1

我的数据集在 teradata 中如下所示：

╔═══════════╦══════════╦══════╗
║ studentid ║   date   ║ days ║
╠═══════════╬══════════╬══════╣
║      1000 ║ 2/1/2017 ║   25 ║
║      1000 ║ 3/8/2017 ║   30 ║
║      1000 ║ 4/4/2017 ║   80 ║
║      1000 ║ 5/1/2017 ║   81 ║
║      1001 ║ 1/1/2017 ║   60 ║
║      1001 ║ 2/1/2017 ║   20 ║
║      1001 ║ 4/1/2017 ║   81 ║
╚═══════════╩══════════╩══════╝

如果最近的两个日期为 80 或 81，我希望有一个新列（标志）应该在行上指示 1。如果不是 0。

对于学生 1001，所有行都应该为 0，因为最后两个日期不是 80 或 81。它需要取最后两个日期。即使 1001 有 81，倒数第二个日期有 20，所以标志需要为 0

期望的输出：

╔═══════════╦══════════╦══════╦══════╗
║ studentid ║   date   ║ days ║ flag ║
╠═══════════╬══════════╬══════╬══════╣
║      1000 ║ 2/1/2017 ║   25 ║    0 ║
║      1000 ║ 3/8/2017 ║   30 ║    0 ║
║      1000 ║ 4/4/2017 ║   80 ║    1 ║
║      1000 ║ 5/1/2017 ║   81 ║    1 ║
║      1001 ║ 1/1/2017 ║   60 ║    0 ║
║      1001 ║ 2/1/2017 ║   20 ║    0 ║
║      1001 ║ 4/1/2017 ║   81 ║    0 ║
╚═══════════╩══════════╩══════╩══════╝
4

2 回答 2

0

对于前两行，您可以应用一个简单的逻辑，这将导致解释中的单个STAT 步骤

如果当前行是第一行：检查此行和下一行是否都包含这些值之一

如果当前行是第二行：检查这行和上一行是否都包含这些值之一

SELECT studentid, date_, Days,
   CASE Row_Number()
        Over (PARTITION BY studentid
              ORDER BY date DESC)
      WHEN 1 
         THEN CASE WHEN Days IN (80,81)
--                    AND Min(Days) Over (PARTITION BY studentid ORDER BY date DESC ROWS BETWEEN 1 Following AND 1 Following) IN (80,81)
                    AND Lead(Days) Over (PARTITION BY studentid ORDER BY date DESC) IN (80,81)
                   THEN 1
                   ELSE 0
              END
      WHEN 2
         THEN CASE WHEN Days IN (80,81) 
--                    AND Min(Days) Over (PARTITION BY studentid ORDER BY date DESC ROWS BETWEEN 1 Preceding AND 1 Preceding) IN (80,81)
                    AND Lag(Days) Over (PARTITION BY studentid ORDER BY date DESC) IN (80,81)
                   THEN 1
                   ELSE 0
              END
      ELSE 0
   END AS flag
FROM tab

如果您的 Teradata 版本不支持lead/lag使用 thenmin语法。

但是，如果您需要将此逻辑应用于 >2 行，则需要一种更通用的方法：

SELECT studentid, date, Days,
   -- check if the first n rows contain only searched values
   CASE WHEN x IS NOT NULL THEN Min(x) Over (PARTITION BY studentid) ELSE 0 END AS flag
FROM
 (
   SELECT studentid, date_, Days,
      CASE
         WHEN Row_Number()
              Over (PARTITION BY studentid
                    ORDER BY date DESC) BETWEEN 1 AND 2   -- only for the first n days
         THEN CASE WHEN Days IN (80,81) THEN 1 ELSE 0 END  -- flag the searched values
      END AS x
   FROM tab AS t
 ) AS dt
于 2018-10-18T08:47:46.417 回答
0

分配行号，row_number然后获取每个学生 ID 的最后 2 行的min和max值。此后，使用case表达式检查条件以分配标志。

select studentid,dt,days
,case when rnum in (1,2) and max_days_latest_2 in (80,81) and min_days_latest_2 in (80,81) then 1 else 0 end as flag
from (select t.*
      ,max(case when rnum in (1,2) then days end) over(partition by studentid) as max_days_latest_2
      ,min(case when rnum in (1,2) then days end) over(partition by studentid) as min_days_latest_2
      from (select t.*,row_number() over(partition by studentid order by dt desc) as rnum
            from tbl t
           ) t
     ) t
于 2018-10-18T00:57:30.033 回答