如何在Impala中使用大量数据解决空白岛问题

问题描述 投票:0回答:1

[Impala表中有类型2维度,该表具有〜500M行,102列:(C1,C2,...,C8,... C100,Eff_DT,EXP_DT)只需要选择具有不同组合值(C1,C2,... C8)的行。对于每个选定的记录,EFF_DT和EXP_DT必须分别是该记录所属组的min(eff_dt)和max(eff_dt)(此处的组由不同的组合(C1,C2,..,C8定义)

简单的分组依据不会在这里解决问题,因为它将忽略同一分组的时滞...

为了简单起见,这是必需的,也是我尝试过的-假设只有2列定义了一个组(而不是8)这是一个输入,所需输出和仅使用group by ...]进行输出的示例

--INPUT                              --DESIRED OUTPUT                     --OUTPUT of SIMPLE GROUP BY
------------------------------------------------------------------------------------------------------------ 
C1  C2  EFF_DT      EXP_DT           C1   C2  Eff_dt      EXP_DT          C1   C2  EFF_DT       EXP_DT
4   8   2013-11-30  2014-01-22       4    8   2013-11-30  2014-01-22      4    8   2013-11-30   2999-12-31
2   8   2014-01-23  2014-01-23       2    8   2014-01-23  2014-01-23      2    8   2014-01-23   2014-01-23
4   8   2014-01-24  2015-12-31       4    8   2014-01-24  2999-12-31
4   8   2016-01-01  2016-12-31
4   8   2017-01-01  2018-03-15
4   8   2018-03-16  2018-07-24
4   8   2018-07-25  2999-12-31

试图在select语句中使用子查询来基于当前行选择max(exp_dt),但由于impala不起作用而无法工作.....

这是我尝试过的查询,它运行正常,但在Impala中不起作用(因为在select语句中不支持子查询)>

SELECT    
     T0.C1,
     T0.C2,
     MIN(T0.EFF_DT) AS MIN_EFF_DT,
     T0.EXP_DT
FROM (
    SELECT 
    T1.C1,
    T1.C2,
    T1.EFF_DT,
    (
        SELECT MAX(T2.EXP_DT)
        FROM (select * from TABLE_NAME ) T2
        WHERE T2.C1 = T1.C1
        AND   T2.C2 = T1.C2
        AND NOT EXISTS (
        SELECT 1 FROM (select * from TABLE_NAME) T3
            WHERE T3.EXP_DT < T2.EXP_DT 
            AND   T3.EXP_DT > T1.EXP_DT
            AND  (T3.C1 <> T2.C1 OR T3.C2 <> T2.C2 )
        )

    ) EXP_DT
    FROM (select * from TABLE_NAME) T1
) T0 
GROUP BY 
T0.C1,
T0.C2,
T0.EXP_DT
ORDER BY MIN_EFF_DT ASC

具有在Impala表中存在类型2维度,该表具有〜500M行,具有102列:(C1,C2,...,C8,... C100,Eff_DT,EXP_DT)仅需要选择具有不同组合的行的值...

sql impala
1个回答
0
投票

[很可能在修改id列后,以前的解决方案将起作用:

select id, c1, c2, min(eff_dt), max(exp_dt)
from (select t.*,
             row_number() over (partition by id order by eff_dt) as seqnum,
             row_number() over (partition by id, c1, c2 order by eff_dt) as seqnum_1
      from t
     ) t
group by id, c1, c2, (seqnum - seqnum_1);
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