Escape regex special characters in a Python string


Does Python have a function that I can use to escape special characters in a regular expression?

For example, I'm "stuck" :\ should become I\'m \"stuck\" :\\.

9/28/2018 5:35:52 PM

Accepted Answer

Use re.escape

>>> import re
>>> re.escape(r'\ a.*$')
'\\\\\\ a\\.\\*\\$'
>>> print(re.escape(r'\ a.*$'))
\\\ a\.\*\$
>>> re.escape('')
>>> print(re.escape(''))

Repeating it here:


Return string with all non-alphanumerics backslashed; this is useful if you want to match an arbitrary literal string that may have regular expression metacharacters in it.

3/28/2019 8:07:22 AM

I'm surprised no one has mentioned using regular expressions via re.sub():

import re
print re.sub(r'([\"])',    r'\\\1', 'it\'s "this"')  # it's \"this\"
print re.sub(r"([\'])",    r'\\\1', 'it\'s "this"')  # it\'s "this"
print re.sub(r'([\" \'])', r'\\\1', 'it\'s "this"')  # it\'s\ \"this\"

Important things to note:

  • In the search pattern, include \ as well as the character(s) you're looking for. You're going to be using \ to escape your characters, so you need to escape that as well.
  • Put parentheses around the search pattern, e.g. ([\"]), so that the substitution pattern can use the found character when it adds \ in front of it. (That's what \1 does: uses the value of the first parenthesized group.)
  • The r in front of r'([\"])' means it's a raw string. Raw strings use different rules for escaping backslashes. To write ([\"]) as a plain string, you'd need to double all the backslashes and write '([\\"])'. Raw strings are friendlier when you're writing regular expressions.
  • In the substitution pattern, you need to escape \ to distinguish it from a backslash that precedes a substitution group, e.g. \1, hence r'\\\1'. To write that as a plain string, you'd need '\\\\\\1' — and nobody wants that.

Licensed under: CC-BY-SA with attribution
Not affiliated with: Stack Overflow