删除重量超过限制的行，直到没有这样的行

Question

我的问题在自然语言中有明确说明，但我无法在（ORACLE）SQL中找到解决方案。

数据具有VALUE（正数）和LIMIT（0到1之间的值表示百分比）列。其任务是删除（或识别）VALUE大于其余行的VALUES总和的LIMIT％的行。另一个公式：删除重量（由VALUE定义）大于总量的LIMIT的行。

请注意，删除某些行后，总和会减少，因此另一行可能会失败并需要删除。

到目前为止，我尝试了解析函数（请参阅下面的示例），递归与连接。一切都没有成功。

该解决方案应该在没有PL/SQL的SQL中。如果共识是没有这样的解决方案存在，那么程序就没问题。

我的错误的解决方法使用分析功能如下：

WITH 
DATA as (
    SELECT 23 as KEY, 100 as VALUE, 0.15 as LIMIT from DUAL 
    UNION ALL 
    SELECT 22 as KEY, 101 as VALUE, 0.05 as LIMIT from DUAL 
    UNION ALL 
    SELECT 21 as KEY, 10 as VALUE, 0.05 as LIMIT from DUAL 
    UNION ALL 
    SELECT 20 as KEY, 1 as VALUE, 0.15 as LIMIT from DUAL 
    UNION ALL 
    SELECT 19 as KEY, 1 as VALUE, 0.15 as LIMIT from DUAL 
    UNION ALL 
    SELECT 18 as KEY, 1 as VALUE, 0.15 as LIMIT from DUAL 
    UNION ALL 
    SELECT 17 as KEY, 1 as VALUE, 0.15 as LIMIT from DUAL 
    UNION ALL 
    SELECT 16 as KEY, 1 as VALUE, 0.15 as LIMIT from DUAL 
    UNION ALL 
    SELECT 15 as KEY, 1 as VALUE, 0.15 as LIMIT from DUAL 
    UNION ALL 
    SELECT 14 as KEY, 1 as VALUE, 0.15 as LIMIT from DUAL 
    UNION ALL 
    SELECT 13 as KEY, 1 as VALUE, 0.15 as LIMIT from DUAL 
    UNION ALL 
    SELECT 12 as KEY, 1 as VALUE, 0.15 as LIMIT from DUAL 
    UNION ALL 
    SELECT 11 as KEY, 1 as VALUE, 0.15 as LIMIT from DUAL 
    UNION ALL 
    SELECT 10 as KEY, 1 as VALUE, 0.15 as LIMIT from DUAL 
    UNION ALL 
    SELECT 9 as KEY, 1 as VALUE, 0.15 as LIMIT from DUAL 
    UNION ALL 
    SELECT 8 as KEY, 1 as VALUE, 0.15 as LIMIT from DUAL 
    UNION ALL 
    SELECT 7 as KEY, 1 as VALUE, 0.15 as LIMIT from DUAL 
    UNION ALL 
    SELECT 6 as KEY, 1 as VALUE, 0.15 as LIMIT from DUAL 
    UNION ALL 
    SELECT 5 as KEY, 1 as VALUE, 0.15 as LIMIT from DUAL 
    UNION ALL 
    SELECT 4 as KEY, 1 as VALUE, 0.15 as LIMIT from DUAL 
    UNION ALL 
    SELECT 3 as KEY, 1 as VALUE, 0.15 as LIMIT from DUAL 
    UNION ALL 
    SELECT 2 as KEY, 1 as VALUE, 0.15 as LIMIT from DUAL 
    UNION ALL 
    SELECT 1 as KEY, 1 as VALUE, 0.15 as LIMIT from DUAL 
), 
REMOVED AS (
    SELECT 
    d.*, 
    CASE 
     WHEN d.VALUE >= d.LIMIT * (SUM(d.VALUE) OVER()) THEN 'N' 
     ELSE 'Y' 
    END as FLAG, 
    d.VALUE/d.LIMIT * (SUM(d.VALUE) OVER()) AS ACT_WEIGHT, 
    SUM(d.VALUE) OVER() AS TOTOAL_SUM 
    FROM DATA d 
) 
SELECT r.KEY 
FROM REMOVED r 
WHERE FLAG='N'; 
-- Wrong: KEY=21 is missing! 

Answer 1

此使用递归子查询保条款（又名CTE）反复产生两套行。第一组消除了高于先前总数百分比限制的值并重新计算新总数，第二组生成的行更新again列，以确定在迭代生成之前是否有任何高于新总数的新行一组新的行。

查询：

WITH cte (key, value, limit, lvl, total, again) AS (
    SELECT key, value, limit, 1, SUM(value) OVER(), 1 
    FROM data 
UNION ALL 
    SELECT key, 
     value, 
     limit, 
     lvl + 1, 
     CASE MOD(lvl, 2) 
      WHEN 1 
      THEN SUM(value) OVER() 
      ELSE total 
     END, 
     CASE MOD(lvl, 2) 
      WHEN 0 
      THEN MAX(CASE WHEN value > limit * total THEN 1 ELSE 0 END) OVER() 
      ELSE 1 
     END 
    FROM cte 
    WHERE (MOD(lvl, 2) = 0 OR value <= limit * total) 
    AND again = 1 
) 
SELECT * 
FROM cte 
WHERE again = 0; 

输出：

 KEY  VALUE  LIMIT  LVL  TOTAL  AGAIN 
---------- ---------- ---------- ---------- ---------- ---------- 
     20   1  .15   5   20   0 
     19   1  .15   5   20   0 
     18   1  .15   5   20   0 
     17   1  .15   5   20   0 
     16   1  .15   5   20   0 
     15   1  .15   5   20   0 
     14   1  .15   5   20   0 
     13   1  .15   5   20   0 
     12   1  .15   5   20   0 
     11   1  .15   5   20   0 
     10   1  .15   5   20   0 
     9   1  .15   5   20   0 
     8   1  .15   5   20   0 
     7   1  .15   5   20   0 
     6   1  .15   5   20   0 
     5   1  .15   5   20   0 
     4   1  .15   5   20   0 
     3   1  .15   5   20   0 
     2   1  .15   5   20   0 
     1   1  .15   5   20   0 

Answer 2

这可以用一个递归CTE来解决，如MT0显示。 MT0的解决方案可以简化 - 这个问题不需要我在MT0评论中提到的“跷跷板递归查询”。

在递归查询中，我从给定数据开始，但添加了三列：值的总和（对于“当前行”），行数和数字1作为占位符。 （任何正数都可以在那里使用。）

然后在递归步骤中，我保持上一级满足限制条件的行（但对于“旧”总和，包括必须丢弃的行）。我再次计算总和（val），以用于下一步，我计算剩下多少行，并计算“this”级别与前一级别之间的计数差异。

如果在某个时候计数差异变为零，那意味着在那个级别上我不需要丢弃任何行。那些是解决问题的答案。我在外部查询中选择它们。

with 
    data (key, value, limit) as (
     select ........... 
    ), 
    r (key, value, limit, tot, cnt, diff) as (
     select key, value, limit, sum(value) over(), count(*) over(), 1 
     from data 
     union all 
     select key, value, limit, sum(value) over(), count(*) over(), 
       cnt - count(*) over() 
     from r 
     where value <= limit * tot and diff > 0 
) 
select key, value, limit 
from r 
where diff = 0 
;