0

我正在尝试从此示例表中获取项目的加权平均值：

物品 价值
项目1 0.10
项目2 0.15

我在这里按照这个例子来计算它：https ://www.wikihow.com/Calculate-Weighted-Average

并尝试了这种说法：

SELECT 
(SELECT CASE
         WHEN value <= 0.1 THEN 'very good'
         WHEN value <= 0.15
              value > 0.1 THEN 'good'
         WHEN valuey <= 0.20
              AND value > 0.15 THEN 'not that good'
         WHEN value <= 0.25
              AND value > 0.20 THEN 'bad'
         ELSE 'very bad'
       END ) AS ratings,
       COUNT (*) AS numberOfItems,
       COUNT (*) /  SUM(DISTINCT ( SELECT COUNT(DISTINCT collateral_id)
FROM   triggered.table
WHERE  id = 386)) as weightingFactor,
(COUNT (*) * (COUNT (*) /  SUM(DISTINCT ( SELECT COUNT(DISTINCT collateral_id)
FROM   triggered.table
WHERE  id = 386)))) as numberXWeightingFactor
FROM   triggered.table
WHERE  id = 386
GROUP BY stars

我试图通过做来获得平均值，SUM(numberXWeightingFactor)但它不起作用。最终给我错误：column "numberxweightingfactor" does not exist。

4

1 回答 1

0

多个问题。最重要的是，您似乎混淆了名称ratings和stars（我将两者都替换为rating），并且窗口函数不允许DISTINCT聚合。

这应该有效：

SELECT t.*
     , t.number_of_items   / c.ct_dist_coll AS weighting_factor
     , t.number_of_items^2 / c.ct_dist_coll AS number_x_weighting_factor

FROM  (
   SELECT CASE
            WHEN                   value <= 0.1  THEN 'very good'
            WHEN value >  0.1  AND value <= 0.15 THEN 'good'
            WHEN value >  0.15 AND value <= 0.20 THEN 'not that good'
            WHEN value >  0.20 AND value <= 0.25 THEN 'bad'
                                                 ELSE 'very bad' END AS rating
        , count(*) AS number_of_items
   FROM   triggered.table
   WHERE  id = 386
   GROUP  BY 1
   ) t
CROSS  JOIN (
   SELECT count(DISTINCT collateral_id) AS ct_dist_coll
   FROM   triggered.table
   WHERE  id = 386
   ) c;
于 2021-10-04T12:18:46.993 回答