SQL - 聚合函数不能在postgresql中嵌套错误

Question

表格1

date_time               | make   | model | miles | reg_no | age_months
----------------------------------------------------------------------
2016-09-28 20:05:03.001 | toyota | prius | 10200 | 1111   | 22
2016-09-28 20:06:03.001 | suzuki | sx4   | 10300 | 1122   | 12
2016-09-28 20:09:03.001 | suzuki | sx4   | 11200 | 1133   | 34
2016-09-28 20:10:03.001 | toyota | prius | 15200 | 1144   | 28
2017-05-28 20:11:03.001 | toyota | prius | 15500 | 1144   | 36

对于上面表1中的数据，我想通过模型如平均值，中位数，q1，q3，iqr等对month miles进行一些聚合。

我的查询如下，但它给出了错误： aggregate functions cannot be nested - 正确的方法是什么？

select
    model
    , COUNT(DISTINCT reg_no) AS distinct_car_count
    , COUNT(*) AS records_count
    , ROUND(AVG(miles/age_months*1.0),2) AS miles_per_month_avg
    , ROUND(PERCENTILE_CONT(0.5) WITHIN GROUP (ORDER BY (miles/age_months*1.0) ASC),2) AS miles_per_month_med
    , ROUND(PERCENTILE_CONT(0.25) WITHIN GROUP (ORDER BY (miles/age_months*1.0) ASC),2) AS miles_per_month_q1
    , ROUND(PERCENTILE_CONT(0.75) WITHIN GROUP (ORDER BY (miles/age_months*1.0) ASC),2) AS miles_per_month_q3
    , miles_per_month_q3 - miles_per_month_q1 as miles_per_month_iqr
    , sum(case when miles/age_months*1.0 <  (miles_per_month_q1 - 1.5*miles_per_month_iqr) then 1 else 0 end) as miles_per_month_num_records_outliers_lower_bound
    , sum(case when miles/age_months*1.0 >  (miles_per_month_q3 + 1.5*miles_per_month_iqr) then 1 else 0 end) as miles_per_month_records_outliers_upper_bound
    , ROUND(stddev_pop(miles/age_months*1.0),2) as miles_per_month_stddev

from table1 a
group by model;

Answer 1

有两个问题：

＃1：您无法嵌套聚合（如错误消息清楚地指示）， miles_per_month_q1是聚合列，并且yu尝试在另一个聚合的miles_per_month_num_records_outliers_lower_bound使用它。

＃2：您尝试在miles_per_month_q1的计算中重用列别名miles_per_month_iqr ，这在标准SQL中是不允许的。

对于这两种情况，您需要添加另一个嵌套级别（即派生表或公用表表达式），在您的情况下，它可能是：

SELECT
    a.model
    , Count(DISTINCT reg_no) AS distinct_car_count
    , Count(*) AS records_count
    , Round(Avg(miles/age_months*1.0),2) AS miles_per_month_avg

    -- now you can use the aliases, but you have to add a dummy (it's always the same value for a given model) aggregation function like MIN or MAX    
    , Min(percentiles.miles_per_month_med)
    , Min(percentiles.miles_per_month_q1)
    , Min(percentiles.miles_per_month_q3)
    , Min(percentiles.miles_per_month_q3 - percentiles.miles_per_month_q1) AS miles_per_month_iqr

    -- now it's no more nested aggregation
    , Sum(CASE WHEN miles/age_months*1.0 <  (percentiles.miles_per_month_q1 - 1.5* (percentiles.miles_per_month_q3 - percentiles.miles_per_month_q1)) THEN 1 ELSE 0 end) AS miles_per_month_num_records_outliers_lower_bound
    , Sum(CASE WHEN miles/age_months*1.0 >  (percentiles.miles_per_month_q3 + 1.5* (percentiles.miles_per_month_q3 - percentiles.miles_per_month_q1)) THEN 1 ELSE 0 end) AS miles_per_month_records_outliers_upper_bound

    , Round(StdDev_Pop(miles/age_months*1.0),2) AS miles_per_month_stddev

FROM table1 a
JOIN
 ( -- calculate the nested aggregates first
   SELECT
       model
       , Round(Percentile_Cont(0.5) Within GROUP (ORDER BY (miles/age_months*1.0) ASC),2) AS miles_per_month_med
       , Round(Percentile_Cont(0.25) Within GROUP (ORDER BY (miles/age_months*1.0) ASC),2) AS miles_per_month_q1
       , Round(Percentile_Cont(0.75) Within GROUP (ORDER BY (miles/age_months*1.0) ASC),2) AS miles_per_month_q3
   FROM table1 a
   GROUP BY model
 ) AS percentiles
ON a.model = percentiles.model
GROUP BY a.model

Answer 2

这就是杀了你的原因：

, sum(case when miles/age_months*1.0 <  (miles_per_month_q1 - 1.5*miles_per_month_iqr) then 1 else 0 end) as miles_per_month_num_records_outliers_lower_bound
, sum(case when miles/age_months*1.0 >  (miles_per_month_q3 + 1.5*miles_per_month_iqr) then 1 else 0 end) as miles_per_month_records_outliers_upper_bound

您正在使用基于另一个聚合函数构建的表达式（mile_per_month_q1，miles_per_month_q3）上的SUM - PERCENTILE_CONT

,ROUND(PERCENTILE_CONT(0.25) WITHIN GROUP (ORDER BY (miles/age_months*1.0) ASC),2) AS miles_per_month_q1
, ROUND(PERCENTILE_CONT(0.75) WITHIN GROUP (ORDER BY (miles/age_months*1.0) ASC),2) AS miles_per_month_q3

使用PERCENTILE_CONT将代码拆分为内部查询，并使用SUM进行外部查询

---

建议解决方案

select
    model
    , COUNT(DISTINCT reg_no) AS distinct_car_count
    , COUNT(*) AS records_count
    , ROUND(AVG(miles/age_months*1.0),2) AS miles_per_month_avg
    , ROUND(PERCENTILE_CONT(0.5) WITHIN GROUP (ORDER BY (miles/age_months*1.0) ASC),2) AS miles_per_month_med
    , min (miles_per_month_q1)  as miles_per_month_q1
    , min (miles_per_month_q3)  as miles_per_month_q3
    , miles_per_month_q3 - miles_per_month_q1 as miles_per_month_iqr
    , sum(case when miles/age_months*1.0 <  (miles_per_month_q1 - 1.5*(miles_per_month_q3 - miles_per_month_q1)) then 1 else 0 end) as miles_per_month_num_records_outliers_lower_bound
    , sum(case when miles/age_months*1.0 >  (miles_per_month_q3 + 1.5*(miles_per_month_q3 - miles_per_month_q1)) then 1 else 0 end) as miles_per_month_records_outliers_upper_bound
    , ROUND(stddev_pop(miles/age_months*1.0),2) as miles_per_month_stddev

from    (select       a.*
                    , ROUND(PERCENTILE_CONT(0.25) WITHIN GROUP (ORDER BY (miles/age_months*1.0) ASC),2) over (partition by model) AS miles_per_month_q1
                    , ROUND(PERCENTILE_CONT(0.75) WITHIN GROUP (ORDER BY (miles/age_months*1.0) ASC),2) over (partition by model) AS miles_per_month_q3

        from        table1 a
        ) a

group by  model
;