Question

我正在尝试分析大量交易数据，并设置了一系列不同的等级来帮助我。我无法理解的是受益人等级。我希望它按时间顺序（而不是按字母顺序）划分受益人发生变化的地方。

从1月至3月支付相同的受益人，然后在6月再次支付，我希望将6月归为单独的“会话”。

如果有帮助，我正在使用Teradata SQL。

我以为解决方案将是DENSE_RANK，但如果我PARTITION BY (CustomerID, Beneficiary) ORDER BY SystemDate，它可以算几个月。如果我PARTITION BY (CustomerID) ORDER BY Beneficiary不是按时间顺序排列的，那么我需要最高的排名才能成为最新的Beneficiary。

SELECT CustomerID, Beneficiary, Amount, SystemDate, Month
  ,RANK() OVER(PARTITION BY CustomerID ORDER BY SystemDate ASC) AS PaymentRank
  ,RANK() OVER(PARTITION BY CustomerID ORDER BY PaymentMonth ASC) AS MonthRank
  ,RANK() OVER(PARTITION BY CustomerID , Beneficiary ORDER BY SystemDate ASC) AS Beneficiary
  ,RANK() OVER(PARTITION BY CustomerID , Beneficiary, ROUND(TRNSCN_AMOUNT, 0) ORDER BY SYSTEM_DATE ASC) AS TransRank
FROM table ORDER BY CustomerID, PaymentRank

CustomerID  Beneficiary Amount  DateStamp   Month   PaymentRank MonthRank   BeneficiaryRank TransactionRank
a   aa  10      Jan 1   1   1   1
a   aa  20      Feb 2   2   2   1
a   aa  20      Mar 3   3   3   2
a   aa  20      Apr 4   4   4   3
a   bb  20      May 5   5   1   1
a   bb  30      Jun 6   6   2   1
a   aa  30      Jul 7   7   5   2
a   aa  30      Aug 8   8   6   1
a   cc  5       Sep 9   9   1   1
a   cc  5       Oct 10  10  2   2
a   cc  5       Nov 11  11  3   3
b   cc  5       Dec 1   1   1   1

到目前为止，这是我想要的，我希望旁边有一列，如下所示

CustomerID  Beneficiary Amount  DateStamp   Month   NewRank
a   aa  10      Jan 1
a   aa  20      Feb 1
a   aa  20      Mar 1
a   aa  20      Apr 1
a   bb  20      May 2
a   bb  30      Jun 2
a   aa  30      Jul 3
a   aa  30      Aug 3
a   cc  5       Sep 4
a   cc  5       Oct 4
a   cc  5       Nov 4
b   cc  5       Dec 1

Answer 1

这是一种空白和岛屿问题。我建议使用lag()和累计金额：

select t.*,
       sum(case when prev_systemdate > systemdate - interval '1' month then 0 else 1 end) over (partition by customerid, beneficiary order by systemdate)            
from (select t.*,
             lag(systemdate) over (partition by customerid, beneficiary order by systemdate) as prev_systemdate
      from t
     ) t

Answer 2

SELECT dt.*,
   -- now do a Cumulative Sum over those 0/1
   SUM(flag)
   OVER(PARTITION BY CustomerID
        ORDER BY SystemDate ASC
                ,flag DESC -- needed if the order by columns are not unique
        ROWS UNBOUNDED PRECEDING) AS NewRank
FROM
 ( 
    SELECT CustomerID, Beneficiary, Amount, SystemDate, Month
      ,RANK() OVER(PARTITION BY CustomerID ORDER BY SystemDate ASC) AS PaymentRank
      ,RANK() OVER(PARTITION BY CustomerID ORDER BY PaymentMonth ASC) AS MonthRank
      ,RANK() OVER(PARTITION BY CustomerID , Beneficiary ORDER BY SystemDate ASC) AS Beneficiary
      ,RANK() OVER(PARTITION BY CustomerID , Beneficiary, ROUND(TRNSCN_AMOUNT, 0) ORDER BY SYSTEM_DATE ASC) AS TransRank
      -- assign a 0 if current & previous Beneficiary are the same, otherwise 1 
      ,CASE WHEN Beneficiary = LAG(Beneficiary) OVER(PARTITION BY CustomerID ORDER BY SystemDate) THEN 0 ELSE 1 END AS flag
    FROM table 
 ) AS dt
ORDER BY CustomerID, PaymentRank

您的Gordon查询问题可能是由您的Teradata版本引起的，LAG仅在16.10+中受支持。但是有一个简单的解决方法：

LAG(Beneficiary) OVER(PARTITION BY CustomerID ORDER BY SystemDate)

--is equivalent to 
MIN(Beneficiary) OVER(PARTITION BY CustomerID ORDER BY SystemDate
                      ROWS BETWEEN 1 PRECEDING AND 1 PRECEDING))

Answer 3

@Gordon和@dnoeth提供了使我步入正轨的构想和代码，以示感谢。

以下内容大部分都是从技巧中删除的，但需要先添加无界的ROWS才能正确进行聚合。没有这个，它只是显示分区的总数。我也将systemdate更改为paymentrank，因为我不得不在一天中重复输入一些东西。

SELECT dt.*,
   -- now do a Cumulative Sum over those 0/1
   SUM(flag) OVER(PARTITION BY CustomerID ORDER BY PaymentRank ASC ROWS UNBOUNDED PRECEDING) AS NewRank
FROM
 ( 
    SELECT CustomerID, Beneficiary, Amount, SystemDate, Month
      -- assign a 0 if current & previous Beneficiary are the same, otherwise 1 
      ,CASE WHEN  Beneficiary = MIN(Beneficiary) OVER (PARTITION BY CustomerID ORDER BY PaymentRank ASC  ROWS BETWEEN 1 PRECEDING AND 1 PRECEDING)  THEN 0 ELSE 1 END  AS Flag ) AS dt
ORDER BY CustomerID, PaymentRank

每当受益人更改时，内部查询都会设置一个标志。然后，外部查询会对它们进行累加。

我不确定前面的内容在做什么，@ dnoeth有一个很好的解释here以下是该解释的来源。

•UNBOUNDED PRECEDING，当前行之前的所有行->已固定

•UNBOUNDED FOLLOWING，当前行之后的所有行->已固定

•x PRECEDING，当前行之前的x行->相对

•y跟随，当前行之后的y行->相对

无法解决具有多个变量的分区上排名

3 个答案: