等同于 Keep in Snowflake

Question

我正在尝试将 oracle 代码转换为雪花，

在甲骨文中：

MIN(salary) KEEP (DENSE_RANK FIRST ORDER BY commission_pct) "Worst",

什么是对应的东西，因为我们在 oracle 中没有“保留”

Answer 1

Snowflake 没有等效的“第一”聚合函数。一种方法是使用条件聚合：

select min(case when seqnum = 1 then salary end) as worst
from (select t.*,
             row_number() over (partition by ? order by commission_pct) as seqnum
      from t
     ) t
group by . . .

? 是用于聚合的列

Answer 2

所以使用这个 CTE 作为示例数据：

WITH data AS (  
    SELECT * FROM VALUES 
        ('a', 2300, 10.1), 
        ('a',4000, 28.7), 
        ('b', 3000, 90.0) 
    AS v(dept, salary, commission_pct)  
)

并应用 Gordon 的代码：

SELECT dept
    ,MIN(CASE WHEN seqnum = 1 THEN salary end) AS worst
FROM (SELECT t.*,
             ROW_NUMBER() OVER (PARTITION BY dept ORDER BY commission_pct) AS seqnum
      FROM data AS t
) 
GROUP BY 1 ORDER BY 1; 

我们根据commission_pct从每个部门（dept）得到第一项，然后我们取这些值的MIN。

2300

如果我们去掉 Min 从而有

WITH data AS (  
    SELECT * FROM values 
        ('a', 2300, 10.1), 
        ('a',4000, 28.7), 
        ('b', 3000, 90.0) 
    AS v(dept, salary, commission_pct)  
)
SELECT 
    CASE WHEN seqnum = 1 THEN salary END AS worst
FROM (SELECT t.*,
             ROW_NUMBER() OVER (PARTITION BY dept ORDER BY commission_pct) as seqnum
      FROM data AS t
) ;

我们得到

WORST
2300
NULL
3000

所以区别在于 QUALIFY 的情况下，非第一行实际上并没有返回。因此其他操作无法访问。

WITH data AS (  
    SELECT * FROM VALUES 
        ('a', 2300, 10.1), 
        ('a',4000, 28.7), 
        ('b', 3000, 90.0) 
    AS v(dept, salary, commission_pct)  
)
SELECT 
    salary AS worst
FROM data 
QUALIFY rOW_NUMBER() OVER (PARTITION BY dept ORDER BY commission_pct) = 1
;

只是给出：

WORST
2300
3000

但是 Snowflake 确实有 FIRST_VALUE，因此有 KEEP 的效果

WITH data AS (  
    SELECT * FROM VALUES 
        ('a', 2300, 10.1), 
        ('a',4000, 28.7), 
        ('b', 3000, 90.0) 
    AS v(dept, salary, commission_pct)  
)
SELECT t.*
    ,first_value(salary) OVER (PARTITION BY dept ORDER BY commission_pct) as same_as_keep
FROM data AS t
;

给予：

DEPT    SALARY    COMMISSION_PCT  SAME_AS_KEEP
a       2300      10.1            2300
a       4000      28.7            2300
b       3000      90.0            3000

因此您（需要一些子选择来消除双窗口功能的歧义）

WITH data AS (  
    SELECT * FROM VALUES 
        ('a', 2300, 10.1), 
        ('a',4000, 28.7), 
        ('b', 3000, 90.0) 
    AS v(dept, salary, commission_pct)  
)
SELECT q.*,
    min(same_as_keep) over (partition by true) as worst
FROM (
    SELECT t.*
        ,first_value(salary) OVER (PARTITION BY dept ORDER BY commission_pct) as same_as_keep
    FROM data AS t
  ) AS q
;

给予：

DEPT    SALARY    COMMISSION_PCT    SAME_AS_KEEP    WORST
a       2300      10.1              2300            2300
a       4000      28.7              2300            2300
b       3000      90.0              3000            2300

但就像很多事情一样，这一切都取决于您如何使用 KEEP 以及您想要的行为方面。

例如，我不知道你是否将 MIN 换成 COUNT，如果 KEEP 对于这个示例数据会给你 2，这就像 Gordon 的 CASE 版本，或者如果它给你 3，这意味着它的行为就像 FIRST VALUE。

Answer 3

您仍然可以使用聚合函数，而无需创建子查询/诉诸窗口函数。

想法是使用支持排序的聚合函数，如ARRAY_AGG 并访问第一个元素：

SELECT sth,
     MIN(salary) KEEP (DENSE_RANK FIRST ORDER BY commission_pct) "Worst"
FROM tab
GROUP BY sth;

应该是：

SELECT sth,
    (ARRAY_AGG(salary) WITHIN GROUP(ORDER BY commission_pct DESC, salary))[0]
FROM tab
GROUP BY sth;