One way is to use re.sub, that's my preferred way.

import re
my_str = "hey th~!ere"
my_new_string = re.sub('[^a-zA-Z0-9 \n\.]', '', my_str)
print my_new_string

Output:

hey there

Another way is to use re.escape:

import string
import re

my_str = "hey th~!ere"

chars = re.escape(string.punctuation)
print re.sub('['+chars+']', '',my_str)

Output:

hey there

Just a small tip about parameters style in python by PEP-8 parameters should be remove_special_chars and not removeSpecialChars

Also if you want to keep the spaces just change [^a-zA-Z0-9 \n\.] to [^a-zA-Z0-9\n\.]

You could use re.escape to escape all the special regex characters in the string, and then enclose the escaped string into [...] so it matches any of them.

>>> re.sub("[%s]" % re.escape('!~@#$%^&*()-+={}[]:;<.>?/\''), '_', table)
'123____________|___"_______\\__,__'

However, as you are not really using that regex as a regex, you might instead just check whether each character is in that string:

>>>''.join("_" if c in '!~@#$%^&*()-+={}[]:;<.>?/\'' else c for c in table)
'123____________|___"_______\\__,__'

Or to make the lookup a bit faster, create a set from the chars in that string first:

>>> bad_chars = set('!~@#$%^&*()-+={}[]:;<.>?/\'')
>>> ''.join("_" if c in bad_chars else c for c in table)

You can use a look-around method to insert the /user/ dynamically:

(?<=url\(')/*(?=(?:.*?Inter\.ttf)'\))

And then use re.sub to replace with /user/:

strings = ["url('Inter.ttf')", "url('/hello/Inter.ttf')"]

p = re.compile(r"(?<=url\(')/?(?=(?:.*?Inter\.ttf)'\))")

for s in strings:
    s = re.sub(p, "/user/", s)
    print(s)

url('user/Inter.ttf')
url('user/hello/Inter.ttf')

Pattern Explanation

(?<=url\('): Positive lookbehind; matches strings that come after a string like url('.

/?: Matches zero or one forward slashes /. This is important for matching paths like /hello/Inter.ttf because it starts with the /. This is going to be selected and replaced with the ending forward slash in the replacement string, /user/.

(?=(?:.*?Inter.ttf)'\): Positive lookahead; matches strings that come before a string that ends with Inter.ttf').

I suggest playing around with it on https://regex101.com, selecting the Substitution method on the left-hand-side.

Edit

If you want to match multiple fonts, you can just remove the Inter.ttf part of the regex:

(?<=url\(')/?(?=(?:.*?)'\))

Alternatively, if you wanted it to append /user/ to paths that had a file extension, you can replace Inter\.ttf with \.\w{3}, which effectively matches 3 of any character in [a-zA-Z0-9_]:

(?<=url\(')/?(?=(?:.*?\.\w{3})'\))

Add periods and dashes to the negated character class.

re.sub('[^a-zA-Z.-]+', ' ', corpus)

This can be done without regex:

>>> string = "Special $#! characters   spaces 888323"
>>> ''.join(e for e in string if e.isalnum())
'Specialcharactersspaces888323'

You can use str.isalnum:

S.isalnum() -> bool

Return True if all characters in S are alphanumeric
and there is at least one character in S, False otherwise.

If you insist on using regex, other solutions will do fine. However note that if it can be done without using a regular expression, that's the best way to go about it.