"Name":"abc"
Expected output: Name
Here in this case, when I have the value "abc", I need to fetch the word Name by using positive look behind and extracting the words between the occurences of ".
You don't need anything fancy like look-behinds for this functionality:
"([A-Za-z]*)":"abc"
Regex101
Edit: Since you've added the python tag, escape your quotes:
\"([A-Za-z]*)\":\"abc\"
New Regex101
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Im using python to extract some info
i wanna get the words/names before the charcter :
but the problem is everythig is tied together
from here
Morgan Stanley.Erik Woodring:
i just wanna extract "Erik Woodring:"
or from here
market.Operator:
i just wanna extract Operator:
sometimes there are questiosn like this
to acquire?Tim Cook:
i just wanna extract "Tim Cook:"
this is what i tried
\w*(?=.*:)
this is not getting what i wanted, its returning a lot of words
This could be the regex you're looking for:
\b[\w\s]+(?=:)
\b world boundary;
[\w\s]+ matches any word or whitespace (at least one character);
(?=:) positive lookahead that specifies the word must be followed by a punctation mark;
https://regex101.com/r/w86oWv/1
If you want to get the ":" too you can simply remove the lookahead:
\b[\w\s]+:
I am new to regexes.
I have the following string : \n(941)\n364\nShackle\n(941)\nRivet\n105\nTop
Out of this string, I want to extract Rivet and I already have (941) as a string in a variable.
My thought process was like this:
Find all the (941)s
filter the results by checking if the string after (941) is followed by \n, followed by a word, and ending with \n
I made a regex for the 2nd part: \n[\w\s\'\d\-\/\.]+$\n.
The problem I am facing is that because of the parenthesis in (941) the regex is taking 941 as a group. In the 3rd step the regex may be wrong, which I can fix later, but 1st I needed help in finding the 2nd (941) so then I can apply the 3rd step on that.
PS.
I know I can use python string methods like find and then loop over the searches, but I wanted to see if this can be done directly using regex only.
I have tried the following regex: (?:...), (941){1} and the make regex literal character \ like this \(941\) with no useful results. Maybe I am using them wrong.
Just wanted to know if it is possible to be done using regex. Though it might be useful for others too or a good share for future viewers.
Thanks!
Assuming:
You want to avoid matching only digits;
Want to match a substring made of word-characters (thus including possible digits);
Try to escape the variable and use it in the regular expression through f-string:
import re
s = '\n(941)\n364\nShackle\n(941)\nRivet\n105\nTop'
var1 = '(941)'
var2 = re.escape(var1)
m = re.findall(fr'{var2}\n(?!\d+\n)(\w+)', s)[0]
print(m)
Prints:
Rivet
If you have text in a variable that should be matched exactly, use re.escape() to escape it when substituting into the regexp.
s = '\n(941)\n364\nShackle\n(941)\nRivet\n105\nTop'
num = '(941)'
re.findall(rf'(?<=\n{re.escape(num)}\n)[\w\s\'\d\-\/\.]+(?=\n)', s)
This puts (941)\n in a lookbehind, so it's not included in the match. This avoids a problem with the \n at the end of one match overlapping with the \n at the beginning of the next.
I'm creating a javascript regex to match queries in a search engine string. I am having a problem with alternation. I have the following regex:
.*baidu.com.*[/?].*wd{1}=
I want to be able to match strings that have the string 'word' or 'qw' in addition to 'wd', but everything I try is unsuccessful. I thought I would be able to do something like the following:
.*baidu.com.*[/?].*[wd|word|qw]{1}=
but it does not seem to work.
replace [wd|word|qw] with (wd|word|qw) or (?:wd|word|qw).
[] denotes character sets, () denotes logical groupings.
Your expression:
.*baidu.com.*[/?].*[wd|word|qw]{1}=
does need a few changes, including [wd|word|qw] to (wd|word|qw) and getting rid of the redundant {1}, like so:
.*baidu.com.*[/?].*(wd|word|qw)=
But you also need to understand that the first part of your expression (.*baidu.com.*[/?].*) will match baidu.com hello what spelling/handle????????? or hbaidu-com/ or even something like lkas----jhdf lkja$##!3hdsfbaidugcomlaksjhdf.[($?lakshf, because the dot (.) matches any character except newlines... to match a literal dot, you have to escape it with a backslash (like \.)
There are several approaches you could take to match things in a URL, but we could help you more if you tell us what you are trying to do or accomplish - perhaps regex is not the best solution or (EDIT) only part of the best solution?
text to capture looks like this..
Policy Number ABCD000012345 other text follows in same line....
My regex looks like this
regex value='(?i)(?:[P|p]olicy\s[N|n]o[|:|;|,][\n\r\s\t]*[\na-z\sA-Z:,;\r\d\t]*[S|s]e\s*[H|h]abla\s*[^\n]*[\n\s\r\t]*|(?i)[P|p]olicy[\s\n\t\r]*[N|n]umber[\s\n\r\t]*)(?P<policy_number>[^\n]*)'
this particular case matches with the second or case.. however it is also capturing everything after the policy number. What can be the stopping condition for it to just grab the number. I know something is wrong but can't find a way out.
