选择满足hive条件的上一行

时间：2018-06-04 20:18:09

标签： hive lag window-functions moving-average

我有这样的产品数据

Product   Date            Sales   Availbility
    xyz      2017-12-31      724.5   6.0
    xyz      2018-01-07      362.25  7.0
    xyz      2018-01-14      281.75  7.0
    xyz      2018-01-21      442.75  7.0
    xyz      2018-01-28      442.75  6.0
    xyz      2018-02-04      402.5   7.0
    xyz      2018-02-11      201.25  3.0
    xyz      2018-02-18      120.75  0.0
    xyz      2018-02-25      40.25   0.0
    xyz      2018-03-11      201.25  0.0
    xyz      2018-03-18      483.0   5.0
    xyz      2018-03-25      322.0   7.0
    xyz      2018-04-01      241.5   7.0
    xyz      2018-04-08      281.75  7.0
    xyz      2018-04-15      523.25  7.0
    xyz      2018-04-22      241.5   7.0
    xyz      2018-04-29      362.25  7.0

数据没有订购（一个小问题），我想要做的是，无论我们在可用性列（第4列）中有0，我想采取前3周（具有完全可用性，即7）平均< / p>

如下所示：

xyz      2017-12-31      724.5   6.0     Null 
xyz      2018-01-07      362.25  7.0     362.25 ( Same value for weeks with availbility = 7) 
xyz      2018-01-14      281.75  7.0     281.75
xyz      2018-01-21      442.75  7.0     442.75 
xyz      2018-01-28      442.75  6.0     361 (362 + 281 + 362/3)the prior fully availble week avg which is avilble)
xyz      2018-02-04      402.5   7.0     402
xyz      2018-02-11      201.25  3.0     375 (402 + 442 + 281 /3)
xyz      2018-02-18      120.75  0.0     375 ( Same since 375 is the most recent 4 fully availble average)
xyz      2018-02-25      40.25   0.0     375
xyz      2018-03-11      201.25  0.0     375
xyz      2018-03-18      483.0   5.0     375
xyz      2018-03-25      322.0   7.0     322
xyz      2018-04-01      241.5   7.0     241
xyz      2018-04-08      281.75  7.0     281
xyz      2018-04-15      523.25  7.0     523
xyz      2018-04-22      241.5   7.0     241
xyz      2018-04-29      362.25  7.0     362

我通过试图找到仅有完全可用周数的3周平均值并将其与剩余的周数联合起来，然后尝试使用滞后函数来检索最近的平均值来接近它。

select a.*,lag(case when a.Full_availble_sales >0 then a.Full_availble_sales end,1) over (partition by a.asin order by a.week_beginning) as Four_wk_avg  from (select asin,week_beginning,avg(sales) as weekly_sales,sum(available_to_purchase) as weekly_availbility,0 as Full_availble_sales from t1 where asin = 'xyz' group by asin,week_beginning having sum(available_to_purchase) < 7
union all
select t.asin,t.week_beginning,t.weekly_sales,t.weekly_availbility,avg(t.weekly_sales) over (partition by t.asin order by t.week_beginning rows between 3 preceding and current row ) as Full_availble_sales from 
(select asin,week_beginning,avg(sales) as weekly_sales,sum(available_to_purchase) as weekly_availbility from t1 where asin = 'xyz' group by asin,week_beginning having sum(available_to_purchase) = 7)t ) a  order by a.week_beginning

O / P 

xyz      2017-12-31      724.5   6.0     0.0     NULL
xyz      2018-01-07      362.25  7.0     362.25  NULL
xyz      2018-01-14      281.75  7.0     322.0   362.25
xyz      2018-01-21      442.75  7.0     362.25  322.0
xyz      2018-01-28      442.75  6.0     0.0     362.25
xyz      2018-02-04      402.5   7.0     372.3125        NULL
xyz      2018-02-11      201.25  3.0     0.0     372.3125
xyz      2018-02-18      120.75  0.0     0.0     NULL
xyz      2018-02-25      40.25   0.0     0.0     NULL
xyz      2018-03-11      201.25  0.0     0.0     NULL
xyz      2018-03-18      483.0   5.0     0.0     NULL
xyz      2018-03-25      322.0   7.0     362.25  NULL
xyz      2018-04-01      241.5   7.0     352.1875        362.25
xyz      2018-04-08      281.75  7.0     311.9375        352.1875
xyz      2018-04-15      523.25  7.0     342.125 311.9375
xyz      2018-04-22      241.5   7.0     322.0   342.125
xyz      2018-04-29      362.25  7.0     352.1875        322.0

