Question

我正在尝试解决工作中的趋势问题，与下面的示例非常相似。我想我有一个方法，但不知道如何在SQL中做到这一点。

输入数据为：

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MTD上的趋势来产生输入的输出，而不依赖于它。这样，当新的月份添加到输入时，输出会更改而不会编辑查询。

TRENDS的逻辑将出现在每个唯一的LOC_ID上。趋势将有三个值：第一个月的“新”RAINED为“Y”，RAINED为“Y”的任何后续月份为“Carryover”，RAINED为“N”的任何月份为“No Rain”。

我想通过引入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，为了在输出中产生“TRENDS”，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

我的问题是我不确定如何为每个唯一的LOC_ID生成PREV_RAINED。我觉得需要通过MTD将LAG（）语句和按LOC_ID顺序的分区组合起来，但我需要做的滞后数量取决于每个月。

是否有一种简单的方法可以生成PREV_RAINED或更简单的方法来解决我的整体问题，同时每个月保留自动化？

感谢您阅读所有这些内容！：）

Answer 1

在下面的SQL中有两部分。

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

通过计算上述两个，我们可以编写CASE WHEN逻辑，根据您的要求计算趋势。

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。

接下来我将处理MATCH_RECOGNIZE解决方案，我可以将其添加到此答案中。

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
;

<强>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Oracle 12c）。测试数据集上的不同解决方案;我被告知MATCH_RECOGNIZE可能比其他解决方案快得多，但这取决于很多因素。

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;

Oracle SQL趋势MTD数据

3 个答案: