And here is my solution !

If we use the FREE-SPACING mode (?x), for the SEARCH part :

SEARCH  (?x-s)  (?: ( \+A (\*)? ) | ( \+B (\*)? ) | (2) )  (?!.*,)
Groups -->      No  1     2         3     4         5     Look-Ahead  

REPLACE (?1(?{2}1:a))(?3(?{4}2:b))?5X

BEWARE that, in the REPLACE part, the FREE-SPACING mode is FORBIDDEN. So, ONLY for INFO :

REPLACE ( ?1 ( ?{2} 1 : a ) )  ( ?3 ( ?{4} 2 : b ) ) ?5 X

and given the data :

12345-01, A+A*2B+B*+A
12345-02, A+AB+B*+AA

it would return :

12345-01, A1XB2a
12345-02, AaB2aA

Notes :

• The first part (?x-s) of the regex search means that :

  • The free-spacing mode is set ( Spaces are not taken in account, except for the [ ] syntax or an escaped space char )

  • Due to (?-s) syntax, the dot regex symbol matches a single standard char only ( not an EOL char )

• Then, the (?:......) syntax defines a non-capturing group

• Now, in this non-capturing group, we have 3 alternatives and the first two contain an optional inner group (\*)? ( Remember that the ? is an other form of the {0,1} quantifier )

• To end, all this regex , so far, will match ONLY IF the final negative look-ahead structure (?!.*,) is verified, that is to say if at current position, reached by the regex engine, there is never a comma, at any further position, in current line

• Now, in the replacement regex :

  • The (?1(?{2}1:a)) syntax means that if  group 1 exists, then if  group 2 exists, then  write 1 else  write a

  • The (?3(?{4}2:b)) syntax means that if  group 3 exists, then if  group 4 exists, then  write 2 else  write b

  • Finally, the ?5X means that if  group 5 exists, then  write an X ( The parentheses are not mandatory as this part ends the regex

  • Note also that it’s not necessary to surround the groups 1, 3 and 5 with braces as these groups are not immediately followed with a digit !

Best Regards,

guy038

I have never really familiarized myself with lookaheads

There are LOTS of wonderful things to try and remember, as @PeterJones just reminded me. I should have made that a non-capture group, then it would not have required a rejig of the replace with code.

As I always say
“The day you stop learning is the day you die”

Terry

find: (^[^,]+,)|(+A*)|(+A)|(+B*)|(+B)|(2)
repl: (?{1}\1)(?{2}1)(?{3}a)(?{4}2)(?{5}b)(?{6}X)

I vote for yours. As an interesting aside, using regex101.com and inputting the 2 example lines and the Find What code, my code took twice as long as @Alan-Kilborn to process. It’s obvious the lookahead is where the extra time is spent.

For a small file to process it may not mean a lot, but sometimes efficiency in coding can be an advantage, hence my vote for @Alan-Kilborn code.

Terry

So as I had to add a negative lookahead the bracket numbering all changed

You could have made the wrapping parentheses a non-capturing group: (?-s)(?:(\+A\*)|(\+A)|(\+B\*)|(\+B)|(2))(?!.*?,), to avoid the renumbering in the replacement.

TIMTOWTDI

I came up with this; seems to work but maybe has holes:

find: (^[^,]+,)|(\+A\*)|(\+A)|(\+B\*)|(\+B)|(2)
repl: (?{1}\1)(?{2}1)(?{3}a)(?{4}2)(?{5}b)(?{6}X)

The result of the replacement with it:

12345-01, A1XB2a
12345-02, AaB2aA

It will involve a bit more thought.

Sorry, about that false start, I think I now have it. We have
FW:(?-s)((\+A\*)|(\+A)|(\+B\*)|(\+B)|(2))(?!.*?,)
RW:(?{2}1)(?{3}a)(?{4}2)(?{5}b)(?{6}X)

So as I had to add a negative lookahead the bracket numbering all changed hence a new replace with code as well.
So basically whenever it finds a character, so long as no , after it on the line it will be changed. As the ID is before the , nothing there should be changed.

Terry

PS should have paid more attention to your statement
I want to make the following replacements but only in the string (after the comma).

So here the 2 must be at a "boundary

Sorry, jumped the gun slightly, it did work on 2nd example, missed that it didn’t work on the first examples. Yes it IS a bit of a poser. It will involve a bit more thought.

Terry

that will leave the ID unchanged

A quick “fix” might be to use:
(\+A\*)|(\+A)|(\+B\*)|(\+B)|(\b2\b)
So here the 2 must be at a "boundary. For the 2 example lines provided it does work, however 2 examples does NOT a book make! It will depend on whether the “2” in the rest of the expression is surrounded by different characters on both sides.

Terry