Assign column value based on the percentage of rows - sql

Question

In DB2 is there a way to assign a column value based on the first x%, then y% and remaining z% of rows?
I've tried using row_number() function but no luck!
Example below
Assuming that the below example count(id) is already arranged in descending order
Input:
ID count(id)
5 10
3 8
1 5
4 3
2 1
Output:
First 30% rows of the above input should be assigned code H, last 30% of the rows will have code L and remaining will have code M. If 30% of rows evaluates to decimal then round up-to 0 decimal place.
ID code
5 H
3 H
1 M
4 L
2 L

Answer 1

You can use window functions:
select t.id,
(case ntile(3) over (order by count(id) desc)
when 1 then 'H'
when 2 then 'M'
when 3 then 'L'
end) as grp
from t
group by t.id;
This puts them into equal sized groups.
For 30-40-30% split with your conditions, you have to be more careful:
select t.id,
(case when (seqnum - 1.0) < 0.3 * cnt then 'H'
when (seqnum + 1.0) > 0.7 * cnt then 'L'
else 'M'
end) as grp
from (select t.id,
count(*) as cnt,
count(*) over () as num_ids,
row_number() over (order by count(*) desc) as seqnum
from t
group by t.id
) t

Answer 2

Try this:
with t(ID, count_id) as (values
(5, 10)
, (3, 8)
, (1, 5)
, (4, 3)
, (2, 1)
)
select t.*
, case
when pst <=30 then 'H'
when pst <=70 then 'M'
else 'L'
end as code
from
(
select t.*
, rownumber() over (order by count_id desc) as rn
, 100*rownumber() over (order by count_id desc)/nullif(count(1) over(), 0) as pst
from t
) t;
The result is:
ID COUNT_ID RN PST CODE
-- -------- -- --- ----
5 10 1 20 H
3 8 2 40 M
1 5 3 60 M
4 3 4 80 L
2 1 5 100 L