replace groups in regex loop - regex

Question

I have these 2 lines:
What is P(output1|cause1=2, cause2=2)
What is P(output2|cause3=2)
I would like to change it to:
method_to_use(model, {"cause1": 2, "cause2": 2}, "output1")
method_to_use(model, {"cause3": 2}, "output2")
this is my regex:
.*P[(]([a-z1-9]+)[|](([a-z1-9]+)=([1-9]),?)+[)]
and I try to replace it like this:
method_to_use(model, {"$3": $4}, "$1")
but I only get the last fit of the group:
method_to_use(model, {"cause2": 2}, "output1")
is it possible to do some kind of a "loop" and change all fits on the way?

Answer 1

You could match the string with the following regular expression.
^.*P\(([^|]+)\|([^=]+)=(\d+)(?:, +([^=]+)=(\d+))?\)$
If capture group 4 is non-empty replace the (whole-string) match with
method_to_use(model, {"$2": $3, "$4": $5}, "$1")
This causes the string
What is P(output1|cause1=2, cause2=2)
to be replaced with
method_to_use(model, {"cause1": 2, "cause2": 2}, "output1")
Demo 1
If capture group 4 is empty replace the match with
method_to_use(model, {"$2": $3}, "$1")
This causes the string
What is P(output2|cause3=2)
to be replaced with
method_to_use(model, {"cause3": 2}, "output2")
Demo 2
Note that the regular expressions at the two links are equivalent, the only difference is that at Demo 1 I expressed the regex in free-spacing mode, which permits it to be self-documenting.
Instead of replacing the entire string one could of course simply form the new string from the values of the capture groups. If that is done ^.*P at the beginning of the regex could be changed to simply P.
The regex engine performs the following operations.
^ # match beginning of line
.*P\( # match 0+ chars then '|('
([^|]+) # save 1+ chars except '|' in cap grp 1 (output)
\| # match ':'
([^=]+) # save 1+ chars except '=' in cap grp 2 (causeA)
= # match '='
(\d+) # save 1+ digits in cap grp 3 (causeAval)
(?: # begin non-cap grp
,\ + # match ',' then 1+ spaces
([^=]+) # match 1+ chars except '=' in cap grp 4 (causeB)
= # match '='
(\d+) # match 1+ digits in cap grp 5 (causeBval)
)? # end non-cap grp and make it optional
\) # match ')'
$ # match end of line

Answer 2

One thing is certain: you can't do that with a single regex.
You may use a 3-step approach:
Replace .*P\( with method_to_use(
Replace \(\K(\w+)\|([^()]+) with model, {$2}, "$1"
Replace (\w+)=(\w+) with "$1": $2
Note that
.*P\( matches any 0 or more chars other than line break chars as many as possible and the P(
\(\K(\w+)\|([^()]+) matches (, then discards it from the match value with \K, and then 1+ word chars are captured into Group 1 ($1 in the replacement pattern), | is matched and then 1+ chars other than ) and ( are captured into Group 2 ($2)
(\w+)=(\w+) - 1+ word chars are captured into Group 1 ($1 in the replacement pattern), = is matched and then 1+ word chars are captured into Group 2 ($2).