I have a regular expression given by a word and a range of words following.
For example:
pattern = 'word \\w+ \\w+ \\w+"
result = [text[match.start():match.end()] for match in re.finditer(pattern, text)]
How could you modify the regular expression so that when there is a smaller number of elements that in the interval also recognize it? For example if the word is in the end of the string I would like it to return that interval too.
Always if possible to return the greatest possible pattern.
Your 'word \\w+ \\w+ \\w+" regex matches a word and then 3 more "words" (space separated). You want to match 0 to 3 of these words. Use
re.findall(r'word(?:\s+\w+){0,3}', s)
Or, to allow any non-word chars in between the "words", replace \s with \W:
re.findall(r'word(?:\W+\w+){0,3}', s)
Details:
word - word string
(?:\s+\w+){0,3} - 0 to 3 sequences (the {0,3} is a greedy version of the limiting quantifier, it will match as many occurrences as possible) of:
\s+ - 1+ whitespaces
\w+ - 1 or more word chars.
See the regex demo.
Related
I have a text example like
0s11 0s12 0s33 my name is 0sgfh 0s1 0s22 0s87
I want to detect the consecutive sequences that start 0s.
So, the expected output should be 0s11 0s12 0s33, 0sgfh 0s1 0s22 0s87
I tried using regex
(0s\w+)
but that would detect each 0s11, 0s12, 0s33, etc. individually.
Any idea on how to modify the pattern?
To get those 2 matches where there are at least 2 consecutive parts:
\b0s\w+(?:\s+0s\w+)+
Explanation
\b A word boundary to prevent a partial word match
0s\w+ Match os and 1+ word chars
(?:\s+0s\w+)+ Repeat 1 or more times whitespace chars followed by 0s and 1+ word chars
Regex demo
If you also want to match a single occurrence:
\b0s\w+(?:\s+0s\w+)*
Regex demo
Note that \w+ matches 1 or more word characters so it would not match only 0s
Should be doable with re.findall(). Your pattern was correct! :)
import re
testString = "0s11 0s12 0s33 my name is 0sgfh 0s1 0s22 0s87"
print(re.findall('0s\w', testString))
['0s11', '0s12', '0s33', '0sgfh', '0s1', '0s22', '0s87']
Hope this helps!
Need to construct a regular expression that counts numbers between alphabets.
schowalte3rguss77ie85 - 2
xyz1zyx - 1
x1y1z1 - 2
I have constructed this . But this doesn't work for case 3.
[[a-z]+[0-9]+[a-z]]*
Any help would be appreciated. Thanks in advance.
Use regx:
(?<=[a-z])\d+(?=[a-z])
Demo: https://regex101.com/r/tpss6x/1
[Javascript]
If you want a count only, the last part should be a lookahead assertion.
If you want to also match uppercase chars, you can make the pattern case insensitive.
[a-z]\d+(?=[a-z])
Explanation
[a-z] Match a single char a-z
\d+ Match 1+ digits
(?=[a-z]) Positive lookahead, assert a char a-z to the right
Regex demo
You can use
(?<=[^\W\d_])\d+(?=[^\W\d_])
See the regex demo. If you want to only support ASCII letters, replace [^\W\d_] (that matches any Unicode letter) with [a-zA-Z].
Details:
(?<=[^\W\d_]) - immediately before the current location, there must be any Unicode letter
\d+ - one or more digits
(?=[^\W\d_]) - immediately after the current location, there must be any Unicode letter.
Counting can be done with len(...), see this Python demo:
import re
text = "schowalte3rguss77ie85"
matches = re.findall(r'(?<=[^\W\d_])\d+(?=[^\W\d_])', text)
print(len(matches)) # => 2
I would like to match a word when it is after a char m or b
So for example, when the word is men, I would like to return en (only the word that is following m), if the word is beetles then return eetles
Initially I tried (m|b)\w+ but it matches the entire men not en
How do I write regex expression in this case?
Thank you!
You could get the match only using a positive lookbehind asserting what is on the left is either m or b using character class [mb] preceded by a word boundary \b
(?<=\b[mb])\w+
(?<= Positive lookbehind, assert what is directly to the left is
\b[mb] Word boundary, match either m or b
) Close lookbehind
\w+ Match 1 + word chars
Regex demo
If there can not be anything after the the word characters, you can assert a whitespace boundary at the right using (?!\S)
(?<=\b[mb])\w+(?!\S)
Regex demo | Python demo
Example code
import re
test_str = ("beetles men")
regex = r"(?<=\b[mb])\w+"
print(re.findall(regex, test_str))
Output
['eetles', 'en']
You may use
\b[mb](\w+)
See the regex demo.
NOTE: When your known prefixes include multicharacter sequences, say, you want to find words starting with m or be, you will have to use a non-capturing group rather than a character class: \b(?:m|be)(\w+). The current solution can thus be written as \b(?:m|b)(\w+) (however, a character class here looks more natural, unless you have to build the regex dynamically).
Details
\b - a word boundary
[mb] - m or b
(\w+) - Capturing group 1: any one or more word chars, letters, digits or underscores. To match only letters, use ([^\W\d_]+) instead.
Python demo:
import re
rx = re.compile(r'\b[mb](\w+)')
text = "The words are men and beetles."
