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
Related
Following code extracts the first sequence of numbers that appear in a string:
num = re.findall(r'^\D*(\d+)', string)
I'd like to add that the regular expression doesn't match numbers preceded by vor V.
Example:
string = 'foobarv2_34 423_wd"
Output: '34'
If you need to get the first match, you need to use re.search, not re.findall.
In this case, you can use a simpler regular expression like (?<!v)\d+ with re.I:
import re
m = re.search(r'(?<!v)\d+', 'foobarv2_34 423_wd', re.I)
if m:
print(m.group()) # => 34
See the Python demo.
Details
(?<!v) - a negative lookbehind that fails the match if there is a v (or V since re.I is used) immediately to the left of the current location
\d+ - one or more digits.
If you cannot use re.search for some reason, you can use
^.*?(?<!v)(\d+)
See this regex demo. Note that \D* (zero or more non-digits) is replaced with .*? that matches zero or more chars other than line break chars as few as possible (with re.S or re.DOTALL, it will also match line breaks) since there is a need to match all digits not preceded with v.
More details:
^ - start of string
.*? - zero or more chars other than line break chars as few as possible
(?<!v) - a negative lookbehind that fails the match if there is a v (or V since re.I is used) immediately to the left of the current location
(\d+) - Group 1: one or more digtis.
I am using python.
The pattern is:
re.compile(r'^(.+?)-?.*?\(.+?\)')
The text like:
text1 = 'TVTP-S2(xxxx123123)'
text2 = 'TVTP(xxxx123123)'
I expect to get TVTP
Another option to match those formats is:
^([^-()]+)(?:-[^()]*)?\([^()]*\)
Explanation
^ Start of string
([^-()]+) Capture group 1, match 1+ times any character other than - ( and )
(?:-[^()]*)? As the - is excluded from the first part, optionally match - followed by any char other than ( and )
\([^()]*\) Match from ( till ) without matching any parenthesis between them
Regex demo | Python demo
Example
import re
regex = r"^([^-()]+)(?:-[^()]*)?\([^()]*\)"
s = ("TVTP-S2(xxxx123123)\n"
"TVTP(xxxx123123)\n")
print(re.findall(regex, s, re.MULTILINE))
Output
['TVTP', 'TVTP']
This regex works:
pattern = r'^([^-]+).*\(.+?\)'
>>> re.findall(pattern, 'TVTP-S2(xxxx123123)')
['TVTP']
>>> re.findall(pattern, 'TVTP(xxxx123123)')
['TVTP']
a quick answer will be
^(\w+)(-.*?)?\((.*?)\)$
https://regex101.com/r/wL4jKe/2/
It is because the first plus is lazy, and the subsequent dash is optional, followed by a pattern that allows any character.
This allows the regex engine to choose the single letter T for the first group (because it is lazy), choose to interpret the dash as just not being there, which is allowed because it is followed by a question mark, and then have the next .* match "VTP-S2".
You can just grab non-dashes to capture, followed by nonparentheses up to the parentheses.
p=re.compile(r'^([^-]*?)[^(]*\(.+?\)')
p.search('TVTP-S2(xxxx123123) blah()').group(1)
The nonparentheses part prevents the second portion from matching 'S2(xxxx123123) blah(' in my modified example above.
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 trying to take off bracketed ends of strings such as version = 10.9.8[35]. I am trying to substitute the integer within brackets pattern
(so all of [35], including brackets) with an empty string using the regex [\[+0-9*\]+] but this also matches with numbers not surrounded by brackets. Am I not using the + quantifier properly?
You could match the format of the number and then match one or more digits between square brackets.
In the replacement using the first capturing group r'\1'
\b([0-9]+(?:\.[0-9]+)+)\[[0-9]+\]
\b Word boundary
( Capture group 1
[0-9]+ Match 1+ digits
(?:\.[0-9]+)+ Match a . and 1+ digits and repeat that 1 or more times
) Close group
\[[0-9]+\] Match 1+ digits between square brackets
Regex demo
For example
import re
regex = r"\b([0-9]+(?:\.[0-9]+)+)\[[0-9]+\]"
test_str = "version = 10.9.8[35]"
result = re.sub(regex, r'\1', test_str)
print (result)
Output
version = 10.9.8
No need for regex
s = '10.9.8[35]'
t = s[:s.rfind("[")]
print(t)
But if you insist ;-)
import re
s = '10.9.8[35]'
t = re.sub(r"^(.*?)[[]\d+[]]$", r"\1", s)
print(t)
Breakdown of regex:
^ - begins with
() - Capture Group 1 you want to keep
.*? - Any number of chars (non-greedy)
[[] - an opening [
\d+ 1+ digit
[]] - closing ]
$ - ends with
\1 - capture group 1 - used in replace part of regex replace. The bit you want to keep.