这不是我想要的。

1 个答案:

答案 0 :(得分：0)

这将完成工作（在窗口上使用聚合函数T& operator=(const T& other)和avg）

max_by

结果与要求完全相同：

WITH
   tt1  (Product,Date_week_beginning,Sales,Availbility) AS 
      ( SELECT * FROM ( VALUES 
      ('xyz','2017-12-31',  724.5   ,6.0),
      ('xyz','2018-01-07',  362.25  ,7.0),
      ('xyz','2018-01-14',  281.75  ,7.0),
      ('xyz','2018-01-21',  442.75  ,7.0),
      ('xyz','2018-01-28',  442.75  ,6.0),
      ('xyz','2018-02-04',  402.5   ,7.0),
      ('xyz','2018-02-11',  201.25  ,3.0),
      ('xyz','2018-02-18',  120.75  ,0.0),
      ('xyz','2018-02-25',  40.25   ,0.0),
      ('xyz','2018-03-11',  201.25  ,0.0),
      ('xyz','2018-03-18',  483.0   ,5.0),
      ('xyz','2018-03-25',  322.0   ,7.0),
      ('xyz','2018-04-01',  241.5   ,7.0),
      ('xyz','2018-04-08',  281.75  ,7.0),
      ('xyz','2018-04-15',  523.25  ,7.0),
      ('xyz','2018-04-22',  241.5   ,7.0),
      ('xyz','2018-04-29',  362.25  ,7.0) )  
   ), tt2 AS (
      SELECT *, avg(sales) OVER (partition by Product order by if(Availbility = 7.0,1),Date_week_beginning ROWS BETWEEN 2 PRECEDING AND CURRENT ROW) avg3
      FROM tt1
   )
   SELECT Product,Date_week_beginning,Sales,Availbility, 
          CASE WHEN Availbility = 7.0 THEN Sales
             ELSE
                 max_by(if(Availbility = 7.0,avg3),(Availbility = 7.0, Date_week_beginning) ) OVER (partition by Product order by Date_week_beginning)
             END new_col   
   FROM tt2
   ORDER BY Product,Date_week_beginning

我假设问题中有两个错别字（可以从示例中验证）：

  1. 问题中的这一行是错误的：Product Date_week_beginning Sales Availbility new_col xyz 2017-12-31 724.5 6.0 NULL xyz 2018-01-07 362.25 7.0 362.25 xyz 2018-01-14 281.75 7.0 281.75 xyz 2018-01-21 442.75 7.0 442.75 xyz 2018-01-28 442.75 6.0 362.25 xyz 2018-02-04 402.5 7.0 402.5 xyz 2018-02-11 201.25 3.0 375.6666666666667 xyz 2018-02-18 120.75 0.0 375.6666666666667 xyz 2018-02-25 40.25 0.0 375.6666666666667 xyz 2018-03-11 201.25 0.0 375.6666666666667 xyz 2018-03-18 483.0 5.0 375.6666666666667 xyz 2018-03-25 322.0 7.0 322.0 xyz 2018-04-01 241.5 7.0 241.5 xyz 2018-04-08 281.75 7.0 281.75 xyz 2018-04-15 523.25 7.0 523.25 xyz 2018-04-22 241.5 7.0 241.5 xyz 2018-04-29 362.25 7.0 362.25 应该是361 (362 + 281 + 362/3)the prior fully availble week avg which is avilble)

  2. 句子：“我想做的是在可用性列（第4列）中我们有0的任何地方。”应为“我想做的是我们不要< / strong>的可用性列（第4列）中有 7.0 。”

相关问题
最新问题