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]+)
Related
I want to capture the digits that follow a certain phrase and also the start and end index of the number of interest.
Here is an example:
text = The special code is 034567 in this particular case and not 98675
In this example, I am interested in capturing the number 034657 which comes after the phrase special code and also the start and end index of the the number 034657.
My code is:
p = re.compile('special code \s\w.\s (\d+)')
re.search(p, text)
But this does not match anything. Could you explain why and how I should correct it?
Your expression matches a space and any whitespace with \s pattern, then \w. matches any word char and any character other than a line break char, and then again \s requires two whitespaces, any whitespace and a space.
You may simply match any 1+ whitespaces using \s+ between words, and to match any chunk of non-whitespaces, instead of \w., you may use \S+.
Use
import re
text = 'The special code is 034567 in this particular case and not 98675'
p = re.compile(r'special code\s+\S+\s+(\d+)')
m = p.search(text)
if m:
print(m.group(1)) # 034567
print(m.span(1)) # (20, 26)
See the Python demo and the regex demo.
Use re.findall with a capture group:
text = "The special code is 034567 in this particular case and not 98675"
matches = re.findall(r'\bspecial code (?:\S+\s+)?(\d+)', text)
print(matches)
This prints:
['034567']
I am new with regex and I would like some help. So I have a string below and I want to make my regex match the first character of the acronym literally + any character[a-z] unlimited times but only for the first character. For the rest of the characters, i would like to just match them as they are. Any help on what to change on my regex line to achieve this, would be highly appreciated.
import re
s = 'nUSA stands for northern USA'
x = (f'({"nUSA"}).+?({" ".join( t[0] + "[a-z]" + t[1:] for t in "nUSA")})(?: )')
print(x)
out: (nUSA).+?(n[a-z]+ U[a-z]+ S[a-z]+ A[a-z]+)(?: )
What i want to achieve with my regex line is something like the pattern below so that it can match for the northern USA.
(nUSA).+?(n[a-z]+ U + S + A)(?: )
instead of the one i get
(nUSA).+?(n[a-z]+ U[a-z]+ S[a-z]+ A[a-z]+)(?: )
I would like it to work for any arbitrary text, not only for the specific one. I am not sure if i have expressed my problem properly.
You may use
import re
s = 'nUSA stands for northern USA'
key='nUSA'
x = rf'\b({key})\b.+?\b({key[0]}[a-z]*\s*{key[1:]})(?!\S)'
# => print(x) => \b(nUSA)\b.+?\b(n[a-z]*\s*USA)(?!\S)
# Or, if the key can contain special chars at the end:
# x = rf'\b({re.escape(key)})(?!\w).+?(?<!\w)({re.escape(key[0])}[a-z]*\s*{re.escape(key[1:])})(?!\S)'
print(re.findall(x, s))
# => [('nUSA', 'northern USA')]
See the Python demo. The resulting regex will look like \b(nUSA)\b.+?\b(n[a-z]*\s*USA)(?!\S), see its demo. Details:
\b - word boundary
(nUSA) - Group 1 capturing the key word
\b / (?!\w) - word boundary (right-hand word boundary)
.+? - any 1+ chars other than linebreak chars as few as possible
\b - word boundary
(n[a-z]*\s*USA) - Group 2: n (first char), then any 0+ lowercase ASCII letters, 0+ whitespaces and the rest of the key string.
(?!\S) - a right-hand whitespace boundary (you may consider using (?!\w) again here).
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.
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 list of words such as:
l = """abca
bcab
aaba
cccc
cbac
babb
"""
I want to find the words that have the same first and last character, and that the two middle characters are different from the first/last character.
The desired final result:
['abca', 'bcab', 'cbac']
I tried this:
re.findall('^(.)..\\1$', l, re.MULTILINE)
But it returns all of the unwanted words as well.
I thought of using [^...] somehow, but I couldn't figure it out.
There's a way of doing this with sets (to filter the results from the search above), but I'm looking for a regex.
Is it possible?
Edit: fixed to use negative lookahead assertions instead of negative lookbehind assertions. Read comments for #AlanMoore and #bukzor explanations.
>>> [s for s in l.splitlines() if re.search(r'^(.)(?!\1).(?!\1).\1$', s)]
['abca', 'bcab', 'cbac']
The solution uses negative lookahead assertions which means 'match the current position only if it isn't followed by a match for something else.' Now, take a look at the lookahead assertion - (?!\1). All this means is 'match the current character only if it isn't followed by the first character.'
There are lots of ways to do this. Here's probably the simplest:
re.findall(r'''
\b #The beginning of a word (a word boundary)
([a-z]) #One letter
(?!\w*\1\B) #The rest of this word may not contain the starting letter except at the end of the word
[a-z]* #Any number of other letters
\1 #The starting letter we captured in step 2
\b #The end of the word (another word boundary)
''', l, re.IGNORECASE | re.VERBOSE)
If you want, you can loosen the requirements a bit by replacing [a-z] with \w. That will allow numbers and underscores as well as letters. You can also restrict it to 4-character words by changing the last * in the pattern to {2}.
Note also that I'm not very familiar with Python, so I'm assuming your usage of findall is correct.
Are you required to use regexes? This is a much more pythonic way to do the same thing:
l = """abca
bcab
aaba
cccc
cbac
babb
"""
for word in l.split():
if word[-1] == word[0] and word[0] not in word[1:-1]:
print word
Here's how I would do it:
result = re.findall(r"\b([a-z])(?:(?!\1)[a-z]){2}\1\b", subject)
This is similar to Justin's answer, except where that one does a one-time lookahead, this one checks each letter as it's consumed.
\b
([a-z]) # Capture the first letter.
(?:
(?!\1) # Unless it's the same as the first letter...
[a-z] # ...consume another letter.
){2}
\1
\b
I don't know what your real data looks like, so chose [a-z] arbitrarily because it works with your sample data. I limited the length to four characters for the same reason. As with Justin's answer, you may want to change the {2} to *, + or some other quantifier.
To heck with regexes.
[
word
for word in words.split('\n')
if word[0] == word[-1]
and word[0] not in word[1:-1]
]
You can do this with negative lookahead or lookbehind assertions; see http://docs.python.org/library/re.html for details.
Not a Python guru, but maybe this
re.findall('^(.)(?:(?!\1).)*\1$', l, re.MULTILINE)
expanded (use multi-line modifier):
^ # begin of line
(.) # capture grp 1, any char except newline
(?: # grouping
(?!\1) # Lookahead assertion, not what was in capture group 1 (backref to 1)
. # this is ok, grab any char except newline
)* # end grouping, do 0 or more times (could force length with {2} instead of *)
\1 # backref to group 1, this character must be the same
$ # end of line