Python regex not matching for “,” or end of string

Question

I want to write a single regular expression in python for below lines to grep the corresponding values:

establishmentCause mo-Signalling,
Freq = 6300
Radio Bearer ID = 0, Physical Cell ID = 396

Here i want to fetch the values for each header, I am using the below regular expression to fetch values and it succeeds for all except "Radio Bearer ID"

pat = re.compile(r'%s\s[=\s]*\b(.*)\b(?:,|\Z)'%items[i])
value = pat.search(line)
print(value.group(1))

This gives the output for "Radio Bearer ID" as 0, Physical Cell ID = 396 where as I want only 0 . Can some one please tell me what the problem is with my regular expression even though I am matching , and \\Z the re engine dose not limit the match till , but continues further.

Answer 1

Quantifier * is greedy. You can use the non-greedy version *? to match as little as possible before the , or end of string ( \\Z ):

pat = re.compile(r'%s\s[=\s]*\b(.*?)\b(?:,|\Z)'%items[i])

Alternatively, you can use a character class excluding , instead:

pat = re.compile(r'%s\s[=\s]*\b([^,]*)\b(?:,|\Z)'%items[i])

Answer 2

You can use Lookbehind & Lookahead

Ex:

import re

s = """establishmentCause mo-Signalling,
Freq = 6300
Radio Bearer ID = 0, Physical Cell ID = 396"""

pat = re.compile(r'(?<=Radio Bearer ID = )(.*)(?=,)')
value = pat.search(s)
print(value.group(1))

Output:

0