How can I select unique values from several columns in Oracle SQL? - sql

Question

Basically, I've got the following table:
ID | Amount
AA | 10
AA | 20
BB | 30
BB | 40
CC | 10
CC | 50
DD | 20
DD | 60
EE | 30
EE | 70
I need to get unique entries in each column as in following example:
ID | Amount
AA | 10
BB | 30
CC | 50
DD | 60
EE | 70
So far following snippet gives almost what I wanted, but first_value() may return some value, which isn't unique in current column:
first_value(Amount) over (partition by ID)
Distinct also isn't helpful, as it returns unique rows, not its values
EDIT:
Selection order doesn't matter

Answer 1

This works for me, even with the problematic combinations mentioned by Dimitri. I don't know how fast that is for larger volumes though
with ids as (
select id, row_number() over (order by id) as rn
from data
group by id
), amounts as (
select amount, row_number() over (order by amount) as rn
from data
group by amount
)
select i.id, a.amount
from ids i
join amounts a on i.rn = a.rn;
SQLFiddle currently doesn't work for me, here is my test script:
create table data (id varchar(10), amount integer);
insert into data values ('AA',10);
insert into data values ('AA',20);
insert into data values ('BB',30);
insert into data values ('BB',40);
insert into data values ('CC',10);
insert into data values ('CC',50);
insert into data values ('DD',20);
insert into data values ('DD',60);
insert into data values ('EE',30);
insert into data values ('EE',70);
Output:
id | amount
---+-------
AA | 10
BB | 20
CC | 30
DD | 40
EE | 50

Answer 2

I suggest using row_number() like this:
select ID ,Amount
from (
select ID ,Amount, row_number() over(partition by id order by 1) as rn
from yourtable
)
where rn = 1
However your expected results don't conform to a discrenable order, some are the first/lowest while some the last/highest so I wasn't sure what to include for the ordering.

Answer 3

My solution implements recursive with and makes following: first - select minival values of ID and amount, then for every next level searches values of ID and amount, which are more than already choosed (this provides uniqueness), and at the end query selects 1 row for every value of recursion level. But this is not an ultimate solution, because it is possible to find a combination of source data, where query will not work (I suppose, that such solution is impossible, at least in SQL).
with r (id, amount, lvl) as (select min(id), min(amount), 1
from t
union all
select t.id, t.amount, r.lvl + 1
from t, r
where t.id > r.id and t.amount > r.amount)
select lvl, min(id), min(amount)
from r
group by lvl
order by lvl
SQL Fiddle

Answer 4

I knew that there is an elegant solution! Thanks to friend of mine for a tip:
select max(ID), mAmount from (
select ID, max(Amount) mAmount from table group by ID
)
group by mAmount;

Answer 5

Maybe something like this can solve:
WITH tx AS
( SELECT ROWNUM ROW_NUMBER,
t.id,
t.amount
FROM test t
INNER JOIN test t2
ON t.id = t2.id
AND t.amount != t2.amount
ORDER BY t.id)
SELECT tx1.id, tx1.amount
FROM tx tx1
LEFT JOIN tx tx2
ON tx1.id = tx2.id
AND tx1.ROW_NUMBER > tx2.ROW_NUMBER
WHERE tx2.ROW_NUMBER IS NULL