python-正则表达式将重复模式放在单个组中

Question

I'm trying to parse a string in regex and am 99% there.

my test string is

 1
  1234 1111 5555 88945
    172.255.255.255 from 172.255.255.255 (1.1.1.1)
      Origin IGP, localpref 300, valid, external, best
      rx pathid: 0, tx pathid: 0x0

my current regex pattern is:

(?P<as_path>(\d{4,10}\s){1,20})\s+(?P<peer_addr>\d{0,3}\.\d{0,3}\.\d{0,3}\.\d{0,3}).*\((?P<peer_rid>\d{0,3}\.\d{0,3}\.\d{0,3}\.\d{0,3})\)\s+.*localpref\s(?P<local_pref>\d+),\s(?P<attribs>\S+,\s{0,4})

im using regex101 to test and have a link to the test here https://regex101.com/r/iGM8ye/1

So currently i have a group2 I don't want this group, could someone tell me why im getting this group and how to remove it?

and the second is, in the attributes I want to match the words, "valid, external, best" currently my pattern only matches "valid," I thought adding the repeat of within the group would of matched all three of those but it hasn't.

How would I achieve matching the repeat of "string, string, string," (string comma space) into one group?

Thanks

EDIT

Desired output

as_path : 1234 1111 5555 88945
peer_addr : 172.255.255.255
peer_rid : 1.1.1.1
local_pref : 300
attribs : valid, external, best 

attiribs may also just be valid, external, or just external, or another entry in the format (stringcommaspace)

Answer 1

Try Regex: (?P<as_path>(?:\\d{4,10}\\s){1,20})\\s+(?P<peer_addr>\\d{0,3}\\.\\d{0,3}\\.\\d{0,3}\\.\\d{0,3}).*\\((?P<peer_rid>\\d{0,3}\\.\\d{0,3}\\.\\d{0,3}\\.\\d{0,3})\\)\\s+.*localpref\\s(?P<local_pref>\\d+),\\s(?P<attribs>[\\S]+,(?: [\\S]+,?)*){0,4}

Demo

Regex in the question had a capturing group (Group 2) for (\\d{4,10}\\s) . it is changed to a non capturing group now (?:\\d{4,10}\\s)

Answer 2

See regex in use here .

(?P<as_path>(?:\d{4,10}\s){1,20})\s+(?P<peer_addr>\d{0,3}(?:\.\d{0,3}){3}).*\((?P<peer_rid>\d{0,3}(?:\.\d{0,3}){3})\)\s+.*localpref\s(?P<local_pref>\d+),\s+(?P<attribs>\S+(?:,\s+\S+){2})

1. You were getting group 2 because your as_path group contained a group. I changed that to a non-capturing group.
2. I changed attribs to \\S+(?:,\\s+\\S+){2}
   • This will match any non-space character one or more times \\S+ , followed by the following exactly twice:
     • ,\\s+\\S+ the comma character, followed by the space character one or more times, followed by any non-space character one or more times
3. I changed peer_addr and peer_rid to \\d{0,3}(?:\\.\\d{0,3}){3} instead of \\d{0,3}\\.\\d{0,3}\\.\\d{0,3}\\.\\d{0,3} . This is a preference, but shortens the expression.

Without that last modification, you can use the following regex (it performs slightly better anyway (as seen here ):

(?P<as_path>(?:\d{4,10}\s){1,20})\s+(?P<peer_addr>\d{0,3}\.\d{0,3}\.\d{0,3}\.\d{0,3}).*\((?P<peer_rid>\d{0,3}\.\d{0,3}\.\d{0,3}\.\d{0,3})\)\s+.*localpref\s(?P<local_pref>\d+),\s+(?P<attribs>\S+(?:,\s+\S+){2})

---

You can also improve the performance by using more specific tokens as the following suggests (notice I also added the x modifier to make it more legible) and as seen here :

(?P<as_path>\d{4,10}(?:\s\d{4,10}){0,19})\s+
(?P<peer_addr>\d{0,3}\.\d{0,3}\.\d{0,3}\.\d{0,3})[^)]*
\((?P<peer_rid>\d{0,3}\.\d{0,3}\.\d{0,3}\.\d{0,3})\)\s+
.*localpref\s(?P<local_pref>\d+),\s+
(?P<attribs>\w+(?:,\s+\w+){2})

Answer 3

You get that separate group because your are repeating a capturing group were the last iteration will be the capturing group, in this case 88945 You could make it non capturing instead (?:

For the second part you could use an alternation to exactly match one of the options (?:valid|external|best)

Your pattern might look like:

(?P<as_path>(?:\d{4,10}\s){1,20})\s+(?P<peer_addr>\d{0,3}\.\d{0,3}\.\d{0,3}\.\d{0,3}).*\((?P<peer_rid>\d{0,3}\.\d{0,3}\.\d{0,3}\.\d{0,3})\)\s+.*localpref\s(?P<local_pref>\d+),\s(?P<attribs>(?:valid|external|best)(?:,\s{0,4}(?:valid|external|best))+)

regex101 demo