I want to remove all the dots in a text that appear after a vowel character. how can I do that?
Here is the code I wish I had:
string = re.sub('[aeuio]\.', '[aeuio]', string)
Meaning like keep whatever vowel you have matched and remove the '.' next to it.
Capture the vowel and replace with a backreference to it:
import re
s = "Se.hi.mo."
s = re.sub(r'([aeuio])\.', r'\1', s)
print(s) # => Sehimo
See the Python demo and a regex demo.
Here, ([aeuio]) forms a capturing group and \1 in the replacement pattern is a numbered backreference referencing the text captured into Group 1.
Mind the usage of raw string literals where a backslash does not form an escape sequence: r'\1' = '\\1'.
Related
How can I replace a substring between page1/ and _type-A with 222.6 in the below-provided l string?
l = 'https://homepage.com/home/page1/222.6 a_type-A/go'
replace_with = '222.6'
Expected result:
https://homepage.com/home/page1/222.6_type-A/go
I tried:
import re
re.sub('page1/.*?_type-A','',l, flags=re.DOTALL)
But it also removes page1/ and _type-A.
You may use re.sub like this:
import re
l = 'https://homepage.com/home/page1/222.6 a_type-A/go'
replace_with = '222.6'
print (re.sub(r'(?<=page1/).*?(?=_type-A)', replace_with, l))
Output:
https://homepage.com/home/page1/222.6_type-A/go
RegEx Demo
RegEx Breakup:
(?<=page1/): Lookbehind to assert that we have page1/ at previous position
.*?: Match 0 or more of any string (lazy)
(?=_type-A): Lookahead to assert that we have _type-A at next position
You can use
import re
l = 'https://'+'homepage.com/home/page1/222.6 a_type-A/go'
replace_with = '222.6'
print (re.sub('(page1/).*?(_type-A)',fr'\g<1>{replace_with}\2',l, flags=re.DOTALL))
Output: https://homepage.com/home/page1/222.6_type-A/go
See the Python demo online
Note you used an empty string as the replacement argument. In the above snippet, the parts before and after .*? are captured and \g<1> refers to the first group value, and \2 refers to the second group value from the replacement pattern. The unambiguous backreference form (\g<X>) is used to avoid backreference issues since there is a digit right after the backreference.
Since the replacement pattern contains no backslashes, there is no need preprocessing (escaping) anything in it.
This works:
import re
l = 'https://homepage.com/home/page1/222.6 a_type-A/go'
pattern = r"(?<=page1/).*?(?=_type)"
replace_with = '222.6'
s = re.sub(pattern, replace_with, l)
print(s)
The pattern uses the positive lookahead and lookback assertions, ?<= and ?=. A match only occurs if a string is preceded and followed by the assertions in the pattern, but does not consume them. Meaning that re.sub looks for a string with page1/ in front and _type behind it, but only replaces the part in between.
I want to split strings like:
(so) what (are you trying to say)
what (do you mean)
Into lists like:
[(so), what, (are you trying to say)]
[what, (do you mean)]
The code that I tried is below. In the site regexr, the regex expression match the parts that I want but gives a warning, so... I'm not a expert in regex, I don't know what I'm doing wrong.
import re
string = "(so) what (are you trying to say)?"
rx = re.compile(r"((\([\w \w]*\)|[\w]*))")
print(re.split(rx, string ))
Using [\w \w]* is the same as [\w ]* and also matches an empty string.
Instead of using split, you can use re.findall without any capture groups and write the pattern like:
\(\w+(?:[^\S\n]+\w+)*\)|\w+
\( Match (
\w+ Match 1+ word chars
(?:[^\S\n]+\w+)* Optionally repeat matching spaces and 1+ word chars
\) Match )
| Or
\w+ Match 1+ word chars
Regex demo
import re
string = "(so) what (are you trying to say)? what (do you mean)"
rx = re.compile(r"\(\w+(?:[^\S\n]+\w+)*\)|\w+")
print(re.findall(rx, string))
Output
['(so)', 'what', '(are you trying to say)', 'what', '(do you mean)']
For your two examples you can write:
re.split(r'(?<=\)) +| +(?=\()', str)
Python regex<¯\(ツ)/¯>Python code
This does not work, however, for string defined in the OP's code, which contains a question mark, which is contrary to the statement of the question in terms of the two examples.
The regular expression can be broken down as follows.
(?<=\)) # positive lookbehind asserts that location in the
# string is preceded by ')'
[ ]+ # match one or more spaces
| # or
[ ]+ # match one or more spaces
(?=\() # positive lookahead asserts that location in the
# string is followed by '('
In the above I've put each of two space characters in a character class merely to make it visible.
