I'm trying to use the following regex \-(.*?)-|\-(.*?)* it seems to work fine on regexr but python says there's nothing to repeat?
I'm trying to match all text in between the first two dashes or if a second dash does not exist after the first all text from the first - onwards.
Also, the regex above includes the dashes, but would preferrably like to exclude these so I don't have to do an extra replace etc.
You can use re.search with this pattern:
-([^-]*)
Note that - doesn't need to be escaped.
An other way consists to only search the positions of the two first dashes, and to extract the substring between these positions. Or you can use split:
>>> 'aaaaa-bbbbbb-ccccc-ddddd'.split('-')[1]
'bbbbbb'
Related
I have an input that is valid if it has this parts:
starts with letters(upper and lower), numbers and some of the following characters (!,#,#,$,?)
begins with = and contains only of numbers
begins with "<<" and may contain anything
example: !!Hel##lo!#=7<<vbnfhfg
what is the right regex expression in python to identify if the input is valid?
I am trying with
pattern= r"([a-zA-Z0-9|!|#|#|$|?]{2,})([=]{1})([0-9]{1})([<]{2})([a-zA-Z0-9]{1,})/+"
but apparently am wrong.
For testing regex I can really recommend regex101. Makes it much easier to understand what your regex is doing and what strings it matches.
Now, for your regex pattern and the example you provided you need to remove the /+ in the end. Then it matches your example string. However, it splits it into four capture groups and not into three as I understand you want to have from your list. To split it into four caputre groups you could use this:
"([a-zA-Z0-9!##$?]{2,})([=]{1}[0-9]+)(<<.*)"
This returns the capture groups:
!!Hel##lo!#
=7
<<vbnfhfg
Notice I simplified your last group a little bit, using a dot instead of the list of characters. A dot matches anything, so change that back to your approach in case you don't want to match special characters.
Here is a link to your regex in regex101: link.
I am attempting to match paragraph numbers inside my block of text. Given the following sentence:
Refer to paragraph C.2.1a.5 for examples.
I would like to match the word C.2.1a.5.
My current code like so:
([0-9a-zA-Z]{1,2}\.)
Only matches C.2.1a. and es., which is not what I want. Is there a way to match the full C.2.1a.5 and not match es.?
https://regex101.com/r/cO8lqs/13723
I have attempted to use ^ and $, but doing so returns no matches.
You should use following regex to match the paragraph numbers in your text.
\b(?:[0-9a-zA-Z]{1,2}\.)+[0-9a-zA-Z]\b
Try this demo
Here is the explanation,
\b - Matches a word boundary hence avoiding matching partially in a large word like examples.
(?:[0-9a-zA-Z]{1,2}\.)+ - This matches an alphanumeric text with length one or two as you tried to match in your own regex.
[0-9a-zA-Z] - Finally the match ends with one alphanumeric character at the end. In case you want it to match one or two alphanumeric characters at the end too, just add {1,2} after it
\b - Matches a word boundary again to ensure it doesn't match partially in a large word.
EDIT:
As someone pointed out, in case your text has strings like A.A.A.A.A.A. or A.A.A or even 1.2 and you don't want to match these strings and only want to match strings that has exactly three dots within it, you should use following regex which is more specific in matching your paragraph numbers.
(?<!\.)\b(?:[0-9a-zA-Z]{1,2}\.){3}[0-9a-zA-Z]\b(?!\.)
This new regex matches only paragraph numbers having exactly three dots and those negative look ahead/behind ensures it doesn't match partially in large string like A.A.A.A.A.A
Updated regex demo
Check these python sample codes,
import re
s = 'Refer to paragraph C.2.1a.5 for examples. Refer to paragraph A.A.A.A.A.A.A for examples. Some more A.A.A or like 1.22'
print(re.findall(r'(?<!\.)\b(?:[0-9a-zA-Z]{1,2}\.){3}[0-9a-zA-Z]\b(?!\.)', s))
Output,
['C.2.1a.5']
Also for trying to use ^ and $, they are called start and end anchors respectively, and if you use them in your regex, then they will expect matching start of line and end of line which is not what you really intend to do hence you shouldn't be using them and like you already saw, using them won't work in this case.
