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我有以下问题...

    Time |  A   |  B |  C  --  Sum should be
      1     a1     b1   c1     a1 + b1 + c1
      2     a2     b2   x      a2 + b1 + c1 
      3     a3     x    x      a3 + b2 + c1
      4     x      b3   c2     a3 + b3 + c2

本质上,总和需要跨越三行中每一行的最新时间值。每个数据列不一定有当前时间的值。

我尝试了几种使用窗口函数的方法,但均未成功。我已经编写了一个存储过程来满足我的需要,但它很慢。

CREATE OR REPLACE FUNCTION timeseries.combine_series(id int[], startTime timestamp, endTime timestamp) 
RETURNS setof RECORD AS $$
DECLARE
    retval double precision = 0;
    row_data timeseries.total_active_energy%ROWTYPE;
    maxCount integer = 0;
    sz integer = 0;
lastVal double precision[];
v_rec RECORD;
BEGIN   
    SELECT INTO sz array_length($1,1);

    FOR row_data IN SELECT * FROM timeseries.total_active_energy  WHERE time >= startTime AND time < endTime AND device_id = ANY($1) ORDER BY time
       LOOP
    retval = row_data.active_power;
    for i IN 1..sz LOOP
        IF $1[i]=row_data.device_id THEN
            lastVal[i] = row_data.active_power;
        ELSE
            retval = retVal + COALESCE(lastVal[i],0);
        END IF;
    END LOOP;

    SELECT row_data.time, retval into v_rec;

    return next v_rec;
     END LOOP;

      return ;
  END;
$$ LANGUAGE plpgsql;

称呼:

select * from timeseries.combine_series('{552,553,554}'::int[], '2013-05-01'::timestamp, '2013-05-02'::timestamp) 
    AS (t timestamp with time zone, val double precision);
4

1 回答 1

1
SELECT ts, a, b, c
       , COALESCE(max(a) OVER (PARTITION BY grp_a), 0)
       + COALESCE(max(b) OVER (PARTITION BY grp_b), 0)
       + COALESCE(max(c) OVER (PARTITION BY grp_c), 0) AS special_sum
FROM  (
   SELECT *
         ,count(a) OVER w AS grp_a
         ,count(b) OVER w AS grp_b
         ,count(c) OVER w AS grp_c
   FROM   t
   WINDOW w AS (ORDER BY ts)
   ) sub
ORDER  BY ts;

NULL首先,使用聚合窗口函数将实际值和后续值放在一个组中count():它不会随NULL值递增。

然后max()从每个组中获取,到达你正在寻找的东西。此时您也可以使用min()or sum(),因为每组只有一个非空值。

COALESCE()如果总的第一个时间值是 ,则捕获NULLNULL

请注意我是如何选择ts列名的,因为我不使用基本类型名称time作为标识符。

测试用例

这也是你们每个人首先应该提供样本数据的方式!

CREATE TEMP TABLE t (ts int, a int, b int, c int);

INSERT INTO t VALUES
  (1, 11,   21,   NULL)
 ,(2, 12,   22,   NULL)
 ,(3, 13,   NULL, NULL)
 ,(4, NULL, 23,   32);
于 2013-06-20T14:15:59.743 回答