Question

我正在尝试在python中编写一个脚本，从Data表中提取数据并填充Max表。使用Data表中的给定数据，脚本应填充Max表，如下所示。

Max_f（f代表未来）是当前项目之后的4个项目中的最大值。

Max_p（p代表过去）是当前之前4个项目中的最大值。

项目2013-08-13 13:19的示例：

max_f最多为21,24,28和30。

max_p最多为25,23,27和26。

前4个max_p项和最后4个max_f应该是n / a，因为数据表中没有足够的值来计算相应的最大值。

          Data                                 Max          
id  datetime          value |   id     datetime       max_f max_p
1   13-Aug-2013 13:15   25  |   1   13-Aug-2013 13:15   27  n/a
2   13-Aug-2013 13:16   23  |   2   13-Aug-2013 13:16   27  n/a
3   13-Aug-2013 13:17   27  |   3   13-Aug-2013 13:17   26  n/a
4   13-Aug-2013 13:18   26  |   4   13-Aug-2013 13:18   28  n/a
5   13-Aug-2013 13:19   25  |   5   13-Aug-2013 13:19   30  27
6   13-Aug-2013 13:20   21  |   6   13-Aug-2013 13:20   31  27
7   13-Aug-2013 13:21   24  |   7   13-Aug-2013 13:21   31  27
8   13-Aug-2013 13:22   28  |   8   13-Aug-2013 13:22   n/a 26
9   13-Aug-2013 13:23   30  |   9   13-Aug-2013 13:23   n/a 28
10  13-Aug-2013 13:24   31  |   10  13-Aug-2013 13:24   n/a 30
11  13-Aug-2013 13:25   29  |   11  13-Aug-2013 13:25   n/a 31

我一直在尝试使用SELECT进行INTERVAL次查询，但我不确定我是否以正确的方式解决问题。

如果有人能指出我正确的方向，那就太好了。

Answer 1

即使它绝对不是最有效的查询类，它也会这样做;它以4分钟的前后间隔为基础;

INSERT INTO `max` (`datetime`, `max_f`, `max_p`) 
  SELECT `data`.datetime, 
    IF(COUNT(DISTINCT f.datetime) < 4, NULL, MAX(f.value)),
    IF(COUNT(DISTINCT p.datetime) < 4, NULL, MAX(p.value))
  FROM data
  LEFT JOIN data f 
    ON f.datetime > data.datetime
   AND f.datetime < DATE_ADD(data.datetime, INTERVAL  5 MINUTE)
  LEFT JOIN data p
    ON p.datetime < data.datetime
   AND p.datetime > DATE_ADD(data.datetime, INTERVAL -5 MINUTE)
  GROUP BY data.datetime

An SQLfiddle to test with

Answer 2

如果您的id值确实是连续的，则可以执行以下操作：

select d.*,
       (case when sum(dnear.id < d.id) = 4
             then max(case when dnear.id <= d.id then dnear.value end)
        end) as max_p,
       (case when sum(dnear.id > d.id) = 4
             then max(case when dnear.id >= d.id then dnear.value end)
        end) as max_p
from data d left outer join
     data dnear
     on dnear.id between d.id - 4 and d.id + 4
group by d.id;

Answer 3

简单子查询：

SELECT Id,
datetime currdatetime, 
(SELECT Max(Value) FROM Data WHERE datetime < currdatetime AND (SELECT COUNT(Value) FROM Data WHERE datetime < currdatetime) > 4) as MaxP, 
(SELECT Max(Value) FROM Data WHERE datetime > currdatetime AND (SELECT COUNT(Value) FROM Data WHERE datetime > currdatetime) > 4) as MaxF 
FROM Data

Answer 4

如果规范要从之前的四行和之后的四行中获取最大值，而不管是否缺少特定分钟的行，则此查询将返回结果集：

SELECT d.id
     , IF( ( SELECT 4 AS count_f 
               FROM `Data` f
              WHERE f.datetime > d.datetime
              ORDER BY f.datetime ASC
              LIMIT 3,1
           )
         , GREATEST(
             ( SELECT f1.value FROM `Data` f1
                WHERE f1.datetime > d.datetime
                ORDER BY f1.datetime ASC LIMIT 0,1
             )
           , ( SELECT f2.value FROM `Data` f2
                WHERE f2.datetime > d.datetime
                ORDER BY f2.datetime ASC LIMIT 1,1
             )
           , ( SELECT f3.value FROM `Data` f3
                WHERE f3.datetime > d.datetime
                ORDER BY f3.datetime ASC LIMIT 2,1
             )
           , ( SELECT f4.value FROM `Data` f4
                WHERE f4.datetime > d.datetime
                ORDER BY f4.datetime ASC LIMIT 3,1
             )
           )
         , 'n/a'
         ) AS max_f
     , IF( ( SELECT 4 AS count_p 
               FROM `Data` p
              WHERE p.datetime < d.datetime
              ORDER BY p.datetime DESC 
              LIMIT 3,1
           )
         , GREATEST(
             ( SELECT p1.value FROM `Data` p1
                WHERE p1.datetime < d.datetime
                ORDER BY p1.datetime DESC LIMIT 0,1
             )
           , ( SELECT p2.value FROM `Data` p2
                WHERE p2.datetime < d.datetime
                ORDER BY p2.datetime DESC LIMIT 1,1
             )
           , ( SELECT p3.value FROM `Data` p3
                WHERE p3.datetime < d.datetime
                ORDER BY p3.datetime DESC LIMIT 2,1
             )
           , ( SELECT p4.value FROM `Data` p4
                WHERE p4.datetime < d.datetime
                ORDER BY p4.datetime DESC LIMIT 3,1
             )
           )
         , 'n/a'
         ) AS max_p
  -- , d.id
  -- , d.datetime
  -- , d.value
  FROM `Data` d
 ORDER BY d.id

但是......由于相关的子查询，这将成为大型集的轻度调光查询。这些子查询将依赖于以datetime为前导列的合适索引。

选择给定日期的过去和未来值

4 个答案: