This question already has an answer here:
Why is Python Regex Wildcard only matching newLine
(1 answer)
Closed 1 year ago.
The following regular expression is not returning any match:
import re
regex = '.*match.*fail.*'
pattern = re.compile(regex)
text = '\ntestmatch\ntestfail'
match = pattern.search(text)
I managed to solve the problem by changing text to repr(text) or setting text as a raw string with r'\ntestmatch\ntestfail', but I'm not sure if these are the best approaches. What is the best way to solve this problem?
Using repr or raw string on a target string is a bad idea!
By doing that newline characters are treated as literal '\n'.
This is likely to cause unexpected behavior on other test cases.
The real problem is that . matches any character EXCEPT newline.
If you want to match everything, replace . with [\s\S].
This means "whitespace or not whitespace" = "anything".
Using other character groups like [\w\W] also works,
and it is more efficient for adding exception just for newline.
One more thing, it is a good practice to use raw string in pattern string(not match target).
This will eliminate the need to escape every characters that has special meaning in normal python strings.
You could add it as an or, but make sure you \ in the regex string, so regex actually gets the \n and not a actual newline.
Something like this:
regex = '.*match(.|\\n)*fail.*'
This would match anything from the last \n to match, then any mix or number of \n until testfail. You can change this how you want, but the idea is the same. Put what you want into a grouping, and then use | as an or.
On the left is what this regex pattern matched from your example.
Related
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.
So I'm writing a Python program that reads lines of serial data, and compares them to a dictionary of line codes to figure out which specific lines are being transmitted. I am attempting to use a Regular Expression in order to filter out the extra garbage line serial read string has on it, but I'm having a bit of an issue.
Every single code in my dictionary looks like this: T12F8B0A22**F8. The asterisks are the two alpha numeric pieces that differentiate each string code.
This is what I have so far as my regex: '/^T12F8B0A22[A-Z0-9]{2}F8$/'
I am getting a few errors with this however. My first error, is that there are some characters are the end of the string I still need to get rid of, which is odd because I thought $/ denoted the end of the line in regex. However when I run my code through the debugger I notice that after running through the following code:
#regexString contains the serial read line data
regexString = re.sub('/^T12F8B0A22[A-Z0-9]{2}F8$/', '', regexString)
My string looks something like this: 'T12F8B0A2200F8\\r'
I need to get rid of the \\r.
If for some reason I can't get rid of this with regex, how in python do you send specific string character through an argument? In this case I suppose it would be length - 3?
Your problem is threefold:
1) your string contains extra \r (Carriage Return character) before \n (New Line character); this is common in Windows and in network communication protocols; it is probably best to remove any trailing whitespace from your string:
regexString = regexString.rstrip()
2) as mentioned by Wiktor Stribiżew, your regexp is unnecessarily surrounded with / characters - some languages, like Perl, define regexp as a string delimited by / characters, but Python is not one of them;
3) your instruction using re.sub is actually replacing the matching part of regexString with an empty string - I believe this is the exact opposite of what you want (you want to keep the match and remove everything else, right?); that's why fixing the regexp makes things "even worse".
To summarize, I think you should use this instead of your current code:
m = re.match('T12F8B0A22[A-Z0-9]{2}F8', regexString)
regexString = m.group(0)
There are several ways to get rid of the "\r", but first a little analysis of your code :
1. the special charakter for the end is just '$' not '$\' in python.
2. re.sub will substitute the matched pattern with a string ( '' in your case) wich would substitute the string you want to get with an empty string and you are left with the //r
possible solutions:
use simple replace:
regexString.replace('\\r','')
if you want to stick to regex the approach is the same
pattern = '\\\\r'
match = re.sub(pattern, '',regexString)
2.2 if you want the acces the different groubs use re.search
match = re.search('(^T12F8B0A22[A-Z0-9]{2}F8)(.*)',regexString)
match.group(1) # will give you the T12...
match.groupe(2) # gives you the \\r
Just match what you want to find. Couple of examples:
import re
data = '''lots of
otherT12F8B0A2212F8garbage
T12F8B0A2234F8around
T12F8B0A22ABF8the
stringsT12F8B0A22CDF8
'''
print(re.findall('T12F8B0A22..F8',data))
['T12F8B0A2212F8', 'T12F8B0A2234F8', 'T12F8B0A22ABF8', 'T12F8B0A22CDF8']
m = re.search('T12F8B0A22..F8',data)
if m:
print(m.group(0))
T12F8B0A2212F8
This question already has answers here:
How do I match any character across multiple lines in a regular expression?
(26 answers)
Closed 3 years ago.
From this URL view-source:https://www.amazon.com/dp/073532753X?smid=A3P5ROKL5A1OLE
I want to get string between var iframeContent = and obj.onloadCallback = onloadCallback;
I have this regex iframeContent(.*?)obj.onloadCallback = onloadCallback;
But it does not work. I am not good at regex so please pardon my lack of knowledge.
I even tried iframeContent(.*?)obj.onloadCallback but it does not work.
It looks like you just want that giant encoded string. I believe yours is failing for two reasons. You're not running in DOTALL mode, which means your . won't match across multiple lines, and your regex is failing because of catastrophic backtracking, which can happen when you have a very long variable length match that matches the same characters as the ones following it.
This should get what you want
m = re.search(r'var iframeContent = \"([^"]+)\"', html_source)
print m.group(1)
The regex is just looking for any characters except double quotes [^"] in between two double quotes. Because the variable length match and the match immediately after it don't match any of the same characters, you don't run into the catastrophic backtracking issue.
I suspect that input string lies across multiple lines.Try adding re.M in search line (ie. re.findall('someString', text_Holder, re.M)).
You could try this regex too
(?<=iframeContent =)(.*)(?=obj.onloadCallback = onloadCallback)
you can check at this site the test.
Is it very important you use DOTALL mode, which means that you will have single-line
I have this expression
:([^"]*) \(([^"]*)\)
and this text
:chkpf_uid ("{4astr-hn389-918ks}")
:"#cert" ("false")
Im trying to match it so that on the first sentence ill get these groups:
chkpf_uid
{4astr-hn389-918ks}
and on the second, ill get these:
#cert
false
I want to avoid getting the quotes.
I can't seem to understand why the expression I use won't match these, especially if I switch the [^"]* to a (.*).
with ([^"]*): wont match
with (.*): does match, but with quotes
This is using the re module in python 2.7
Sidenote: your input may require a specific parser to handle, especially if it may have escape sequences.
Answering the question itself, remember that a regex is processed from left to right sequentially, and the string is processed the same here. A match is returned if the pattern matches a portion/whole string (depending on the method used).
If there are quotation marks in the string, and your pattern does not let match those quotes, the match will be failed, no match will be returned.
A possible solution can be adding the quotes as otpional subpatterns:
:"?([^"]*)"? \("?([^"]*)"?\)
^^ ^^ ^^ ^^
See the regex demo
The parts you need are captured into groups, and the quotes, present or not, are just matched, left out of your re.findall reach.
This question already has answers here:
How do I match any character across multiple lines in a regular expression?
(26 answers)
Closed 2 years ago.
I'm trying to extract a simple sentence from a string delimited with a # character.
str = "#text text text \n text#"
with this pattern
pattern = '#(.+)#'
now, the funny thing is that regular expression isn't matched when the string contains newline character
out = re.findall(pattern, str) # out contains empty []
but if I remove \n from string it works fine.Any idea how to fix this ?
Also pass the re.DOTALL flag, which makes the . match truly everything.
Make the '.' special character match any character at all, including a newline; without this flag, '.' will match anything except a newline.
Use re.DOTALL if you want your . to match newline also: -
>>> out = re.findall('#(.+)#', my_str, re.DOTALL)
>>> out
['text text text \n text']
Also, it's not a good idea to use built-in names as your variable names. Use my_str instead of str.
Try this regex "#([^#]+)#"
It will match everything between the delimiters.
Add the DOTALL flag to your compile or match.