = re.match(reg,t1).groups()
print(keep+end)
keep,lose,end = re.match(reg,t2).groups()
print(keep+end)
keep,lose,end = re.match(reg,t3).groups()
print(keep+end)
Produces
>>>
func1(paramKeep)
func1(paramKeep)
func1(func3(paramKeep))
func1(func3(paramKeep))
Apply these two regexp in this order
s/(func1)([^,]*)(, )?(paramIDontWant)(.)/func2$2$5/;
s/(func2\()(func3\()(paramIWantToKeep).*/$1$3)/;
These cope with the two examples you gave. I guess that the real world code you are editing is slightly more complicated but the general idea of applying a series of regexps might be helpful
From the following text...
Acme Inc.<SPACE>12345<SPACE or TAB>bla bla<CRLF>
... I need to extract company name + zip code + rest of the line.
Since either a TAB or a SPACE character can separate the second from the third tokens, I tried using the following regex:
FIND:^(.+) (\d{5})(\t| )(.+)$
REPLACE:\1\t\2\t\3
However, the contents of the alternative part is put in the \3 part, so the result is this:
Acme Inc.<TAB>12345<TAB><TAB or SPACE here>$
How can I tell the (Perl) regex engine that (\t| ) is an alternative instead of a token to be saved in RAM?
Thank you.
You want:
^(.+?) (\d{5})[\t ](.+)$
Since you are matching one character or the other, you can use a character class instead. Also, I made your first quantifier non-greedy (+? instead of +) to reduce the amount of backtracking the engine has to do to find the match.
In general, if you want to make capture groups not capture anything, you can add ?: to it, like:
^(.+?) (\d{5})(?:\t| )(.+)$
Use non-capturing parentheses:
^(.+) (\d{5})(?:\t| )(.+)$
One way is to use \s instead of ( |\t) which will match any whitespace char.
See Backslash-sequences for how Perl defines "whitespace".