Empty string instead of unmatched group error

Question

I have this piece of code:

for n in (range(1,10)):
    new = re.sub(r\'(regex(group)regex)?regex\', r\'something\'+str(n)+r\'\\1\', old, count=1)
         


        

Answer 1

To simplify:

Problem

1. You are getting the error "sre_constants.error: unmatched group" from a Python 2.7 regex.
2. You have any regex pattern with optional groups (with or without nested expressions) and are trying to use those groups in your sub replacement argument (re.sub(pattern, *repl*, string) or compiled.sub(*repl*, string))

Solution:

For results, return match.group(1) instead of \1 (or 2, 3, etc.). That's it; there is no or needed. The group result(s) can be returned with a function or a lambda.

Example

You are using a common regex to strip C-style comments. Its design uses an optional group 1 to pass through pseudo-comments which should not be deleted (if they exist).

pattern = r'//.*|/\*[\s\S]*?\*/|("(\\.|[^"])*")'
regex = re.compile(pattern)

Using \1 fails with the error: "sre_constants.error: unmatched group":

return regex.sub(r'\1', string)

Using .group(1) succeeds:

return regex.sub(lambda m: m.group(1), string)

For those not familiar with lambda, this solution is equivalent to:

def optgroup(match):
    return match.group(1)
return regex.sub(optgroup, string)

---

See the accepted answer for an excellent discussion of why \1 fails due to Bug 1519638. While the accepted answer is authoritative, it has two shortcomings: 1) the example from the original question is so convoluted that it makes the example solution difficult reading, and 2) it suggests returning a group or empty string -- that is not required, you may merely call .group() on each match.