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 working on a Chinese NLP project. I need to remove all punctuation characters except those characters between numbers and remain only Chinese character(\u4e00-\u9fff),alphanumeric characters(0-9a-zA-Z).For example,the
hyphen in 12-34 should be kept while the equal mark after 123 should be removed.
Here is my python script.
import re
s = "中国，中，。》％国foo中¥国bar#中123=国％中国12-34中国"
res = re.sub(u'(?<=[^0-9])[^\u4e00-\u9fff0-9a-zA-Z]+(?=[^0-9])','',s)
print(res)
the expected output should be
中国中国foo中国bar中123国中国12-34中国
but the result is
中国中国foo中国bar中123=国中国12-34中国
I can't figure out why there is an extra equal sign in the output?
Your regex will first check "=" against [^\u4e00-\u9fff0-9a-zA-Z]+. This will succeed. It will then check the lookbehind and lookahead, which must both fail. Ie: If one of them succeeds, the character is kept. This means your code actually keeps any non-alphanumeric, non-Chinese characters which have numbers on any side.
You can try the following regex:
u'([\u4e00-\u9fff0-9a-zA-Z]|(?<=[0-9])[^\u4e00-\u9fff0-9a-zA-Z]+(?=[0-9]))'
You can use it as such:
import re
s = "中国，中，。》％国foo中¥国bar#中123=国％中国12-34中国"
res = re.findall(u'([\u4e00-\u9fff0-9a-zA-Z]|(?<=[0-9])[^\u4e00-\u9fff0-9a-zA-Z]+(?=[0-9]))',s)
print(res.join(''))
I suggest matching and capturing these characters in between digits (to restore them later in the output), and just match them in other contexts.
In Python 2, it will look like
import re
s = u"中国，中，。》％国foo中¥国bar#中123=国％中国12-34中国"
pat_block = u'[^\u4e00-\u9fff0-9a-zA-Z]+';
pattern = u'([0-9]+{0}[0-9]+)|{0}'.format(pat_block)
res = re.sub(pattern, lambda x: x.group(1) if x.group(1) else u"" ,s)
print(res.encode("utf8")) # => 中国中国foo中国bar中123国中国12-34中国
See the Python demo
If you need to preserve those symbols inside any Unicode digits, you need to replace [0-9] with \d and pass the re.UNICODE flag to the regex.
The regex will look like
([0-9]+[^\u4e00-\u9fff0-9a-zA-Z]+[0-9]+)|[^\u4e00-\u9fff0-9a-zA-Z]+
It will works like this:
([0-9]+[^\u4e00-\u9fff0-9a-zA-Z]+[0-9]+) - Group 1 capturing
[0-9]+ - 1+ digits
[^\u4e00-\u9fff0-9a-zA-Z]+ - 1+ chars other than those defined in the specified ranges
[0-9]+ - 1+ digits
| - or
[^\u4e00-\u9fff0-9a-zA-Z]+ - 1+ chars other than those defined in the specified ranges
In Python 2.x, when a group is not matched in re.sub, the backreference to it is None, that is why a lambda expression is required to check if Group 1 matched first.
I have to replace text with text which was found. Smth like this:
regex = u'barbar'
oldstring = u'BarBaR barbarian BarbaRONt'
pattern = re.compile(regex, re.UNICODE | re.DOTALL | re.IGNORECASE)
newstring = pattern.sub(.....)
print(newstring) # And here is what I want to see
>>> u'TEXT1BarBaRTEXT2 TEXT1barbarTEXT2ian TEXT1BarbaRTEXT2ONt'
So I want to receive my original text, where each word that matches 'barbar' (with ignored case) will be surrounded by two words, TEXT1 and TEXT2. Return value must be a unicode string.
How can I realize it? Thanks!
You can use capturing group for that:
regex = u'(barbar)'
...
pattern.sub('TEXT1\\1TEXT2', oldstring)
# => u'TEXT1BarBaRTEXT2 TEXT1barbarTEXT2ian TEXT1BarbaRTEXT2ONt'
Taking barbar into parenthesis makes regexp to capture every part of the string that matches this part of the regexp into a group. As it's the first (and the only one) capturing group you can refer to it as \1 anywhere in the replacement string or in the regexp itself.
For more explanation see (...) and \number sections in the docs.
Btw, if you don't like escaping of the slash before group number you can use raw string instead:
pattern.sub(r'TEXT1\1TEXT2', oldstring)