Question

I want to match words that contain special characters or that begin with ‘http://’

So this sentence

%he#llo, my website is: http://www.url.com/abcdef123

should turn into this

my website

So far, i have this

re.sub(r"^[^\w]", " ", "%he#llo, my website is: http://www.url.com/abcdef123")

This just removes the symbols, but it doesn’t remove the words associated with the symbol (it also doesn’t remove ‘:’ and ‘,’), nor does it remove the URL.

Answer 1

For the example string you give, the following regular expression works OK:

>>> a = '%he#llo, my website is: http://www.url.com/abcdef123'
>>> re.findall('(http://\S+|\S*[^\w\s]\S*)',a)
['%he#llo,', 'is:', 'http://www.url.com/abcdef123']

… or you can remove those words with re.sub

>>> re.sub('(http://\S+|\S*[^\w\s]\S*)','',a)
' my website  '

The | means alternation and will match the expression on either side within the group. The part on the left matches http:// followed by one or more non-space characters. The part on the right matches zero or more non-space characters, followed by anything that isn’t a word or space character, followed by zero or more non-space characters — that ensures that you have a string with at least one non-word character and no spaces.

Updated: Of course, as the other answers implicitly suggest, since the http:// prefix contains a non-word character (/) you don’t need to have that as an alternative – you could simplify the regular expression to \S*[^\w\s]\S*. However, perhaps the example above with alternation is still useful.