獲取不同的最小值

Question

我在Hive中擁有此表test。

+----------+-------+-------+ 
| name | price | notes | 
+----------+-------+-------+ 
| product1 | 100 |  | 
| product1 | 200 | note1 | 
| product2 | 10 | note2 | 
| product2 |  5 | note2 | 
+----------+-------+-------+ 

，我希望得到這個結果（不同的產品，最低價格）

+----------+-------+-------+ 
| name | price | notes | 
+----------+-------+-------+ 
| product1 | 100 |  | 
| product2 |  5 | note2 | 
+----------+-------+-------+ 

我不能使用，因爲不同notes的下面的查詢中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

在組被刪除notes，然後再試一次： -

SELECT name, MIN(price), notes 
FROM test 
GROUP BY name 

Run Code

Answer 2

試試這個

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

你可以用windowing functions爲此在蜂巢。

查詢：

select distinct name 
    , min_price 
    , notes 
from (
    select * 
    , min(price) over (partition by name) num_price 
    from db.table) x 
where min_price = price 

輸出：

product1 100   
product2 5  note2 

Answer 4

這可以使用子查詢，以及被發現。

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; 

上述查詢將給出輸出爲：

product1  100 
product2  5  note2 

但是，該suquery方法有2次迭代在相同的表，並且不建議用於較大的表。

對於較大的表，請參見@GoBrewers14答案：

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;