How to match a new line character in Python raw string

Question

I got a little confused about Python raw string. I know that if we use raw string, then it will treat \'\\\' as a normal backslash (ex. r\'\\n\' wo

Answer 1

def clean_with_puncutation(text):    
    from string import punctuation
    import re
    punctuation_token={p:'' for p in punctuation}
    punctuation_token['
']=""
    punctuation_token['\n']=""
    punctuation_token['']=''
    punctuation_token['']=''
  #punctuation_token



    regex = r"(
)|()|()|[\n\!\@\#\$\%\^\&\*\(\)\[\]\
           {\}\;\:\,\.\/\?\|\`\_\\+\\\=\~\-\<\>]"

###Always put new sequence token at front to avoid overlapping results
 #text = '!@#$%^&*()[]{};:,./<>?\|`~-= _+\
\n \ '
    text_=""

    matches = re.finditer(regex, text)

    index=0

    for match in matches:
     #print(match.group())
     #print(punctuation_token[match.group()])
     #print ("Match at index: %s, %s" % (match.start(), match.end()))
        text_=text_+ text[index:match.start()] +" " 
              +punctuation_token[match.group()]+ " "
        index=match.end()
    return text_