How to calculate leave-out-median using PL/SQL analytic function - sql

Question

I'm trying to write an analytic function in PL/SQL that, when applied to a column within a table, returns for each row in the table, the median of the column excluding the given row.
An example to clarify: Suppose I have a table TABLE consisting of one column X that takes on the following values:
1
2
3
4
5
I want to define an analytic function LOOM() such that:
SELECT LOOM(X)
FROM TABLE
delivers the following:
3.5
3.5
3
2.5
2.5
i.e., for each row, the median of X, excluding the given row. I've been struggling to build the desired LOOM() function.

Answer 1

I'm not sure if there is a "clever" way to do this. You can do the calculation with a correlated subquery.
Assuming the x values are unique -- as in your example --
with t as (
select 1 as x from dual union all
select 2 as x from dual union all
select 3 as x from dual union all
select 4 as x from dual union all
select 5 as x from dual
)
select t.*,
(select median(x)
from t t2
where t2.x <> t.x
) as loom
from t;
EDIT:
A more efficient method uses analytic functions but requires more direct calculation of the median. For instance:
with t as (
select 1 as x from dual union all
select 2 as x from dual union all
select 3 as x from dual union all
select 4 as x from dual union all
select 5 as x from dual
)
select t.*,
(case when mod(cnt, 2) = 0
then (case when x <= candidate_1 then candidate_2 else candidate_1 end)
else (case when x <= candidate_1 then (candidate_2 + candidate_3)/2
when x = candidate_2 then (candidate_1 + candidate_3)/2
else (candidate_1 + candidate_2) / 2
end)
end) as loom
from (select t.*,
max(case when seqnum = floor(cnt / 2) then x end) over () as candidate_1,
max(case when seqnum = floor(cnt / 2) + 1 then x end) over () as candidate_2,
max(case when seqnum = floor(cnt / 2) + 2 then x end) over () as candidate_3
from (select t.*,
row_number() over (order by x) as seqnum,
count(*) over () as cnt
from t
) t
) t