Get distinct of minimum value - hive

Question

I have this table test in Hive.
+----------+-------+-------+
| name | price | notes |
+----------+-------+-------+
| product1 | 100 | |
| product1 | 200 | note1 |
| product2 | 10 | note2 |
| product2 | 5 | note2 |
+----------+-------+-------+
and I expect to get this result (distinct of products with minimum price)
+----------+-------+-------+
| name | price | notes |
+----------+-------+-------+
| product1 | 100 | |
| product2 | 5 | note2 |
+----------+-------+-------+
I can't use the following query because of different notes in product1.
SELECT name, MIN(price), notes
FROM test
GROUP BY name, notes;
+----------+-------+-------+
| name | price | notes |
+----------+-------+-------+
| product1 | 100 | |
| product1 | 200 | note1 |
| product2 | 5 | note2 |
+----------+-------+-------+

Answer 1

Remove notes in group by and try again:-
SELECT name, MIN(price), notes
FROM test
GROUP BY name
Run Code

Answer 2

Try this
SELECT name,
SUBSTRING_INDEX(GROUP_CONCAT(price ORDER BY price DESC),',',1) AS min_price,
SUBSTRING_INDEX(GROUP_CONCAT(notes ORDER BY price DESC),',',1) AS note_value
FROM test
GROUP BY name;

Answer 3

You can do this in Hive with windowing functions.
Query:
select distinct name
, min_price
, notes
from (
select *
, min(price) over (partition by name) num_price
from db.table ) x
where min_price = price
Output:
product1 100
product2 5 note2

Answer 4

This can be found using a subquery as well.
hive> select A.name,A.price,B.notes from (select name,min(price) as price from products group by name) as A
inner join (select name,price,notes from products) as B
on a.name = b.name and a.price = b.price;
The above query will give the output as:
product1 100
product2 5 note2
But, the suquery approach has 2 iterations over the same table and is not suggested for larger tables.
For larger tables,see #GoBrewers14 answer:
hive> select name,price,notes from (select *, min(price)over(partition by name) as min_price from products) as a
> where a.price = a.min_price;