I have the following string:
the quick brown fox abc(1)(x)
with the following regex:
(?i)(\s{1})(abc\(1\)\([x|y]\))
and the output is
abc(1)(x)
which is expected, however, I can't seem to:
use \W \w \d \D etc to extract more than 1 space
combine the quantifier to add more spaces.
I would like the following output:
the quick brown fox abc(1)(x)
from the primary lookup "abc(1)(x)" I would like up to 5 words on either side of the lookup. my assumption is that spaces would demarcate a word.
Edit 1:
The 5 words on either side would be unknown for future examples. the string may be:
cat with a black hat is abc(1)(x) the quick brown fox jumps over the
lazy dog.
In this case, the desired output would be:
with a black hat is abc(1)(x) the quick brown fox jumps
Edit 2:
edited the expected output in the first example and added "up to" 5 words
(?:[0-9A-Za-z_]+[^0-9A-Za-z_]+){0,5}abc\(1\)\([xy]\)(?:[^0-9A-Za-z_]+[0-9A-Za-z_]+){0,5}
Note that I've changed \w+ to [0-9A-Za-z_]+ and \W+ to [^0-9A-Za-z_]+ because depending on your locale / Unicode settings \W and \w might not act the way you expect in Python.
Also note I don't specifically look for spaces, just "non-word characters" this probably handles edge cases a little better for quote characters etc.
But regardless this should get you most of the way there.
BTW: You calling this "lookaround" - really it has nothing to do with "regex lookaround" the regex feature.
If I understand your requirements correctly, you want to do something like this:
(?:\w+[ ]){0,5}(abc\(1\)\([xy]\))(?:[ ]\w+){0,5}
Demo.
BreakDown:
(?: # Start of a non-capturing group.
\w+ # Any word character repeated one or more times (basically, a word).
[ ] # Matches a space character literally.
) # End of the non-capturing group.
{0,5} # Match the previous group between 0 and 5 times.
( # Start of the first capturing group.
abc\(1\) # Matches "abc(1)" literally.
\([xy]\) # Matches "(x)" or "(y)". You don't need "|" inside a character class.
) # End of the capturing group.
(?:[ ]\w+){0,5} # Same as the non-capturing group above but the space is before the word.
Notes:
To make the pattern case insensitive, you may start it with (?i) as you're doing already or use the re.IGNORECASE flag.
If you want to support words not separated by a space, you may replace [ ] with either \W+ (which means non-word characters) or with a character class which includes all the punctuation characters that you want to support (e.g., [.,;?! ]).
Related
I'm using re to take the questions from a text. I just want the sentence with the question, but it's taking multiple sentences before the question as well. My code looks like this:
match = re.findall("[A-Z].*\?", data2)
print(match)
an example of a result I get is:
'He knows me, and I know him. Do YOU know me? Hey?'
the two questions should be separated and the non question sentence shouldn't be there. Thanks for any help.
The . character in regex matches any text, including periods, which you don't want to include. Why not simply match anything besides the sentence ending punctuation?
questions = re.findall(r"\s*([^\.\?]+\?)", data2)
# \s* sentence beginning space to ignore
# ( start capture group
# [^\.\?]+ negated capture group matching anything besides "." and "?" (one or more)
# \? question mark to end sentence
# ) end capture group
You could look for letters, digits, and whitespace that end with a '?'.
>>> [i.strip() for i in re.findall('[\w\d\s]+\?', s)]
['Do YOU know me?', 'Hey?']
There would still be some edge cases to handle, like there could be punctuation like a ',' or other complexities.
You can use
(?<!\S)[A-Z][^?.]*\?(?!\S)
The pattern matches:
(?<!\S) Negative lookbehind, assert a whitespace boundary to the left
[A-Z] Match a single uppercase char A-Z
[^?.]*\? Match 0+ times any char except ? and . and then match a ?
(?!\S) Negative lookahead, assert a whitespace boundary to the right
Regex demo
You should use the ^ at the beginning of your expression so your regex expression should look like this: "^[A-Z].*\?".
"Matches the beginning of the string, or the beginning of a line if the multiline flag (m) is enabled. This matches a position, not a character."
If you have multiple sentences in your line you can use the following regex:
"(?<=.\s+)[A-Z].*\?"
?<= is called positive lookbehind. We try to find sentences which either start in a new line or have a period (.) and one or more whitespace characters before them.
