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).
Related
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 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'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".
From a string such as
70849 mozilla/5.0(linux;u;android4.2.1;zh-cn)applewebkit/534.30(khtml,likegecko)version/4.0mobilesafari/534.30
I want to get the first parenthesized content linux;u;android4.2.1;zh-cn.
My code looks like this:
s=r'70849 mozilla/5.0(linux;u;android4.2.1;zh-cn)applewebkit/534.30(khtml,likegecko)version/4.0mobilesafari/534.30'
re.search("(\d+)\s.+\((\S+)\)", s).group(2)
but the result is the last brackets' contents khtml,likegecko.
How to solve this?
The main issue you have is the greedy dot matching .+ pattern. It grabs the whole string you have, and then backtracks, yielding one character from the right at a time, trying to accommodate for the subsequent patterns. Thus, it matches the last parentheses.
You can use
^(\d+)\s[^(]+\(([^()]+)\)
See the regex demo. Here, the [^(]+ restricts the matching to the characters other than ( (so, it cannot grab the whole line up to the end) and get to the first pair of parentheses.
Pattern expalantion:
^ - string start (NOTE: If the number appears not at the start of the string, remove this ^ anchor)
(\d+) - Group 1: 1 or more digits
\s - a whitespace (if it is not a required character, it can be removed since the subsequent negated character class will match the space)
[^(]+ - 1+ characters other than (
\( - a literal (
([^()]+) - Group 2 matching 1+ characters other than ( and )
\)- closing ).
Debuggex Demo
Here is the IDEONE demo:
import re
p = re.compile(r'^(\d+)\s[^(]+\(([^()]+)\)')
test_str = "70849 mozilla/5.0(linux;u;android4.2.1;zh-cn)applewebkit/534.30(khtml,likegecko)version/4.0mobilesafari/534.30"
print(p.findall(test_str))
# or using re.search if the number is not at the beginning of the string
m = re.search(r'(\d+)\s[^(]+\(([^()]+)\)', test_str)
if m:
print("Number: {0}\nString: {1}".format(m.group(1), m.group(2)))
# [('70849', 'linux;u;android4.2.1;zh-cn')]
# Number: 70849
# String: linux;u;android4.2.1;zh-cn
You can use a negated class \(([^)]*)\) to match anything between ( and ):
>>> s=r'70849 mozilla/5.0(linux;u;android4.2.1;zh-cn)applewebkit/534.30(khtml,likegecko)version/4.0mobilesafari/534.30'
>>> m = re.search(r"(\d+)[^(]*\(([^)]*)\)", s)
>>> print m.group(1)
70849
>>> print m.group(2)
linux;u;android4.2.1;zh-cn
myreg = r"\babcb\"
mystr = "sdf ddabc"
mystr1 = "sdf abc"
print(re.findall(myreg,mystr))=[]
print(re.findall(myreg,mystr1))=[abc]
Until now everything works as expected but if i change my reg and my str to.
myreg = r"\b\+abcb\"
mystr = "sdf +abc"
print(re.findall(myreg,mystr)) = [] but i would like to get [+abc]
I have noticed that using the following works as expected.
myreg = "^\\+abc$"
mystr = "+abc"
mystr1 = "-+abc"
My question: Is it possible to achieve the same results as above without splitting the string?
Best regards,
Gabriel
Your problem is the following:
\b is defined as the boundary between a \w and a \W character
(or vice versa).
\w contains the character set [a-zA-Z0-9_]
\W contains the character set [^a-zA-Z0-9_], which means all characters except [a-zA-Z0-9_]
'+' is not contained in \w so you won't match the boundary between the whitespace and the '+'.
To get what you want, you should remove the first \b from your pattern:
import re
string = "sdf +abc"
pattern = r"\+abc\b"
matches = re.findall(pattern, string)
print matches
['+abc']
There are two problems
Before your + in +abc, there is no word boundary, so \b cannot match.
Your regex \b\+abcb\ tries to match a literal b character after abc (typo).
Word Boundaries
The word boundary \b matches at a position between a word character (letters, digits and underscore) and a non-word character (or a line beginning or ending). For instance, there is a word boundary between the + and the a
Solution: Make your Own boundary
If you want to match +abc but only when it is not preceded by a word character (for instance, you don't want it inside def+abc), then you can make your own boundary with a lookbehind:
(?<!\w)\+abc
This says "match +abc if it is not preceded by a word character (letter, digit, underscore)".