python3: extract IP address from compiled pattern

You can consider installing the excellent regex module, which supports many advanced regex features, including branch reset groups, designed to solve exactly the problem you outlined in this question. Branch reset groups are denoted by (?|...). All capture groups of the same positions or names in different alternative patterns within a branch reset grouop share the same capture groups for output.

Notice that in the example below the matching capture group becomes the named capture group, so that you don't need to iterate over multiple groups searching for a non-empty group:

import regex

ip_pattern = r'(?P<ip>\d{1,3}\.\d{1,3}\.\d{1,3}\.\d{1,3})'
patterns = [
    'Registration from {ip} - Wrong password',
    'Call from {ip} rejected because extension not found'
]
pattern = regex.compile('(?|%s)' % '|'.join(patterns).format(ip=ip_pattern))
for line in sys.stdin:
    match = regex.search(pattern, line)
    if match:
        print(match['ip'])

Demo: https://repl.it/@blhsing/RegularEmbellishedBugs

why don't you check which regex matched?

if 'ip1' in match :
    print match['ip1']
if 'ip2' in match :
    print match['ip2']

or something like:

names = [ 'ip1', 'ip2', 'ip3' ]
for n in names :
    if n in match :
        print match[n]

or even

num = 1000   # can easily handle millions of patterns =)
for i in range(num) :
    name = 'ip%d' % i
    if name in match :
        print match[name]