Input is a two-sentence string:
s = 'Sentence 1 here. This sentence contains 1 fl. oz. but is one sentence.'
I'd like to .split s into sentences based on the logic that:
sentences end with one or more periods, exclamation marks, questions marks, or period+quotation mark
and are then followed by 1+ whitespace characters and a capitalized alpha character.
Desired result:
['Sentence 1 here.', 'This sentence contains 1 fl. oz. but is one sentence.']
Also okay:
['Sentence 1 here', 'This sentence contains 1 fl. oz. but is one sentence.']
But I currently chop off the 0th element of each sentence because the uppercase character is captured:
import re
END_SENT = re.compile(r'[.!?(.")]+[ ]+[A-Z]')
print(END_SENT.split(s))
['Sentence 1 here', 'his sentence contains 1 fl. oz. but is one sentence.']
Notice the missing T. How can I tell .split to ignore certain elements of the compiled pattern?
((?<=[.!?])|(?<=\.\")) +(?=[A-Z])
Try it here.
Although I would suggest the below to allow quotes to be followed by any of .!? to be a split condition
((?<=[.!?])|(?<=[.!?]\")) +(?=[A-Z])
Try it here.
Explanation
The common stuff in both +(?=[A-Z])
' +' #One or more spaces(The actual splitting chars used.)
(?= #START positive look ahead check if it followed by this, but do not consume
[A-Z] #Any capitalized alphabet
) #END positive look ahead
The conditions for what comes before the space
For Solution1
( #GROUP START
(?<= #START Positive look behind, Make sure this comes before but do not consume
[.!?] #any one of these chars should come before the splitting space
) #END positive look behind
| #OR condition this is also the reason we had to put all this in GROUP
(?<= #START Positive look behind,
\.\" #splitting space could precede by .", covering a condition that is not by the previous set of . or ! or ?
) #END positive look behind
) #END GROUP
For Solution2
( #GROUP START
(?<=[.!?]) #Same as the previous look behind
| #OR condition
(?<=[.!?]\") #Only difference here is that we are allowing quote after any of . or ! or ?
) #GROUP END
It's easier to describe the sentence than trying to identify the delimiter. So instead of re.split try with re.findall:
re.findall(r'([^.?!\s].*?[.?!]*)\s*(?![^A-Z])', s)
To preserve the next uppercase letter, the pattern uses a lookahead that is only a test and doesn't consume characters.
details:
( # capture group: re.findall return only the capture group content if any
[^.?!\s] # the first character isn't a space or a punctuation character
.*? # a non-greedy quantifier
[.?!]* # eventual punctuation characters
)
\s* # zero or more white-spaces
(?![^A-Z]) # not followed by a character that isn't a uppercase letter
# (this includes an uppercase letter and the end of the string)
Obviously, for more complicated cases with abbreviations, names, etc., you have to use tools like nltk or any other nlp tools trained with dictionaries.
I have the following regex:
res = re.finditer(r'(?:\w+[ \t,]+){0,4}my car',txt,re.IGNORECASE|re.MULTILINE)
for item in res:
print(item.group())
When I use this regex with the following string:
"my house is painted white, my car is red.
A horse is galloping very fast in the road, I drive my car slowly."
I am getting the following results:
house is painted white, my car
the road, I drive my car
My question is about the quantifier {0,4} that should apply to the whole group. The group collects words with the expression \w+ and some separation symbols with the [ ]. Does the the quantifier apply only to the "words" defined by \w+? In the results I am getting 4 words plus space and comma. It's unclear to me.
So, here's what's happening. You're using ?: to make a non capture group, which collects 1 or more "words", followed by a [ \t,] (a space, tab char, or comma), match one or more of the preceeding. {0,4} matches between 0-4 of the non-capturing group. So it looks at the word "my car" and captures the 4 words before it, since all 4 of them match the \w+ and the , and space get eaten by the character set you specified.
Broken apart more succinctly
(?: -- Non capturing group
\w+ Grab all words
[ \t,]+ -- Grab all spaces, comma, or tab characters
) -- End capture group
{0,4} -- Match the previous capture group 0-4 times
my car -- Based off where you find the words "my car"
As a result this will match 0-4 words / spaces / commas / tabs before the appearance of "my car"
This is working as written
I want to have a regex to find a phrase and two words preceding it if there are two words.
For example I have the string (one sentence per line):
Chevy is my car and Rusty is my horse.
My car is very pretty my dog is red.
If i use the regex:
re.finditer(r'[\w+\b|^][\w+\b]my car',txt)
I do not get any match.
If I use the regex:
re.finditer(r'[\S+\s|^][\S+\s]my car',txt)
I am getting:
's my car' and '. My car' (I am ignoring case and using multi-line)
Why is the regex with \w+\b not finding anything? It should find two words and 'my car'
How can I get two complete words before 'my car' if there are two words. If there is only one word preceding my car, I should get it. If there are no words preceding it I should get only 'my car'. In my string example I should get: 'Chevy is my car' and 'My car' (no preceding words here)
In your r'[\w+\b|^][\w+\b]my car regex, [\w+\b|^] matches 1 symbol that is either a word char, a +, a backdpace, |, or ^ and [\w+\b] matches 1 symbol that is either a word char, or +, or a backspace.
The point is that inside a character class, quantifiers and a lot (but not all) special characters match literal symbols. E.g. [+] matches a plus symbol, [|^] matches either a | or ^. Since you want to match a sequence, you need to provide a sequence of subpatterns outside of a character class.
It seems as if you intended to use \b as a word boundary, however, \b inside a character class matches only a backspace character.
To find two words and 'my car', you can use, for example
\S+\s+\S+\s+my car
See the regex demo (here, \S+ matches one or more non-whitespace symbols, and \s+ matches 1 or more whitespaces, and the 2 occurrences of these 2 consecutive subpatterns match these symbols as a sequence).
To make the sequences before my car optional, just use a {0,2} quantifier like this:
(?:\S+[ \t]+){0,2}my car
See this regex demo (to be used with the re.IGNORECASE flag). See Python demo:
import re
txt = 'Chevy is my car and Rusty is my horse.\nMy car is very pretty my dog is red.'
print(re.findall(r'(?:\S+[ \t]+){0,2}my car', txt, re.I))
Details:
(?:\S+[ \t]+){0,2} - 0 to 2 sequences of 1+ non-whitespaces followed with 1+ space or tab symbols (you may also replace it with [^\S\r\n] to match any horizontal space or \s if you also plan to match linebreaks).
my car - a literal text my car.
I have a regex that matches all three characters words in a string:
\b[^\s]{3}\b
When I use it with the string:
And the tiger attacked you.
this is the result:
regex = re.compile("\b[^\s]{3}\b")
regex.findall(string)
[u'And', u'the', u'you']
As you can see it matches you as a word of three characters, but I want the expression to take "you." with the "." as a 4 chars word.
I have the same problem with ",", ";", ":", etc.
I'm pretty new with regex but I guess it happens because those characters are treated like word boundaries.
Is there a way of doing this?
Thanks in advance,
EDIT
Thaks to the answers of #BrenBarn and #Kendall Frey I managed to get to the regex I was looking for:
(?<!\w)[^\s]{3}(?=$|\s)
If you want to make sure the word is preceded and followed by a space (and not a period like is happening in your case), then use lookaround.
(?<=\s)\w{3}(?=\s)
If you need it to match punctuation as part of words (such as 'in.') then \w won't be adequate, and you can use \S (anything but a space)
(?<=\s)\S{3}(?=\s)
As described in the documentation:
A word is defined as a sequence of alphanumeric or underscore characters, so the end of a word is indicated by whitespace or a non-alphanumeric, non-underscore character.
So if you want a period to count as a word character and not a word boundary, you can't use \b to indicate a word boundary. You'll have to use your own character class. For instance, you can use a regex like \s[^\s]{3}\s if you want to match 3 non-space characters surrounded by spaces. If you still want the boundary to be zero-width (i.e., restrict the match but not be included in it), you could use lookaround, something like (?<=\s)[^\s]{3}(?=\s).
This would be my approach. Also matches words that come right after punctuations.
import re
r = r'''
\b # word boundary
( # capturing parentheses
[^\s]{3} # anything but whitespace 3 times
\b # word boundary
(?=[^\.,;:]|$) # dont allow . or , or ; or : after word boundary but allow end of string
| # OR
[^\s]{2} # anything but whitespace 2 times
[\.,;:] # a . or , or ; or :
)
'''
s = 'And the tiger attacked you. on,bla tw; th: fo.tes'
print re.findall(r, s, re.X)
output:
['And', 'the', 'on,', 'bla', 'tw;', 'th:', 'fo.', 'tes']