So for example I have this string
var = 'column1;column2;column3\r\nval1;val2;val3\r\n;val4;val5;val6\r\n'
I want to be able to find all \r\n and replace it with temp\r\n, but I want to ignore column3\r\n
Tried to do ^(?!.*column3).*$\r\n but the \r\n syntax does not work
You want to use a negative lookbehind, that is make the substitution when \r\n is not preceded by column3:
re.sub(r'(?<!column3)\r\n', r'temp\r\n', var)
For example:
>>> import re
>>>
>>> var = 'column1;column2;column3\r\nval1;val2;val3\r\n;val4;val5;val6\r\n'
>>> new_text = re.sub(r'(?<!column3)\r\n', r'temp\r\n', var)
>>> new_text
'column1;column2;column3\r\nval1;val2;val3temp\r\n;val4;val5;val6temp\r\n'
>>>
Related
I have a string named
Set-Cookie: BIGipServerApp_Pool_SSL=839518730.47873.0000; path=/
I am trying to extract 839518730.47873.0000 from it. For exact string I am fine with my regex but If I include any digit before 1st = then its all going wrong.
No Digit
>>> m=re.search('[0-9.]+','Set-Cookie: BIGipServerApp_Pool_SSL=839518730.47873.0000; path=/')
>>> m.group()
'839518730.47873.0000'
With Digit
>>> m=re.search('[0-9.]+','Set-Cookie: BIGipServerApp_Pool_SSL2=839518730.47873.0000; path=/')
>>> m.group()
'2'
Is there any way I can extract `839518730.47873.0000' only but doesnt matter what else lies in the string.
I tried
>>> m=re.search('=[0-9.]+','Set-Cookie: BIGipServerApp_Pool_SSL=839518730.47873.0000; path=/')
>>> m.group()
'=839518730.47873.0000'
As well but its starting with '=' in the output and I dont want it.
Any ideas.
Thank you.
If your substring always comes after the first =, you can just use capture group with =([\d.]+) pattern:
import re
result = ""
m = re.search(r'=([0-9.]+)','Set-Cookie: BIGipServerApp_Pool_SSL2=839518730.47873.0000; path=/')
if m:
result = m.group(1) # Get Group 1 value only
print(result)
See the IDEONE demo
The main point is that you match anything you do not need and match and capture (with the unescaped round brackets) the part of pattern you need. The value you need is in Group 1.
You can use word boundaries:
\b[\d.]+
RegEx Demo
Or to make match more targeted use lookahead for next semi-colon after your matched text:
\b[\d.]+(?=\s*;)
RegEx Demo2
Update :
>>> m.group(0)
'839518730.47873.0000'
>>> m=re.search(r'\b[\d.]+','Set-Cookie: BIGipServerApp_Pool_SSL2=839518730.47873.0000; path=/')
>>> m.group(0)
'839518730.47873.0000'
>>>
I have a text file with the following content:
aaaaaaCRLF
bbbbbb'CRLF
ccccccCRLF
I want the remove the CRLF in the lines where there is no ' before the CRLF. The destination text should be:
aaaaaa
bbbbbb'CRLF
cccccc
Any idea on how to do this with RegEx?
I'm using Python, but this should work with any decent Regex engine. Use a negative lookbehind!
>>> import re
>>> s = "aaaaaaa\r\nasdasdasd'\r\nasdasdas\r\n"
>>> p = r"(?<!')\r\n"
>>> re.sub(p, '', s)
"aaaaaaaasdasdasd'\r\nasdasdas"
edit: Oh, you mean \r\n. Pattern adapted.
I have a regex pattern:
import regex as re
re.sub(r'(.*)\bHello (.*) BGC$\b', "OTR", 'Hello People BGC')
This will replace to give OTR, but how do I find out what the matched characters are within the (.*)?
Using regex==2016.1.10, Python 3.5.1
Compile the pattern and then call match() and sub() separately:
>>> pattern = re.compile(r'^Hello (.*?) BGC$')
>>> s = 'Hello People BGC'
>>> pattern.match(s).group(1)
'People'
>>> pattern.sub("OTR", s)
'OTR'
I have tested this Regex
(?<=\))(.+?)(?=\()|(?<=\))(.+?)\b|(.+?)(?=\()
but it doesn't work for strings like this pattern (ef)abc(gh).
I got a result like this "(ef)abc".
But these 3 regexes (?<=\))(.+?)(?=\() , (?<=\))(.+?)\b, (.+?)(?=\()
do work separately for "(ef)abc(gh)", "(ef)abc" ,"abc(ef)" .
can anyone tell me where the problem is or how can I get the expected result?
Assuming you are looking to match the text from between the elements in parenthesis, try this:
^(?:\(\w*\))?([\w]*)(?:\(\w*\))?$
^ - beginning of string
(?:\(\w*\))? - non-capturing group, match 0 or more alphabetic letters within parens, all optional
([\w]*) - capturing group, match 0 or more alphabetic letters
(?:\(\w*\))? - non-capturing group, match 0 or more alphabetic letters within parens, all optional
$ - end of string
You haven't specified what language you might be using, but here is an example in Python:
>>> import re
>>> string = "(ef)abc(gh)"
>>> string2 = "(ef)abc"
>>> string3 = "abc(gh)"
>>> p = re.compile(r'^(?:\(\w*\))?([\w]*)(?:\(\w*\))?$')
>>> m = re.search(p, string)
>>> m2 = re.search(p, string2)
>>> m3 = re.search(p, string3)
>>> print m.groups()[0]
'abc'
>>> print m2.groups()[0]
'abc'
>>> print m3.groups()[0]
'abc'
\([^)]+\)|([^()\n]+)
Try this.Just grab the capture or group.See demo.
https://regex101.com/r/tX2bH4/6
Your problem is that (.+?)(?=\() matches "(ef)abc" in "(ef)abc(gh)".
The easiest solution to this problem is be more explicit about what you are looking for. In this case by exchanging "any character" ., with "any character that is not a parenthesis" [^\(\)].
(?<=\))([^\(\)]+?)(?=\()|(?<=\))([^\(\)]+?)\b|([^\(\)]+?)(?=\()
A cleaner regexp would be
(?:(?<=^)|(?<=\)))([^\(\)]+)(?:(?=\()|(?=$))
I have a string like "some_{abcd_etc}_text"
eveything between { } should be removed, including {} itself.
I need only the string "some_text" at the end.
How can this been done by regex?
You could use this expression:
{.*?}
Sure, just replace this with an empty string:
{[^}]+}
Here is a Python example:
>>> from re import sub
>>> s = r'some_{abcd_etc}_text'
>>> sub(r'{[^}]+}', '', s)
'some__text'