How to match a newline character in a 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' would be \ and n). However, I was wondering what if I want to match a new line character in raw string. I tried r'\\n', but it didn't work.

Anybody has some good idea about this?

Answer 1

The simplest answer is to simply not use a raw string. You can escape backslashes by using \\.

If you have huge numbers of backslashes in some segments, then you could concatenate raw strings and normal strings as needed:

r"some string \ with \ backslashes" "\n"

(Python automatically concatenates string literals with only whitespace between them.)

Remember if you are working with paths on Windows, the easiest option is to just use forward slashes - it will still work fine.

Answer 2

you also can use [\r\n] for matching to new line

Answer 3

def clean_with_puncutation(text):    
    from string import punctuation
    import re
    punctuation_token={p:'<PUNC_'+p+'>' for p in punctuation}
    punctuation_token['<br/>']="<TOKEN_BL>"
    punctuation_token['\n']="<TOKEN_NL>"
    punctuation_token['<EOF>']='<TOKEN_EOF>'
    punctuation_token['<SOF>']='<TOKEN_SOF>'
  #punctuation_token



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

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