There already is an approaching answer in R gsub("[^[:alnum:]['-]", " ", my_string), but it does not work in Python:
my_string = 'compactified on a calabi-yau threefold # ,.'
re.sub("[^[:alnum:]['-]", " ", my_string)
gives 'compactified on a calab yau threefold # ,.'
So not only does it remove the intra-word dash, it also removes the last letter of the word preceding the dash. And it does not remove punctuation
Expected result (string without any punctuation but intra-word dash): 'compactified on a calabi-yau threefold'
R uses TRE (POSIX) or PCRE regex engine depending on the perl option (or function used). Python uses a modified, much poorer Perl-like version as re library. Python does not support POSIX character classes, as [:alnum:] that matches alpha (letters) and num (digits).
In Python, [:alnum:] can be replaced with [^\W_] (or ASCII only [a-zA-Z0-9]) and the negated [^[:alnum:]] - with [\W_] ([^a-zA-Z0-9] ASCII only version).
The [^[:alnum:]['-] matches any 1 symbol other than alphanumeric (letter or digit), [, ', or -. That means the R question you refer to does not provide a correct answer.
You can use the following solution:
import re
p = re.compile(r"(\b[-']\b)|[\W_]")
test_str = "No - d'Ante compactified on a calabi-yau threefold # ,."
result = p.sub(lambda m: (m.group(1) if m.group(1) else " "), test_str)
print(result)
The (\b[-']\b)|[\W_] regex matches and captures intraword - and ' and we restore them in the re.sub by checking if the capture group matched and re-inserting it with m.group(1), and the rest (all non-word characters and underscores) are just replaced with a space.
If you want to remove sequences of non-word characters with one space, use
p = re.compile(r"(\b[-']\b)|[\W_]+")
Related
I am cleaning a text and I would like to remove all the hyphens and special characters. Except for the hyphens between two words such as: tic-tacs, popcorn-flavoured.
I wrote the below regex but it removes every hyphen.
text='popcorn-flavoured---'
new_text=re.sub(r'[^a-zA-Z0-9]+', '',text)
new_text
I would like the output to be:
popcorn-flavoured
You can replace matches of the regular expression
-(?!\w)|(?<!\w)-
with empty strings.
Regex demo <¯\_(ツ)_/¯> Python demo
The regex will match hyphens that are not both preceded and followed by a word character.
Python's regex engine performs the following operations.
- match '-'
(?!\w) the previous character is not a word character
|
(?<!\w) the following character is not a word character
- match '-'
(?!\w) is a negative lookahead; (?<!\w) is a negative lookbehind.
As an alternative, you could capture a hyphen between word characters and keep that group in the replacement. Using an alternation, you could match the hyphens that you want to remove.
(\w+-\w+)|-+
Explanation
(\w+-\w+) Capture group 1, match 1+ word chars, hyphen and 1+ word chars
| Or
-+ Match 1+ times a hyphen
Regex demo | Python demo
Example code
import re
regex = r"(\w+-\w+)|-+"
test_str = ("popcorn-flavoured---\n"
"tic-tacs")
result = re.sub(regex, r"\1", test_str)
print (result)
Output
popcorn-flavoured
tic-tacs
You can use findall() to get that part that matches your criteria.
new_text = re.findall('[\w]+[-]?[\w]+', text)[0]
Play around with it with other inputs.
You can use
p = re.compile(r"(\b[-]\b)|[-]")
result = p.sub(lambda m: (m.group(1) if m.group(1) else ""), text)
Test
With:
text='popcorn-flavoured---'
Output (result):
popcorn-flavoured
Explanation
This pattern detects hyphens between two words:
(\b[-]\b)
This pattern detects all hyphens
[-]
Regex substitution
p.sub(lambda m: (m.group(1) if m.group(1) else " "), text)
When hyphen detected between two words m.group(1) exists, so we maintain things as they are
else "")
Occurs when the pattern was triggered by [-] then we substitute a "" for the hyphen removing it.
I have a decent familiarity with regex but this is tricky. I need to find instances like this from a SQL case statement:
when col_name = 'this can be a word or sentence'
I can match the above when it's just one word, but when it's more than one word it's not working.
s = """when col_name = 'a sentence of words'"""
x = re.search("when\s(\w+)\s*=\s*\'(\w+)", s)
if x:
print(x.group(1)) # this returns "col_name"
print(x.group(2)) # this returns "a"
I want group(2) to return "a sentence of words" but I'm just getting the first word. That part could either be one word or several. How to do it?
When I add in the second \', then I get no match:
x = re.search("when\s(\w+)\s*=\s*\'(\w+)\'", s)
You may match all characters other than single quotation mark rather than matching letters, digits and connector punctuation ("word" chars) with the Group 2 pattern:
import re
s = """when col_name = 'a sentence of words'"""
x = re.search(r"when\s+(\w+)\s*=\s*'([^']+)", s)
if x:
print(x.group(1)) # this returns "col_name"
print(x.group(2)) # this returns "a sentence of words"
See the Python demo
The [^'] is a negated character class that matches any char but a single quotation mark, see the regex demo.
In case the string can contain escaped single quotes, you may consider replacing [^'] with
If the escape char is ': ([^']*(?:''[^']*)*)
If the escape char is \: ([^\\']*(?:\\.[^'\\]*)*).
Note the use of the raw string literal to define the regex pattern (all backslashes are treated as literal backslashes inside it).
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 this code for removing all punctuation from a regex string:
import regex as re
re.sub(ur"\p{P}+", "", txt)
How would I change it to allow hyphens? If you could explain how you did it, that would be great. I understand that here, correct me if I'm wrong, P with anything after it is punctuation.
[^\P{P}-]+
\P is the complementary of \p - not punctuation. So this matches anything that is not (not punctuation or a dash) - resulting in all punctuation except dashes.
Example: http://www.rubular.com/r/JsdNM3nFJ3
If you want a non-convoluted way, an alternative is \p{P}(?<!-): match all punctuation, and then check it wasn't a dash (using negative lookbehind).
Working example: http://www.rubular.com/r/5G62iSYTdk
Here's how to do it with the re module, in case you have to stick with the standard libraries:
# works in python 2 and 3
import re
import string
remove = string.punctuation
remove = remove.replace("-", "") # don't remove hyphens
pattern = r"[{}]".format(remove) # create the pattern
txt = ")*^%{}[]thi's - is - ###!a !%%!!%- test."
re.sub(pattern, "", txt)
# >>> 'this - is - a - test'
If performance matters, you may want to use str.translate, since it's faster than using a regex. In Python 3, the code is txt.translate({ord(char): None for char in remove}).
You could either specify the punctuation you want to remove manually, as in [._,] or supply a function instead of the replacement string:
re.sub(r"\p{P}", lambda m: "-" if m.group(0) == "-" else "", text)