(?i)[P|p]olicy[\s\n\t\r]*[N|n]umber[\s\n\r\t]*)
current output
ABCD000012345othertextfollowsinsameline....
expected output
ABCD000012345
You may use a more simple regex, just finding from the beginning "[P|p]olicy\s*[N|n]umber\s*\b([A-Z]{4}\d+)\b.*" and use the word boundary \b
pattern = re.compile(r"[P|p]olicy\s*[N|n]umber\s*\b([A-Z0-9]+)\b.*")
line = "Policy Number ABCD000012345 other text follows in same line...."
matches = pattern.match(line)
id_res = matches.group(1)
print(id_res) # ABCD000012345
And if there's always 2 words before you can use (?:\w+\s+){2}\b([A-Z0-9]+)\b.*
Also \s is for [\r\n\t\f\v ] so no need to repeat them, your [\n\r\s\t] is just \s
you don't need the upper and lower case p and n specified since you're already specifying case insensitive.
Also \s already covers \n, \t and \r.
(?i)policy\s+number\s+([A-Z]{4}\d+)\b
for verification purpose: Regex
Another Solution:
^[\s\w]+\b([A-Z]{4}\d+)\b
for verification purpose: Regex
I like this better, in case your text changes from policy number
I have huge string like this dsdasdludocid=15878284988193842600#lrd=0x3be04dcc5b5ac513:0xdc5b0011ebb625a8,2
I want to get the number after ludocid, only consecutive numbers.
I have tried this regex (ludocid).*(?=\d+\d+) and many more but no luck.
You can try ludocid=(\d+):
s = "dsdasdludocid=15878284988193842600#lrd=0x3be04dcc5b5ac513:0xdc5b0011ebb625a8,2"
import re
re.findall(r"ludocid=(\d+)", s)
# ['15878284988193842600']
You can use this regex:
ludocid\D*(\d+)
RegEx Demo
This will match literal ludocid followed by 0 or more non-digits and then it will match 1 or more digits in captured group #1
Code:
>>> s = 'dsdasdludocid=15878284988193842600#lrd=0x3be04dcc5b5ac513:0xdc5b0011ebb625a8,2'
>>> print re.search(r'ludocid\D*(\d+)', s).group(1)
15878284988193842600
It looks like you just threw a bunch of regex bits together... Let's work through that.
First, this is the correct regex: ludocid.(\d+)
(You would want to use it with re.search instead of re.match, by the way. Match requires the regex to match the entire string.)
But let's look at yours and see what went wrong and how we can get to the correct regex.
(ludocid).*(?=\d+\d+)
Imagine a regex as a function. You pass it the right things, and it gives you the appropriate result. When you wrap things in parentheses, you're saying "Find this and give it back to me." You don't need the ludocid given back to you, I'm guessing... so remove those paren.
ludocid.*(?=\d+\d+)
Now you've got a .*. This is dangerous in regular expressions because it literally says "Grab as many of anything as you possibly can!" Often I use the non-greedy version (.*?), but in this case it looks like we're just expecting a single extra character there. If you know the literal character you can use that, but to be safe I'll leave it as ., which says "Grab any one character."
ludocid.(?=\d+\d+)
Now let's go inside the parentheses. You've got \d+\d+, which says "Find a sequence of one or more digits, and then find another sequence of one or more digits." This equates to "Find a sequence of two or more digits." I don't think this is what you wanted (it's not how you described the problem, anyway), so let's reduce that:
ludocid.(?=\d+)
Okay, great. Now... what is (?=...) for? It's called a lookahead assertion. It says "If you find this string, match things in front of it." The example given in the Python 2.7 documentation is:
(?=...)
Matches if ... matches next, but doesn’t consume any of the string. This is called a lookahead assertion. For example, Isaac (?=Asimov) will match 'Isaac ' only if it’s followed by 'Asimov'.
Essentially this means that your regex will never return the digits. Instead, it looks to see if digits exist, and then it returns things from the rest of the regex. Remove the lookahead assertion and we're there:
ludocid.(\d+)
When you use this with re.search, you'll get the group you want:
>>> s = "dsdasdludocid=15878284988193842600#lrd=0x3be04dcc5b5ac513:0xdc5b0011ebb625a8,2"
>>> import re
>>> re.search(r"ludocid.(\d+)", s).group(1)
'15878284988193842600'
To match only the digits that follow, stopping at the first non-numeric char, try a positive look behind:
(?<=ludocid=)(\d+)
So:
re.findall(r"(?<=ludocid=)(\d+)", s)
The positive look behind will look for what you want, and only match if it is preceded by the 'flag' string.
**Note: **You may need to escape that second = sign like this: (?<=ludocid\=)(\d+)