In python, I want to match substring containing two terms with up to a certain number of words in between but not when it is equal to a certain substring.
I have this regular expression (regex101) that does the first part of the job, matching two terms with up to a certain number of words in between.
But I want to add a part or condition with AND operator to exclude a specific sentence like "my very funny words"
my(?:\s+\w+){0,2}\s+words
Expected results for this input:
I'm searching for my whatever funny words inside this text
should match with my whatever funny words
while for this input:
I'm searching for my very funny words inside this text
there should be no match
Thank you all for helping out
You may use the following regex pattern:
my(?! very funny)(?:\s+\w+){0,2}\s+words
This inserts a negative lookahead (?! very funny) into your existing pattern to exclude the matches you don't want. Here is a working demo.
Related
I am building a tweet classification model, and I am having trouble finding a regex pattern that fits what I am looking for.
what I want the regex pattern to pick up:
Any hashtags used in the tweet but without the hash mark (example - #omg to just omg)
Any mentions used in the tweet but without the # symbol (example - #username to just username)
I don't want any numbers or any words containing numbers returned ( this is the most difficult task for me)
Other than that, I just want all words returned
Thank you in advance if you can help
Currently I am using this pattern:** r"(?u)\b\w\w+\b"** but it is failing to remove numbers.
This regex should work.
(#|#)?(?![^ ]*\d[^ ]*)([^ ]+)
Explanation:
(#|#)?: A 'hash' or 'at' character. Match 0 or 1 times.
(?!): Negative lookahead. Check ahead to see if the pattern in the brackets matches. If so, negate the match.
[^ ]*\d[^ ]*: any number of not space characters, followed by a digit, followed by any number of space characters. This is nested in the negative lookahead, so if a number is found in the username or hashtag, the match is negated.
([^ ]+): One or more not space characters. The negative lookahead is a 0-length match, so if it passes, fetch the rest of the username/hashtag (Grouped with brackets so you can replace with $2).
I am quite new to regex, working on string verification where I want both conditions to be met. I am matching text containing 7digit numbers starting with 4 or 7 + string needs to contain one of the provided words.
What I managed so far:
\b((4|7)\d{6})\b|(\border|Order|Bestellung|bestellung|commande|Commande|ordine|Ordine|objednavku|Objednavku|objednavka|Objednavka)
Regex above correctly finds numbers but words are after OR statement which I would need to follow AND logic instead.
Could you please help me implement a change that would work as AND statement between digits and words?
You can use
(?s)^(?=.*\b(?:order|Order|Bestellung|bestellung|commande|Commande|ordine|Ordine|objednavku|Objednavku|objednavka|Objednavka)\b).*\b([47]\d{6})\b
If you can and want use a case insensitive matching with re.I, you can use
(?si)^(?=.*\b(?:order|bestellung|commande|ordine|objednavk[ua])\b).*\b([47]\d{6})\b
See the regex demo.
This matches
^ - start of string
(?=.*\b(?:order|Order|Bestellung|bestellung|commande|Commande|ordine|Ordine|objednavku|Objednavku|objednavka|Objednavka)\b) - a positive lookahead that matches any zero or more chars, as many as possible, up to any of the whole words listed in the group
.* - zero or more chars, as many as possible
\b([47]\d{6})\b - a 7-digit number as a whole word that starts with 4 or 7.
Do not forget to use a raw string literal to define a regex in Python code:
pattern = r'(?si)^(?=.*\b(?:order|bestellung|commande|ordine|objednavk[ua])\b).*\b([47]\d{6})\b'
By default, everything in regex is AND
if you do
abc,
it means "a" AND "b" AND "c"
so there is no need for an AND in regex
just remove the | between the numbers match and the words
\b(4|7)\d{6}(border|Order|Bestellung|bestellung|commande|Commande|ordine|Ordine|objednavku|Objednavku|objednavka|Objednavka)\b
I assume the backslash with the first word \border was a mistake.
This can match stuff like : "4958374border"
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 am attempting to extract some some raw strings using re module in python. The end of a to-be-extracted section is identified by a repeating word (repeated multiple times), Current efforts always captures the last match of the repeating word. How can I modify this behavior?
A textfile has been extracted from a pdf. The entire PDF is stored as one string. A general formatting of the string is as below:
*"***Start of notes: Collection of alphanumeric words and characters EndofsectionTopic A: string of words Endofsection"*
The intended string to be captured is: "Collection of alphanumeric words and characters"
The attempted solution used in this situation was: "
re.compile(r"*{3}Start of notes:(.+)\sEndofsection")
This attempt tends to match the whole string rather than just "Collection of alphanumeric words and characters" as intended.
One possible approach is to split with Endofsection and then extract the string from the first section only - this works, but I was hoping to find a more elegant solution using re.compile.
Two problems in your regex,
You need to escape * as it is a meta character as \*
Second, you are using (.+) which is a greedy quantifier and will try matching as much as possible, but since you want the shortest match, you need to just change it to (.+?)
Fixing these two issues, gives you the correct intended match.
Regex Demo
Python code,
import re
s = "***Start of notes: Collection of alphanumeric words and characters EndofsectionTopic A: string of words Endofsection"
m = re.search(r'\*{3}Start of notes:(.+?)\sEndofsection', s)
if m:
print(m.group(1))
Prints,
Collection of alphanumeric words and characters
I am having a regular expression to match a particular pattern. Say, a pattern that will match all three letter words. But i want it to not match words like 'and','got' etc... What would be the best way to do it ,in Python.
My pattern is
r'\b\w{3}\b'
I tried
r'(\b\w{3}\b)(?!and)'
but fails.
Regexes match left to right, and lookaheads are no exception. Your expression will match three letters that are not followed by and (which is impossible because of the \b, by the way).
Move the lookahead before the \w to make it work:
r'(\b(?!and)\w{3}\b)'
You can add more words there --
r'(\b(?!and|got|may)\w{3}\b)'
but for more non-matches it may be more effective to match all three letter words and use code to strip the result of them.