How to replace an optional suffix in sed

Question

I am using for testing purposes 2 lines with and without the optional suffix (last 2 enumerated elements ,D2,E2).

echo -e "A1,B2,C2\nA2,B2,C2,D2,E2" | sed -E 's/^(.*),(.*),(.*)((,.*)(,.*)){0,1}$/[\1],\2,\3\5\6/'

[A1],B2,C2
[A2,B2,C2],D2,E2

where I would like to obtain

[A1],B2,C2
[A2],B2,C2,D2,E2

I simply want to make the last 2 components optional and append them at the end in case they are found.

I tried with repetitions {x,y}, with question mark (?), but I don't manage to make really optional the 2 last elements on the sed find pattern. Looks like the sed greediness is extremely greedy because I am wrapping the pattern with '^' and '$' and instead of attempting first the unoptional pattern, it just takes the optional one with very greedy first component..

NOTE: The regex is a simplification of my real need so overcoming by different approaches might be futile. I really want to do it with sed and this particular approach, since this is just to improve my knowledgement on the tricks of the tool

EDIT: Thanks for the answers, this was my goal

sed -E 's/^([^,]*),([^,]*),([^,]*)(,[^,]*)?(,[^,]*)?$/[\1],\2,\3\4\5/'

Answer 1

I'm not sure about the purpose of your sed construct, but I can tell you about your mistake in it.
In your regexp you try to match the first three letter-number combinations with (.*),(.*),(.*). Since regexp in sed are 'greedy' the first (.*) will already match the three combinations as letters, numbers and commas are matched with .. To match single combinations you better match for none commas (e.g ([^,]*), witch matches any amount of none comma characters. Your command would then look like this:

echo -e "A1,B2,C2\nA2,B2,C2,D2,E2" | sed -E 's/^([^,]*),([^,]*),([^,]*)((,[^,]*)(,[^,]*)){0,1}$/[\1],\2,\3\5\6/'

If it is always a three or five combination input, you could also 'shorten' the regex to

echo -e "A1,B2,C2\nA2,B2,C2,D2,E2" | sed -E 's/^([^,]*)(((,[^,]*){2}){1,2})$/[\1]\2/'

but that actually depends on your use case.
The ^([^,]*) matches the first letter-number combination,
the ((,[^,]*){2})matches the next two or four combinations with a comma in front of it (,B2,C2 or ,B2,C2,D2,E2 in your input examples).

Answer 2

Still not sure what OP is trying to achieve, but to make a non-greedy match then you can use a negated character class, in this case [^,].

echo -e "A1,B2,C2\nA2,B2,C2,D2,E2" | 
    sed -E 's/([^,]+)((,[^,]+){2})((,[^,]+){0,2})/[\1]\2\4/'

[A1],B2,C2
[A2],B2,C2,D2,E2

echo -e "A1,B2,C2\nA2,B2,C2,D2,E2" | 
    sed -E 's/([^,]+)((,[^,]+){2})((,[^,]+){0,2})/[\1]\2\5/'

[A1],B2,C2
[A2],B2,C2,E2

Though catching each match separately to embellish is a different matter...

If OP would care to give a fuller example....perhaps we can all go round again! The exercise is good for us ;)

Answer 3

$ echo -e "A1,B2,C2\nA2,B2,C2,D2,E2" | sed -e 's/[^,]*/[&]/' -e 's/[^,]*/(&)/2' -e 's/[^,]*/{&}/3'
[A1],(B2),{C2}
[A2],(B2),{C2},D2,E2

Rather than trying to do it all in one single substitution, you can use the fact that you can pick which one of the matches to work with by using a numerical flag at the end (here /2 and /3 for the 2nd and 3rd match).

The sed command used above, for clarity:

sed -e 's/[^,]*/[&]/' \
    -e 's/[^,]*/(&)/2' \
    -e 's/[^,]*/{&}/3'