Python regular expression escaping

Question

I am trying to use (?<!\\)# to match #s without a \ in the front (the task is to escape the unescaped #s in a string). This regex works on several online regex validators. However it doesn't work with the python re module. I also tried escaping the symbols in the regex, but it either produces errors or does not produce the expected output.

re.sub("(?<!\\)#","\#",'asd\#fh## #')

How can I modify this regex so it can produce the output asd\\#fh\\#\\# \\#(The output has \s escaped so there are double \\)?

Use it like this: re.sub(r"(?<!\\)#", r"\\#", r'asd\#fh## #') — anubhava
– anubhava, Commented Jul 8, 2022 at 17:40
@anubhava thanks! didn't know about raw strings. Can you post it as an answer so that I can accept it ? — Nick
– Nick, Commented Jul 8, 2022 at 17:42
See Confused about backslashes in regular expressions. Also, How python and the regex module handle backslashes. Replacing with backslashes: Escape special characters in a Python string. — Wiktor Stribiżew
– Wiktor Stribiżew, Commented Jul 8, 2022 at 17:43

anubhava · Accepted Answer · 2022-07-08 17:56:06Z

You have few issues in your code:

'asd\#fh## #' is same as 'asd#fh## #' in normal string (unless you use raw string mode)
Likewise "\#" in replacement is same as just #
Similarly "(?<!\\)#" will generate regex syntax error as it will become (?<!\)# without matching ) for negative lookahead

You need to use raw string mode or use double escaping to get it right:

repl = re.sub(r"(?<!\\)#", r"\#", r'asd\#fh## #') # repl = 'asd\\#fh\\#\\# \\#'

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