Python regexp with optional groups

Question

With Python I need to parse the following string, that can be written in four different ways:

:param MyParam: My description [MyValue] {MyGroup}  
:param MyParam: My description [MyValue]  
:param MyParam: My description {MyGroup}  
:param MyParam: My description

The expectation is to extract MyParam, My description, MyValue and MyGroup

I tried with the following pattern:

^\:param(?P<param>.*)\:(?P<desc>.*)\s*(\[(?P<value>.*)\])?\s*(\{(?P<group>.*)\})?$

but in all cases it is too greedy.

Using the following one:

^\:param(?P<param>.*)\:(?P<desc>.*) (\[(?P<value>.*)\])? (\{(?P<group>.*)\})?$

I get valid results only if there are spaces according to the pattern expectations.
Any idea how to fix this ?

Use the non-greedy variants, i.e. .*? instead of .*.

inof
– inof

2021-04-22 14:43:38 +00:00
Commented Apr 22, 2021 at 14:43 — inof
– inof, Commented Apr 22, 2021 at 14:43
BTW, you don’t have to escape the colons.

inof
– inof

2021-04-22 14:45:10 +00:00
Commented Apr 22, 2021 at 14:45 — inof
– inof, Commented Apr 22, 2021 at 14:45

Arvind Kumar Avinash · Accepted Answer · 2021-04-22 19:01:48Z

3

You may use this regex to parse all the possible combination mentioned in your question:

^:param\s+(?P<param>[^:]+):\s*(?P<desc>.*?)\s*(?:\[(?P<value>[^]]*)\])?\s*(?:\{(?P<group>[^}]*)\})?$

RegEx Demo

RexEx Details:

^: Start
:: Match a colon
param\s+: Match text param followed by 1+ whitespace characters
(?P<param>[^:]+): Match 1+ non-colon characters in named group param
:\s*: Match a : followed by 0+ whitespace characters
(?P<desc>.*?): Match 0+ of any characters in named group desc
\s*: Match 0+ whitespace characters
(?:\[(?P<value>[^]]*)\])?: Optionally match [...] and capture inner text in in named group value
\s*: Math 0+ whitespace characters
(?:\{(?P<group>[^}]*)\})?: Optionally match {...} and capture inner text in in named group group
$: End

edited Apr 22, 2021 at 19:01

Arvind Kumar Avinash

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answered Apr 22, 2021 at 14:47

anubhava

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