Eliminate duplicates in ListAgg (Oracle)

You can use regular expressions and regexp_replace to remove the duplicates after concatenation with listagg:

SELECT Num1, 
       RTRIM(
         REGEXP_REPLACE(
           (listagg(Num2,'-') WITHIN GROUP (ORDER BY Num2) OVER ()), 
           '([^-]*)(-\1)+($|-)', 
           '\1\3'),
         '-') Num2s 
FROM ListAggTest;

This could be tidier if Oracle's regex flavour supported lookahead or non-capturing groups, but it doesn't.

However this solution does avoid scanning the source more than once.

DBFiddle here

As far as I can see, with the currently available language specification, this is the shortest to achieve what you want if it must be done with listagg.

select distinct
       a.Num1, 
       b.num2s
  from listaggtest a cross join (
       select listagg(num2d, '-') within group (order by num2d) num2s 
       from (
         select distinct Num2 num2d from listaggtest
       )
      ) b;

What was your solution that was worse than the custom aggregate solution?

Although this is an old post with an accepted answer, I think the LAG() analytic function works well in this case and is noteworthy:

• LAG() removes duplicate values in column num2 with minimal expense
• No need for non-trivial regular expression to filter results
• Just one full table scan (cost=4 on simple example table)

Here is the proposed code:

with nums as (
SELECT 
    num1, 
    num2, 
    decode( lag(num2) over (partition by null order by num2), --get last num2, if any
            --if last num2 is same as this num2, then make it null
            num2, null, 
            num2) newnum2
  FROM ListAggTest
) 
select 
  num1, 
  --listagg ignores NULL values, so duplicates are ignored
  listagg( newnum2,'-') WITHIN GROUP (ORDER BY Num2) OVER () num2s
  from nums;

The results below appear to be what the OP desires:

NUM1  NUM2S       
1   2-3-4-5-6
2   2-3-4-5-6
3   2-3-4-5-6
4   2-3-4-5-6
5   2-3-4-5-6
6   2-3-4-5-6