Oracle SQL Trending MTD数据

Question

我正试图解决工作中的一个趋势问题，非常类似于下面的例子。我想我有一个方法，但不知道如何在SQL中实现它。

输入数据是：

MTD         LOC_ID  RAINED
1-Apr-16    1       Y
1-Apr-16    2       N
1-May-16    1       N
1-May-16    2       N
1-Jun-16    1       N
1-Jun-16    2       N
1-Jul-16    1       Y
1-Jul-16    2       N
1-Aug-16    1       N
1-Aug-16    2       Y

所需的输出是：

MTD         LOC_ID  RAINED  TRENDS
1-Apr-16    1       Y       New
1-May-16    1       N       No Rain
1-Jun-16    1       N       No Rain
1-Jul-16    1       Y       Carryover
1-Aug-16    1       N       No Rain
1-Apr-16    2       N       No Rain
1-May-16    2       N       No Rain
1-Jun-16    2       N       No Rain
1-Jul-16    2       N       No Rain
1-Aug-16    2       Y       New

我试图通过不依赖MTD的趋势从输入中产生输出。这样，当将新的月份添加到输入中时，输出就会发生变化，而无需编辑查询。

趋势的逻辑将出现在每一个独特的LOC_ID上。趋势将有三个值:第一个月下雨的“新”是"Y"，接下来的几个月里“下雨”是"Y"，“没有雨”在任何月份都是"N“。

我想通过引入一个带有listagg的中间步骤来自动化这个问题。例如，对于LOC_ID = "1"：

MTD         LOC_ID  RAINED  PREV_RAINED
1-Apr-16    1       Y       (null) / 0 / (I don't care)
1-May-16    1       N       Y
1-Jun-16    1       N       Y;N
1-Jul-16    1       Y       Y;N;N
1-Aug-16    1       N       Y;N;N;Y

这样，在产出上产生“趋势”，我可以说：

case when RAINED = 'Y' then
    case when not regexp_like(PREV_RAINED, 'Y', 'i') then
        'New'
    else
        'Carryover'
    end
else
    'No Rain'
end as TRENDS

我的问题是，我不知道如何为每个唯一的PREV_RAINED生成LOC_ID，我觉得它需要按照MTD的LOC_ID顺序组合LOC_ID()语句和分区，但是我需要做的延迟的数量取决于每个月。

是否有一种简单的方法来生成PREV_RAINED，还是一个更简单的方法来解决我的整体问题，同时每个月保持自动化？

谢谢你阅读所有这些！:)

Answer 1

在下面的SQL中有两个部分。

(i) Calculating the ROWNUMBER value for rained attribute at loc_id,rained level.
(ii) Get the count at partition level loc_id,rained.

通过计算上述两种情况，我们可以编写逻辑时根据您的需求计算趋势的情况。

SELECT mtd,
       loc_id,
       rained,
       CASE WHEN rained = 'N' THEN 'No Rain'
            WHEN rained = 'Y' AND rn = 1 THEN 'New'
            ELSE 'Carry Over'    
        END AS Trends       
  FROM
        ( 
            SELECT mtd,
                   loc_id,
                   rained,                   
                   ROW_NUMBER() OVER ( PARTITION BY loc_id,rained ORDER BY mtd ) AS rn,
                   COUNT(*) OVER ( PARTITION BY loc_id,rained ) AS count_locid_rained               
              FROM INPUT
              ORDER BY loc_id,mtd,rained,rn
         ) X;

Answer 2

以下是旧版本的解决方案。WITH子句用于输入数据；解决方案在WITH子句之后立即开始。

接下来我将研究一个MATCH_RECOGNIZE解决方案，我可以将它添加到这个答案中。

with
     input_data ( mtd, loc_id, rained ) as (
       select to_date('1-Apr-16', 'dd-Mon-rr'), 1, 'Y' from dual union all
       select to_date('1-Apr-16', 'dd-Mon-rr'), 2, 'N' from dual union all
       select to_date('1-May-16', 'dd-Mon-rr'), 1, 'N' from dual union all
       select to_date('1-May-16', 'dd-Mon-rr'), 2, 'N' from dual union all
       select to_date('1-Jun-16', 'dd-Mon-rr'), 1, 'N' from dual union all
       select to_date('1-Jun-16', 'dd-Mon-rr'), 2, 'N' from dual union all
       select to_date('1-Jul-16', 'dd-Mon-rr'), 1, 'Y' from dual union all
       select to_date('1-Jul-16', 'dd-Mon-rr'), 2, 'N' from dual union all
       select to_date('1-Aug-16', 'dd-Mon-rr'), 1, 'N' from dual union all
       select to_date('1-Aug-16', 'dd-Mon-rr'), 2, 'Y' from dual
     )
select mtd, loc_id, rained,
       case rained when 'N' then 'No Rain'
                   else case when rn = 1 then 'New' 
                                         else 'Carryover' end
                   end  as trends
from ( select mtd, loc_id, rained, 
              row_number() over (partition by loc_id, rained order by mtd) rn
       from   input_data
)
order by loc_id, mtd
;

输出

MTD                     LOC_ID RAINED TRENDS  
------------------- ---------- ------ ---------
01/04/2016 00:00:00          1      Y New      
01/05/2016 00:00:00          1      N No Rain  
01/06/2016 00:00:00          1      N No Rain  
01/07/2016 00:00:00          1      Y Carryover
01/08/2016 00:00:00          1      N No Rain  
01/04/2016 00:00:00          2      N No Rain  
01/05/2016 00:00:00          2      N No Rain  
01/06/2016 00:00:00          2      N No Rain  
01/07/2016 00:00:00          2      N No Rain  
01/08/2016 00:00:00          2      Y New      

 10 rows selected

Answer 3

使用MATCH_RECOGNIZE的解决方案(仅适用于Oracle12c)。在您的数据集中测试不同的解决方案；我被告知MATCH_RECOGNIZE可能比其他解决方案快得多，但这取决于许多因素。

select loc_id, mtd, rained, trends
from input_data
  match_recognize (
    partition by loc_id, rained
    order by     mtd
    measures     mtd as mtd,
                 case when rained = 'N' then 'No Rain'
                      else case when match_number() = 1 then 'New' else 'Carryover' end
                      end as trends
    pattern (a)
    define a as 0 = 0
  )
order by loc_id, mtd;