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3.8 组内复杂判断

 
SPL性能优化基础练习(38)

按月统计出当月发生过连续三天都有订单的用户数
SQL

WITH a AS(
    SELECT count (1) AS num, day (OrderDate) days,
        CustomerID,year(OrderDate) Years,month(OrderDate) Months
    FROM Orders
    WHERE OrderDate>='2022-01-01'
    GROUP BY CustomerID, day (OrderDate), year(OrderDate),month(OrderDate)), 
b AS (
    SELECT
        lag(days,1) OVER(PARTITION BY CustomerID, Years,Months ORDER BY days) pd, 
        days, CustomerID,Years,Months
    FROM a),
c AS (
    SELECT *,
        CASE WHEN days-pd=1 THEN 0 ELSE 1 END AS tmp 
    FROM b),
d AS (
    SELECT CustomerID,Years,Months,sum (tmp) OVER (
            PARTITION BY CustomerID,Years,Months ORDER BY days) NoOrderDays
    FROM c),
e AS (
    SELECT CustomerID, Years,Months,NoOrderDays,count (*) ContinuousDays 
    FROM d 
    GROUP BY CustomerID,Years,Months,NoOrderDays),
f AS (
    SELECT CustomerID,Years,Months,max (ContinuousDays) mCon 
    FROM e 
    GROUP BY CustomerID,Years,Months 
    HAVING max (ContinuousDays)>=3)
SELECT Years,Months,count (*) 
FROM f 
GROUP BY Years,Months

SPL

A
1 >st=date(“2022-01-01”),start=days@o(st)
2 =file(“Orders_Account.ctx”).open().cursor@x(CustomerID,OrderDate;OrderDate>=start)
3 =A2.group(CustomerID,month@y(OrderDate))
4 =A3.(~.select@1((a=case(OrderDate-OrderDate[-1],0:a,1:a+1;1))==3))
5 =A4.groups(year(OrderDate):Years,month(OrderDate):Months; count(1))

A3 有序读出每用户每月的订单数据形成小集合
A4 针对小集合计算出是否发生过连续三天下单的情况
A5 A4 返回数据已对 CustomerID 有序了，只要简单 count 就可以得到去重计数了

 

SPL性能优化基础练习(38)