# First occurrence:
m = rx.search(text)
if m:
print(m.group(1)) # => en
# All occurrences
print( rx.findall(text) ) # => ['en', 'eetles']
(?<=[mb])\w+/
You can use this above regex. The regex means "Any word starts with m or b".
(?<=[mb]): positive lookbehind
\w+: matches any word character (equal to [a-zA-Z0-9]+)
I'm trying to create a regex that matches a third person form of a verb created using the following rule:
If the verb ends in e not preceded by i,o,s,x,z,ch,sh, add s.
So I'm looking for a regex matching a word consisting of some letters, then not i,o,s,x,z,ch,sh, and then "es". I tried this:
\b\w*[^iosxz(sh)(ch)]es\b
According to regex101 it matches "likes", "hates" etc. However, it does not match "bathes", why doesn't it?
You may use
\b(?=\w*(?<![iosxz])(?<![cs]h)es\b)\w*
See the regex demo
Since Python re does not support variable length alternatives in a lookbehind, you need to split the conditions into two lookbehinds here.
Pattern details:
\b - a leading word boundary
(?=\w*(?<![iosxz])(?<![cs]h)es\b) - a positive lookahead requiring a sequence of:
\w* - 0+ word chars
(?<![iosxz]) - there must not be i, o, s, x, z chars right before the current location and...
(?<![cs]h) - no ch or sh right before the current location...
es - followed with es...
\b - at the end of the word
\w* - zero or more (maybe + is better here to match 1 or more) word chars.
See Python demo:
import re
r = re.compile(r'\b(?=\w*(?<![iosxz])(?<![cs]h)es\b)\w*')
s = 'it matches "likes", "hates" etc. However, it does not match "bathes", why doesn\'t it?'
print(re.findall(r, s))
If you want to match strings that end with e and are not preceded by i,o,s,x,z,ch,sh, you should use:
(?<!i|o|s|x|z|ch|sh)e
Your regex [^iosxz(sh)(ch)] consists of character group, the ^ simply negates, and the rest will be exactly matched, so it's equivalent to:
[^io)sxz(c]
which actually means: "match anything that's not one of "io)sxz(c".
I have a working regex that matches ONE of the following lines:
A punctuation from the following list [.,!?;]
A word that is preceded by the beginning of the string or a space.
Here's the regex in question ([.,!?;] *|(?<= |\A)[\-'’:\w]+)
What I need it to do however is for it to match 3 instances of this. So, for example, the ideal end result would be something like this.
Sample text: "This is a test. Test"
Output
"This" "is" "a"
"is" "a" "test"
"a" "test" "."
"test" "." "Test"
I've tried simply adding {3} to the end in the hopes of it matching 3 times. This however results in it matching nothing at all or the occasional odd character. The other possibility I've tried is just repeating the whole regex 3 times like so ([.,!?;] *|(?<= |\A)[\-'’:\w]+)([.,!?;] *|(?<= |\A)[\-'’:\w]+)([.,!?;] *|(?<= |\A)[\-'’:\w]+) which is horrible to look at but I hoped it would work. This had the odd effect of working, but only if at least one of the matches was one of the previously listed punctuation.
Any insights would be appreciated.
I'm using the new regex module found here so that I can have overlapping searches.
What is wrong with your approach
The ([.,!?;] *|(?<= |\A)[\-'’:\w]+) pattern matches a single "unit" (either a word or a single punctuation from the specified set [.,!?;] followed with 0+ spaces. Thus, when you fed this pattern to the regex.findall, it only could return just the chunk list ['This', 'is', 'a', 'test', '. ', 'Test'].
Solution
You can use a slightly different approach: match all words, and all chunks that are not words. Here is a demo (note that C'est and AUX-USB are treated as single "words"):
>>> pat = r"((?:[^\w\s'-]+(?=\s|\b)|\b(?<!')\w+(?:['-]\w+)*))\s*((?1))\s*((?1))"
>>> results = regex.findall(pat, text, overlapped = True)
>>> results
[("C'est", 'un', 'test'), ('un', 'test', '....'), ('test', '....', 'aux-usb')]
Here, the pattern has 3 capture groups, and the second and third one contain the same pattern as in Group 1 ((?1) is a subroutine call used in order to avoid repeating the same pattern used in Group 1). Group 2 and Group 3 can be separated with whitespaces (not necessarily, or the punctuation glued to a word would not be matched). Also, note the negative lookbehind (?<!') that will ensure that C'est is treated as a single entity.
Explanation
The pattern details:
((?:[^\w\s'-]+(?=\s|\b)|\b(?<!')\w+(?:['-]\w+)*)) - Group 1 matching:
(?:[^\w\s'-]+(?=\s|\b) - 1+ characters other than [a-zA-Z0-9_], whitespace, ' and - immediately followed with a whitespace or a word boundary
| - or
\b(?<!')\w+(?:['-]\w+)*) - 1+ word characters not preceded with a ' (due to (?<!')) and preceded with a word boundary (\b) and followed with 0+ sequences of - or ' followed with 1+ word characters.
\s* - 0+ whitespaces
((?1)) - Group 2 (same pattern as for Group 1)
\s*((?1)) - see above