Sql 查找不同行中2个数字之间的差异

我有一个Postgres表，记录每分钟的int值请求计数。 我在一些服务器上有一些请求类型，它们都在同一个表中：

time                |  key1    | key2      | key3   | value     
-----------------------------------------------------------------------
2017-01-16 18:00:53 | server1  | webpage1  | type1  | 30
2017-01-16 18:00:55 | server1  | webpage2  | type1  | 31
2017-01-16 18:00:58 | server1  | webpage3  | type1  | 32
2017-01-16 18:00:59 | server1  | webpage4  | type1  | 33
2017-01-16 18:01:00 | server1  | webpage5  | type1  | 34
2017-01-16 18:01:01 | server1  | webpage6  | type1  | 35
2017-01-16 18:01:02 | server1  | webpage7  | type1  | 36
2017-01-16 18:01:03 | server1  | webpage8  | type1  | 37
2017-01-16 18:01:04 | server1  | webpage1  | type1  | 56
2017-01-16 18:01:06 | server1  | webpage2  | type1  | 35
2017-01-16 18:01:07 | server1  | webpage3  | type1  | 43
2017-01-16 18:01:10 | server1  | webpage4  | type1  | 64
2017-01-16 18:01:13 | server1  | webpage5  | type1  | 44
2017-01-16 18:01:14 | server1  | webpage6  | type1  | 66
2017-01-16 18:01:16 | server1  | webpage7  | type1  | 56
2017-01-16 18:01:18 | server1  | webpage8  | type1  | 22

假设key1和key3也有不同的值。为了这个示例，我发出了一些数据

我需要的结果是组key1，key2，key3上的最新值减去最新值的1偏移量的差值[我需要每分钟的速率]

我成功地在同一个表中获得了按键分组的最新偏移量和1偏移量的结果：

SELECT * FROM 
(SELECT ROW_NUMBER() 
        OVER(PARTITION BY key1, key2, key3 ORDER BY time DESC) as rnum,
 time, key1, key2, key3, value FROM test ORDER BY time DESC) a
WHERE rnum < 3;

现在，我想我可以取MINtime和MAXtime的value列并计算diff，但是我不能合并这些行

在@HartCO评论之后，我能够做到：

select time, new_val-last_val, key1, key2, key3 from
  (select distinct max(time)  over(partition by key1, key2, key3) as time,
          max(value) over(partition by key1, key2, key3) as new_val,
          min(value) over(partition by key1, key2, key3) as last_val,
          key1, key2, key3
   from (select row_number() over(partition by key1, key2, key3 order by time desc) as rnum, 
                time, key1, key2, key3, value from test order by time desc) a 
where rnum < 3) b;

---

但是在网页8上，所需的输出应该是-15，而不是22。

这些行之间的差异被一定量的偏移，最好使用窗口函数来处理。如果您的表不是很大，则可以获取最新的值。请注意，DISTINCT ON是一个Postgresql扩展

SELECT DISTINCT ON (key1, key2, key3)
       time,
       key1,
       key2,
       key3,
       value - lag(value) OVER (PARTITION BY key1, key2, key3 ORDER BY time) 
FROM test
ORDER BY key1, key2, key3, time DESC;

这给了我们

        time         |    key1    |    key2     |   key3   | ?column? 
---------------------+------------+-------------+----------+----------
 2017-01-16 18:01:04 |  server1   |  webpage1   |  type1   |       26
 2017-01-16 18:01:06 |  server1   |  webpage2   |  type1   |        4
 2017-01-16 18:01:07 |  server1   |  webpage3   |  type1   |       11
 2017-01-16 18:01:10 |  server1   |  webpage4   |  type1   |       31
 2017-01-16 18:01:13 |  server1   |  webpage5   |  type1   |       10
 2017-01-16 18:01:14 |  server1   |  webpage6   |  type1   |       31
 2017-01-16 18:01:16 |  server1   |  webpage7   |  type1   |       20
 2017-01-16 18:01:18 |  server1   |  webpage8   |  type1   |      -15
(8 rows)

当然，您可以使用well解决方案，例如左连接

WITH diffs AS (
    SELECT time,
           key1,
           key2,
           key3,
           value - lag(value) OVER (PARTITION BY key1, key2, key3 ORDER BY time)
    FROM test)
SELECT d1.*
FROM diffs d1
LEFT JOIN diffs d2
  ON (d1.key1, d1.key2, d1.key3) = (d2.key1, d2.key2, d2.key3)
 -- This allows us to single out the greatest row
 AND d1.time < d2.time
WHERE d2.time IS NULL
-- Ordering is just for show
ORDER BY d1.key1, d1.key2, d1.key3;

---

使用Postgresql 9.5，规划人员识别出了这种模式，并使用反连接作为最终的查询计划。使用NOT EXISTS也可以得到类似的结果。

您似乎很接近，就像您使用ROW\U NUMBER OVER一样，您可以使用MINtime OVER。。。对于MAX`来说也是一样的，以获取那些与PARTITION by子句定义的行组相关的值，这两个子句不使用ORDER by。@HartCO-你能给出MAX和MIN的例子吗？@HartCO-请看我的编辑。但这还不够好。如果计数器减少，我将无法捕捉到它，我正在执行maxval和minval，并且它与maxtime和MinTime不相关。您是否可以获取样本数据并显示所需的输出以帮助澄清？请参阅。这将更容易使用。

        time         |    key1    |    key2     |   key3   | ?column? 
---------------------+------------+-------------+----------+----------
 2017-01-16 18:01:04 |  server1   |  webpage1   |  type1   |       26
 2017-01-16 18:01:06 |  server1   |  webpage2   |  type1   |        4
 2017-01-16 18:01:07 |  server1   |  webpage3   |  type1   |       11
 2017-01-16 18:01:10 |  server1   |  webpage4   |  type1   |       31
 2017-01-16 18:01:13 |  server1   |  webpage5   |  type1   |       10
 2017-01-16 18:01:14 |  server1   |  webpage6   |  type1   |       31
 2017-01-16 18:01:16 |  server1   |  webpage7   |  type1   |       20
 2017-01-16 18:01:18 |  server1   |  webpage8   |  type1   |      -15
(8 rows)

WITH diffs AS (
    SELECT time,
           key1,
           key2,
           key3,
           value - lag(value) OVER (PARTITION BY key1, key2, key3 ORDER BY time)
    FROM test)
SELECT d1.*
FROM diffs d1
LEFT JOIN diffs d2
  ON (d1.key1, d1.key2, d1.key3) = (d2.key1, d2.key2, d2.key3)
 -- This allows us to single out the greatest row
 AND d1.time < d2.time
WHERE d2.time IS NULL
-- Ordering is just for show
ORDER BY d1.key1, d1.key2, d1.key3;