I am trying to break a line on all non-word patterns except .(dot)
Usually I guess it can be done as [\W ^[.]] in java, but how to I do in python?
>>> import re
>>> the_string="http://hello-world.com"
>>> re.findall(r'[\w.]+',the_string)
['http', 'hello', 'world.com']
A very good reference for Python's regular expression module is available here. Following should do the trick for you.
import re
re.split(r'[\w.]+', text_string)
Or,
import re
re.findall('[^\w.]+', text_string)
Your Java syntax is off, to begin with. This is what you were trying for:
[\W&&[^.]]
That matches a character from the intersection of the sets described by "any non-word character" and "any character except ." But that's overkill when you can just use:
[^\w.]
...or, "any character that's not a word character or .". It's the same in Python (and in most other flavors, too), though you probably want to match one or more of the characters:
re.split(r'[^\w.]+', the_string)
But it's probably simpler to use #gnibbler's approach of matching the parts that you want to keep, not the ones you want to throw away:
re.findall(r'[\w.]+', the_string)
I'm assuming that you want to split a string on all non-word patterns except a dot.
Edit: Python doesn't support the Java-style regex syntax that you are using. I'd suggest first replacing all dots with a long string, then splitting the string, then putting the dots back in.
import re
long_str = "ABCDEFGH"
str = str.replace('.', long_str)
result = re.split(r'\W', str)
Then as you are using result, replace all the long_str sequences with a dot again.
This is a very bad solution, but it works.
Python has a convenience function for that
>>> s = "ab.cd.ef.gh"
>>> s.split(".")
['ab', 'cd', 'ef', 'gh']
Related
I was trying to write regex in Python to replace all non-ascii with an underscore, but if one of the characters is "S̄" (an 'S' with a line on the top), it adds an extra 'S'... Is there a way to account for this character as well? I believe it's a valid utf-8 character, but not ascii
Here's there code:
import re
line = "ra*ndom wordS̄"
print(re.sub('[\W]', '_', line))
I would expect it to output:
ra_ndom_word_
But instead I get:
ra_ndom_wordS__
The reason Python works this way is that you are actually looking at two distinct characters; there's an S and then it's followed by a combining macron U+0304
In the general case, if you want to replace a sequence of combining characters and the base character with an underscore, try e.g.
import unicodedata
def cleanup(line):
cleaned = []
strip = False
for char in line:
if unicodedata.combining(char):
strip = True
continue
if strip:
cleaned.pop()
strip = False
if unicodedata.category(char) not in ("Ll", "Lu"):
char = "_"
cleaned.append(char)
return ''.join(cleaned)
By the by, \W does not need square brackets around it; it's already a regex character class.
Python's re module lacks support for important Unicode properties, though if you really want to use specifically a regex for this, the third-party regex library has proper support for Unicode categories.
"Ll" is lowercase alphabetics and "Lu" are uppercase. There are other Unicode L categories so maybe tweak this to suit your requirements (unicodedata.category(char).startswith("L") maybe?); see also https://www.fileformat.info/info/unicode/category/index.htm
You can use the following script to get the desired output:
import re
line="ra*ndom wordS̄"
print(re.sub('[^[-~]+]*','_',line))
Output
ra_ndom_word_
In this approach, it works with other non-ascii characters as well :
import re
line="ra*ndom ¡¢£Ä wordS̄. another non-ascii: Ä and Ï"
print(re.sub('[^[-~]+]*','_',line))
output:
ra_ndom_word_another_non_ascii_and_
I have a string = "msdjdgf(^&%*(Aroha Technologies&^$^&*^CHJdjg" with special characters.
what i am trying is to remove all special charecters in the string and then display the word 'Aroha Technologies'
i was able to do with hard coding using lstrip() function but can anyone help me out how can i display string 'Aroha Technologies' in a single line using regular expressions.
edit suggested:-
by using this lstrip() and rstrip() functions i was able to remove characters from the string.
str = "msdjdgf(^&%*(Aroha Technologies&^$^&*^CHJdjg"
str=str.lstrip('msdjdgf(^&%*(')
str=str.rstrip('&^$^&*^CHJdjg')
here, A bit more dirty approach
import re # A module in python for String matching/operations
a = "msdjdgf(^&%*(Aroha Technologies&^$^&*^CHJdjg"
stuff = re.findall('\W(\w+\s\w+)\W', a)
print(stuff[0]) # Aroha Technologies
hope this helps ;)
You don't provide a lot of information, so this may or may not be close to what you want:
import re
origstr = "msdjdgf(^&%(Aroha Technologies&^$^&^CHJdjg"
match = re.search("[A-Z][a-z]*(?: [A-Z][a-z]*)*", origstr)
if match:
newstr = match.group()
(looks for a series of capitalized words with spaces between them)
I have a string in python and I want to replace multiple consecutive repeating character into 1.