Output in both cases:
10.9.8
Use regex101.com to familiarise yourself more. If you click on any of the regex samples at bottom right of the website, it will give you more info. You can also use it to generate regex code in a variety of languages too. (not good for Java though!).
There's also a great series of Python regex videos on Youtube by PyMoondra.
A simpler regex solution:
import re
pattern = re.compile(r'\[\d+\]$')
s = '10.9.8[35]'
r = pattern.sub('', s)
print(r) # 10.9.8
The pattern matches square brackets at the end of a string with one or more number inside. The sub then replaces the square brackets and number with an empty string.
If you wanted to use the number in the square brackets just change the sub expression such as:
import re
pattern = re.compile(r'\[(\d+)\]$')
s = '10.9.8[35]'
r = pattern.sub(r'.\1', s)
print(r) # 10.9.8.35
Alternatively as said by the other answer you can just find it and splice to get rid of it.
How does one replace a pattern when the substitution itself is a variable?
I have the following string:
s = '''[[merit|merited]] and [[eat|eaten]] and [[go]]'''
I would like to retain only the right-most word in the brackets ('merited', 'eaten', 'go'), stripping away what surrounds these words, thus producing:
merited and eaten and go
I have the regex:
p = '''\[\[[a-zA-Z]*\[|]*([a-zA-Z]*)\]\]'''
...which produces:
>>> re.findall(p, s)
['merited', 'eaten', 'go']
However, as this varies, I don't see a way to use re.sub() or s.replace().
s = '''[[merit|merited]] and [[eat|eaten]] and [[go]]'''
p = '''\[\[[a-zA-Z]*?[|]*([a-zA-Z]*)\]\]'''
re.sub(p, r'\1', s)
? so that for [[go]] first [a-zA-Z]* will match empty (shortest) string and second will get actual go string
\1 substitutes first (in this case the only) match group in a pattern for each non-overlapping match in the string s. r'\1' is used so that \1 is not interpreted as the character with code 0x1
well first you need to fix your regex to capture the whole group:
>>> s = '[[merit|merited]] and [[eat|eaten]] and [[go]]'
>>> p = '(\[\[(?:[a-zA-Z]*\|)*([a-zA-Z]*)\]\])'
>>> [('[[merit|merited]]', 'merited'), ('[[eat|eaten]]', 'eaten'), ('[[go]]', 'go')]
[('[[merit|merited]]', 'merited'), ('[[eat|eaten]]', 'eaten'), ('[[go]]', 'go')]
This matches the whole [[whateverisinhere]] and separates the whole match as group 1 and just the final word as group 2. You can than use \2 token to replace the whole match with just group 2:
>>> re.sub(p,r'\2',s)
'merited and eaten and go'
or change your pattern to:
p = '\[\[(?:[a-zA-Z]*\|)*([a-zA-Z]*)\]\]'
which gets rid of grouping the entire match as group 1 and only groups what you want. you can then do:
>>> re.sub(p,r'\1',s)
to have the same effect.
POST EDIT:
I forgot to mention that I actually changed your regex so here is the explanation:
\[\[(?:[a-zA-Z]*\|)*([a-zA-Z]*)\]\]
\[\[ \]\] #literal matches of brackets
(?: )* #non-capturing group that can match 0 or more of whats inside
[a-zA-Z]*\| #matches any word that is followed by a '|' character
( ... ) #captures into group one the final word
I feel like this is stronger than what you originally had because it will also change if there are more than 2 options:
>>> s = '[[merit|merited]] and [[ate|eat|eaten]] and [[go]]'
>>> p = '\[\[(?:[a-zA-Z]*\|)*([a-zA-Z]*)\]\]'
>>> re.sub(p,r'\1',s)
'merited and eaten and go'