If simple version is required, you can use this easy to understand and modify regex ([A-Z]{1}\.[0-9]{1,3}\.[0-9]{1,3}[a-z]{1}\.[0-9]{1,3})
I think we should keep the regex expression simple and readable.
You can use the regex
**(?:[a-zA-Z]+\.){3}[a-zA-Z]+**
Explanation -
The expression (?:[a-zA-Z]+.){3} ensures that the group (?:[a-zA-Z]+.) is to be repeated 3 times within the word. The group contains an alphabetic character followed a dot.
The word would end with an alphabetic character.
Output:
['C.2.1a.5']
I'm trying to use regex to find proxy address on a website. Currently I'm using this piece of regex (\d{1,3}\.){3}\d{1,3}:(\d+). It works on regexr.com and in sublime text, but when I try to use it in Python it doesn't work as expected.
This is the piece of code I'm using:
p = re.compile("(\d{1,3}\.){3}\d{1,3}:(\d+)")
ipCandidates = p.findall(soupString)
It should return proxies like this 120.206.182.172:8123 but it returns tuples like this ('44.', '3128'). What can I do to fix this?
Thank you.
re.findall() only returns the contents of capturing groups instead of the whole match (if you have such groups in your regex).
Then, you're repeating a capturing group three times, which means that only the third repetition is preserved (the other two are overwritten).
Change your regex to
p = re.compile(r"(?:\d{1,3}\.){3}\d{1,3}:\d+")
and you'll get whole matches.
If you do want tuples of the separate submatches (without the dots and colon), you can do that, too, but you can't use repetition then:
p = re.compile(r"(\d{1,3})\.(\d{1,3})\.(\d{1,3})\.(\d{1,3}):(\d+)")
Also, always use raw strings for regexes, so regex escape sequences and string escape sequences can't be confused.
I need to build a program that can read multiple lines of code, and extract the right information from each line.
Example text:
no matches
one match <'found'>
<'one'> match <found>
<'three'><'matches'><'found'>
For this case, the program should detect <'found'>, <'one'>, <'three'>, <'matches'> and <'found'> as matches because they all have "<" and "'".
However, I cannot work out a system using regex to account for multiple matches on the same line. I was using something like:
re.search('^<.*>$')
But if there are multiple matches on one line, the extra "'<" and ">'" are taken as part of the .*, without counting them as separate matches. How do I fix this?
This works -
>>> r = re.compile(r"\<\'.*?\'\>")
>>> r.findall(s)
["<'found'>", "<'one'>", "<'three'>", "<'matches'>", "<'found'>"]
Use findall instead of search:
re.findall( r"<'.*?'>", str )
You can use re.findall and match on non > characters inside of the angle brackets:
>>> re.findall('<[^>]*>', "<'three'><'matches'><'found'>")
["<'three'>", "<'matches'>", "<'found'>"]
Non-greedy quantifier '?' as suggested by anubhava is also an option.
Say I have a string like {{ComputersRule}} and a regex like: [^\}]+. How would I get regular expressions to start at a specified point in the string, i.e. Once it has reached the third character in the string. If it's relevant, and I doubt it is, I'm working in Python version 2.7.3. Thank you.
I'd recommend using Python to grab the substring from the third character onwards, and then apply the regex to the rest.
Otherwise, you could just use the regex . (any character except newline) to gobble up the first n characters:
^.{3}([^\}]+)
Notice the ^.{3} which forces the [^\}]+ to not include the first three characters of the string (the ^ anchors to the start of the string/line). The brackets capture the bit you want to extract (so get capturing group 1).
In your particular case, if it's just a case of "I want the text inside the {{ and }}" you could do \{\{([^\}]+)\}\} or [^\{\}]+.
It appears that what you want to do is to match text within the double braces.
The trick is to specify the braces in the regex but capture the part within. In this case try
\{\{([^}]+)\}\}