For example:
st = "UUUURRGGGEENNTTT"
print(st.replace(r'(\w){2,}',r'\1'))
But this command doesn't seems to be working, please can anybody help in finding what's wrong with this command?
There is one more way to solve this but wanted to understand why the above command fails and is there any way to correct it:
print(re.sub(r"([a-z])\1+",r"\1",st)) -- print URGENT
you need to use regex.
so you can do this:
import re
re.sub(r'[^\w\s]|(.)(?=\1)', '', 'UUURRRUU')
the result is UR.
this is a snapshot of what I have got:
for this regex: (.)(?=.*\1)
(.) means: match any char except new lines (line breaks)
?=. means: lookahead every char except new line (.)
* means: match a preceding token
\1 means: to mach the result of captured group, which is the U or R ...
then replace all matches with ''
also you can check this:
lookahead
also check this tool I solve my regex using it,
it describe everything and you can learn a lot from it:
regexer
The reason for why your code does not work is because str.replace does not support regex, you can only replace a substring with another string. You will need to use the re module if you want to replace by matching a regex pattern.
Secondly, your regex pattern is also incorrect, (\w){2,} will match any characters that occurs 2 or more times (doesn’t have to be the same character though), so it will not work. You will need to do something like this:
import re
st = "UUUURRGGGEENNTTT"
print(re.sub(r'(\w)\1+',r'\1', st)))
# URGENT
Now this will only match the same character 2 or more times.
An alternative, “unique” solution to this is that you can use the unique_justseen recipe that itertools provides:
from itertools import groupby
from operator import itemgetter
st = "UUUURRGGGEENNTTT"
new ="".join(map(next, map(itemgetter(1), groupby(st))))
print(new)
# URGENT
string.replace(s, old, new[, maxreplace]) only does substring replacement:
>>> '(\w){2,}'.replace(r'(\w){2,}',r'\1')
'\\1'
That's why it fails and it can't work with regex expression so no way to correct the first command.
I met some problems when I'd like to check if the input regex if correct or not.
I'd like to check is there one backslash before every symbol, but I don't know how to implement using Python.
For example:
number: 123456789. (return False)
phone\:111111 (return True)
I try to use (?!) and (?=) in Python, but it doesn't work.
Update:
I'd like to match the following string:
\~, \!, \#, \$, \%, \^, \&, \*, \(, \), \{, \}, \[, \], \:, \;, \", \', \>, \<, \?
Thank you very much.
import re
if re.seach(r"\\\W", "phone\:111111") is not None:
print("OK")
Does it work?
Reading between the lines a bit, it sounds like you are trying to pass a string to a regex and you want to make sure it has no special characters in it that are unescaped.
Python's re module has an inbuilt re.escape() function for this.
Example:
>>> import re
>>> print(re.escape("phone:111111"))
"phone\\:111111"
Check that the entire string is composed of single characters or pairs of backslash+symbol:
import re
def has_backslash_before_every_symbol(s):
return re.match(r"^(\\[~!#$%^&*(){}\[\]:;"'><?]|[^~!#$%^&*(){}\[\]:;"'><?])*$", s) is not None
Python regex reference: https://docs.python.org/3/library/re.html
I have this string:
'Is?"they'
I want to find the question mark (?) in the string, and put it at the end of the string. The output should look like this:
'Is"they?'
I am using the following regular expression in python 2.7. I don't know why my regex is not working.
import re
regs = re.sub('(\w*)(\?)(\w*)', '\\1\\3\\2', 'Is?"they')
print regs
Is?"they # this is the output of my regex.
Your regex doesn't match because " is not in the \w character class. You would need to change it to something like:
regs = re.sub('(\w*)(\?)([^"\w]*)', '\\1\\3\\2', 'Is?"they')
As shown here, " is not captured by \w. Hence, it would probably be best to just use a .:
>>> import re
>>> re.sub("(.*)(\?)(.*)", r'\1\3\2', 'Is?"they')
'Is"they?'
>>>
. captures anything/everything in Regex (except newlines).
Also, you'll notice that I used a raw-string for the second argument of re.sub. Doing so is cleaner than having all those backslashes.