SQL 查询挑战 - 在列中找到最常见的项目并将结果汇总到一个数据透视表 table

Question

我正在寻找一个查询来进行以下转换。

基本上我想在每个网站上，每天找到前 3 个频繁 SELL_COUNTRY 和前 3 个频繁类别。（比如网站1，日期6-5-2017，SELL_COUNTRY有2*US，1*JP，1*UK，所以TOP1_SELL_COUNTRY是US，JP和UK去TOP2_SELL_COUNTRY 和 TOP3_SELL_COUNTRY。CATEGORY 列的思路相同）

我现在的解决方案涉及到很多子查询，这种方法可行，但我觉得它太复杂了。我很感兴趣 sql master 是如何优雅地做到这一点的。

目前我知道如何使用

来自

到

Answer 1

WITH a1 AS
(
SELECT *,
       COUNT(*) OVER( PARTITION BY website,SUBSTRING(visit_date,1,8),sell_country ) AS sell_cntry,
       COUNT(*) OVER( PARTITION BY website,SUBSTRING(visit_date,1,8),pur_country ) AS pur_cntry
  FROM Yourtable
),
a2 AS 
(
SELECT website,
       visit_date,
       sell_country,
       RANK() OVER ( PARTITION BY website,SUBSTRING(visit_date,1,8) ORDER BY sell_cntry DESC ) AS sell_cntry_rnk
  FROM a1
),
a3 AS
(
SELECT website,
       visit_date,
       pur_country,
       RANK() OVER ( PARTITION BY website,SUBSTRING(visit_date,1,8) ORDER BY pur_cntry DESC ) AS pur_cntry_rnk
  FROM a1
),
a4 AS
(
SELECT a2.website AS company,
       a2.v_date,
       CASE WHEN a2.sell_cntry_rn = 1 THEN a2.sell_country  END  AS TOP1_SELL_COUNTRY,
       CASE WHEN a2.sell_cntry_rn = 2 THEN a2.sell_country  END  AS TOP2_SELL_COUNTRY,
       CASE WHEN a2.sell_cntry_rn = 3 THEN a2.sell_country  END  AS TOP3_SELL_COUNTRY,
       CASE WHEN a3.pur_cntry_rn = 1 THEN a3.pur_country  END  AS TOP1_PUR_COUNTRY,
       CASE WHEN a3.pur_cntry_rn = 2 THEN a3.pur_country  END  AS TOP2_PUR_COUNTRY,
       CASE WHEN a3.pur_cntry_rn = 3 THEN a3.pur_country  END  AS TOP3_PUR_COUNTRY 
  FROM (
          SELECT Z.*,
                 ROW_NUMBER() OVER( PARTITION BY website,v_date ORDER BY sell_cntry_rnk,sell_country ) AS sell_cntry_rn
            FROM 
          (
            SELECT DISTINCT website,
                            SUBSTRING(visit_date,1,8) AS v_date,
                            sell_cntry_rnk,
                            sell_country
              FROM a2 
            ) Z
          WHERE Z.sell_cntry_rnk <= 3 
        ) a2
 INNER JOIN
       (
         SELECT *,
                ROW_NUMBER() OVER( PARTITION BY website,v_date ORDER BY pur_cntry_rnk,pur_country ) AS pur_cntry_rn
           FROM
          ( SELECT DISTINCT website,
                            SUBSTRING(visit_date,1,8) AS v_date,
                            pur_cntry_rnk,
                            pur_country
              FROM a3
           ) Z
          WHERE Z.pur_cntry_rnk <= 3 
       ) a3 
    ON a2.website =  a3.website
   AND a2.v_date = a3.v_date
),
a5 AS
(
SELECT company,
       v_date,
       MAX(TOP1_SELL_COUNTRY) AS TOP1_SELL_COUNTRY,
       MAX(TOP2_SELL_COUNTRY) AS TOP2_SELL_COUNTRY,
       MAX(TOP3_SELL_COUNTRY) AS TOP3_SELL_COUNTRY,
       MAX(TOP1_PUR_COUNTRY) AS TOP1_PUR_COUNTRY,
       MAX(TOP2_PUR_COUNTRY) AS TOP2_PUR_COUNTRY,
       MAX(TOP3_PUR_COUNTRY) AS TOP3_PUR_COUNTRY
  FROM a4
GROUP BY company,
         v_date
)
SELECT company,
       v_date,
       CASE WHEN TOP1_SELL_COUNTRY IS NULL THEN 'NA' ELSE TOP1_SELL_COUNTRY END AS TOP1_SELL_COUNTRY,
       CASE WHEN TOP2_SELL_COUNTRY IS NULL THEN 'NA' ELSE TOP2_SELL_COUNTRY END AS TOP2_SELL_COUNTRY,
       CASE WHEN TOP3_SELL_COUNTRY IS NULL THEN 'NA' ELSE TOP3_SELL_COUNTRY END AS TOP3_SELL_COUNTRY,
       CASE WHEN TOP1_PUR_COUNTRY IS NULL THEN 'NA' ELSE TOP1_PUR_COUNTRY END AS TOP1_PUR_COUNTRY,
       CASE WHEN TOP2_PUR_COUNTRY IS NULL THEN 'NA' ELSE TOP2_PUR_COUNTRY END AS TOP2_PUR_COUNTRY,
       CASE WHEN TOP3_PUR_COUNTRY IS NULL THEN 'NA' ELSE TOP3_PUR_COUNTRY END AS TOP3_PUR_COUNTRY
  FROM a5
 ORDER BY company,v_date;

Answer 2

我会分 3 个步骤完成：

按国家/地区分组并按计数排名
按类别分组并按计数排名
使用条件聚合混合结果（这只会将值放在必要的单元格中，因为 CASE 的结果将只是您的值和许多 NULL 值，所以 min() 输出值)

像这样：

WITH
countries as (
    SELECT *, row_number() over (partition by website,date order by count desc)
    FROM (
        SELECT
         website
        ,date::date
        ,sell_country
        ,count(1)
        FROM your_table
        GROUP BY 1,2,3
    )
)
,categories as (
    SELECT *, row_number() over (partition by website,date order by count desc)
    FROM (
        SELECT
         website
        ,date::date
        ,category
        ,count(1)
        FROM your_table
        GROUP BY 1,2,3
    )
)
SELECT
 website
,date
,coalesce(min(case when t1.row_number=1 then t1.sell_country end),'NA') as top1_sell_country
,coalesce(min(case when t1.row_number=2 then t1.sell_country end),'NA') as top2_sell_country
,coalesce(min(case when t1.row_number=3 then t1.sell_country end),'NA') as top3_sell_country
,coalesce(min(case when t2.row_number=1 then t2.category end),'NA') as top1_sell_category
,coalesce(min(case when t2.row_number=2 then t2.category end),'NA') as top2_sell_category
,coalesce(min(case when t2.row_number=3 then t2.category end),'NA') as top3_sell_category
FROM countries t1 
FULL JOIN categories t2
USING (website,date)
GROUP BY 1,2
ORDER BY 1,2

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