I wish to chop some text into sentences.
I wish to match all text up until: a period followed by a space, a question mark followed by a space or an exclamation mark followed by a space, in an non greedy fashion.
Additionally, the punctuation might be found at the very end of the string or followed by a /r/n for example.
This will almost do it:
([^\.\?\!]*)
But I'm missing the space in the expression. How do I fix this?
Example:
I' a.m not. So? Sure about this! Actually. Should give:
I' a.m not
So
Sure about this
Actually
You can achieve such conditions by using positive lookahead assertions.
[^.?!]+(?=[.?!] )
See it here on Regexr.
When you look at the demo, The sentences at the end of a row with no following space are not matched. You can fix this by adding an alternation with the Anchor $ and using the modifier m (makes the $ match the end of a row):
[^.?!]+(?=[.?!](?: |$))
See it here on Regexr
Try this:
(.*?[!\.\?] )
.* gives all,
[] is any of these characters
then the () gives you a group to reference so you can get the match out.
Use a non-greedy match with s look ahead:
^.*?(?=[.!?]( |$))
Note how you don't have to escape those chars when they are in a character class [...].
This should do it:
^.*?(?=[!.?][\s])
Related
I am trying to grab fary_trigger_post in the code below using Regex. However, I don't understand why it always includes " in the end of the matched pattern, which I don't expect.
Any idea or suggestion?
re.match(
r'-instance[ "\']*(.+)[ "\']*$',
'-instance "fary_trigger_post" '.strip(),
flags=re.S).group(1)
'fary_trigger_post"'
Thank you.
The (.+) is greedy and grabs ANY character until the end of the input. If you modified your input to include characters after the final double quote (e.g. '-instance "fary_trigger_post" asdf') you would find the double quote and the remaining characters in the output (e.g. fary_trigger_post" asdf). Instead of .+ you should try [^"\']+ to capture all characters except the quotes. This should return what you expect.
re.match(r'-instance[ "\']*([^"\']+)[ "\'].*$', '-instance "fary_trigger_post" '.strip(), flags=re.S).group(1)
Also, note that I modified the end of the expression to use .* which will match any characters following the last quote.
Here's what I'd use in your matching string, but it's hard to provide a better answer without knowing all your cases:
r'-instance\s+"(.+)"\s*$'
When you try to get group 1 (i.e. (.+)) regex will follow this match to the end of string, as it can match . (any character) 1 or more times (but it will take maximum amount of times). I would suggest use the following pattern:
'-instance[ "\']*(.+)["\']+ *$'
This will require regex to match all spaces in the end and all quoutes separatelly, so that it won't be included into group 1
In Python regex, how would I match only the facebook.com...777 substrings given either string? I don't want the ?sfnsn=mo at the end.
I have (?<=https://m\.)([^\s]+) to match everything after the https://m.. I also have (?=\?sfnsn) to match every thing in front of ?sfnsn.
How do I combine the regex to only return the facebook.com...777 part for either string.
have: https://m.facebook.com/story.php?story_fbid=123456789&id=7777777777?sfnsn=mo
want: facebook.com/story.php?story_fbid=123456789&id=7777777777
have: https://m.facebook.com/story.php?story_fbid=123456789&id=7777777777
want: facebook.com/story.php?story_fbid=123456789&id=7777777777
Here's what I was messing around with https://regex101.com/r/WYz5dn/2
(?<=https://m\.)([^\s]+)(?=\?sfnsn)
You could use a capturing group instead of a positive lookbehind and match either ?sfnsn or the end of the string.
https://m\.(\S*?)(?:\?sfnsn|$)
Regex demo
Using the lookarounds, the pattern could be:
(?<=https://m\.)\S*?(?=\?sfnsn|$)
Regex demo
Putting a ? at the end works, since the last grouped lookahead may or may not exist, we put a question mark after it:
(?<=https://m\.)([^\s]+)(?=\?sfnsn)?
I am trying to use python re to match a string with a specific pattern.
The problem I met is, I have this expected sentence:
"It is X. not X`
X can be anything; A word, or a bunch of word, or number, or digits.
The pattern I build is:
It is \w+. not \w+
just using
string.replace("X", "\w+")
It works if X is a word, or bunch of words, or int, but not for digits. How can I build my pattern in order to match everything in this pattern?
The . is a special character in a regular expression that will match any character. So .+ will match one or more characters.
r"It is .+\. not .+"
Not that the period is escaped \., this is because in that case, you want to match an actual period.
Because .+ won't work in some cases, for example
It is quote. not a double-quote
It is a dog. not a cat
I would use this one instead :
(?<=It is ).+(?=\.)|(?<=not ).+$
Explanation
(?<=It is ).+(?=\.) Any consecutive characters precedeed by It is and followed by a point
| OR
(?<=not ).*$ Any consecutive characters precedeed by not and followed by end of line anchor
(?<=It is ).*(?=\.)|(?<=not ).*$
Demo
I have figured out, can use str.replace("X", "(\w+|\d+\.\d+)") to approach the problem. Hope can help others having the same issue.
I have ONE string as plain text and want to extract phone numbers of any format from it.
Here is my regex:
r = re.compile(r"(\d{3}[-\.\s]??\d{3}[-\.\s]??\d{4}|\(\d{3}\)[-\s*]\d{3}[-\.\s]??\d{4})")
It extracts the following matches correctly:
617.933.6444
(880)-567-4565
(880) 567-4565
222-333-8888
555 666 4444
9999999999
But how can I avoid getting 7986815059 when I have 798681505951 in the text?
How to make an ending for my regex? (it should not contain letters and digits after and before, exact number count must be 10)
!!!!
Decision
If somebody needs to find US phone numbers in string, use link from the last Wiktor Stribiżew comment.
You need to use word boundaries, but placing them into your pattern is not obvious. It is due to the fact that the second alternative starts with a non-word char, \(. Thus, the first \b must be added at the beginning of the first alternative, and the trailing one at the very end of the pattern:
r'(\b\d{3}[-.\s]?\d{3}[-.\s]?\d{4}|\(\d{3}\)[-\s*]\d{3}[-.\s]?\d{4})\b'
^^ ^^
See the regex demo
You may also require a non-word char or start of string before (. Then add \B at the second alternative start:
r'(\b\d{3}[-.\s]?\d{3}[-.\s]?\d{4}|\B\(\d{3}\)[-\s*]\d{3}[-.\s]?\d{4})\b'
^^
See another demo
Also, note that there is no need escaping a . inside a character class, it is already parsed as a literal dot in [.]. And no need using a lazy ?? quantifier, it does not make sense here and a greedy version, ?, will work equally well and will look "cleaner".
Given is the following python script:
text = '<?xml version="1.24" encoding="utf-8">'
mu = (".??[?]?[?]", "....")
for item in mu:
print item,":",re.search(item, text).group()
Can someone please explain why the first hit with the regex .??[?]?[?] returns <? instead of just ?.
My explaination:
.?? should match nothing as .? can match or not any char and the second ? makes it not greedy.
[?]? can match ? or not, so nothing is good, too
[?] just matches ?
That should result in ? and not in <?
For the same reason o*?bar matches oobar in foobar. Even if the quantifier is non-greedy the regex will try to match from the first char in all possible ways, before moving on to the next.
First the .?? matches an empty string, but when the regex engine backtracks to it, it matches <, thus making the rest of the regex match, without moving the start position of the match to the next character.
Regex "greediness" only affects backtracking; it doesn't mean that the regex engine will skip earlier potential match points — a regex always takes the first possible match. In this case, that means <? because it starts farther to the left than ?.