SQL平均值和零点

我有用于传感器记录数据的应用程序，我希望能够以从多个传感器的平均值，可以是一个，两个，三个或大量...SQL平均值和零点

编辑：这些是温度传感器，因此0是传感器可能存储为数据库中值的值。

我最初的出发点是这样的SQL查询：

SELECT grid.t5||'.000000' as ts, 
avg(t.sensorvalue) sensorvalue1 
, avg(w.sensorvalue)AS sensorvalue2 
FROM 
(SELECT generate_series(min(date_trunc('hour', ts))       
,max(ts), interval '5 min') AS t5 FROM device_history_20865735 where  
ts between '2015/05/13 09:00' and '2015/05/14 09:00' ) grid 

LEFT JOIN device_history_20865735 t ON t.ts >= grid.t5 AND t.ts < grid.t5 + interval '5 min' 
LEFT JOIN device_history_493417852 w ON w.ts >= grid.t5 AND w.ts < grid.t5 + interval '5 min' 
--WHERE t.sensorvalue notnull 
GROUP BY grid.t5 ORDER BY grid.t5

我得到5点分钟的平均值，因为它是我的应用程序更好。

结果如预期具有用于任一sensorvalue1或2 NULL值：

ts;sensorvalue1;sensorvalue2 
"2015-05-13 09:00:00.000000";19.9300003051758; 
"2015-05-13 09:05:00.000000";20; 
"2015-05-13 09:10:00.000000";; 
"2015-05-13 09:15:00.000000";20.0599994659424; 
"2015-05-13 09:20:00.000000";; 
"2015-05-13 09:25:00.000000";20.1200008392334;

我的目标是从所有可用的传感器计算，每次5分钟间隔的平均，从而空值是我想的问题使用CASE语句，这样如果有一个NULL，以获得其他传感器的值...

SELECT grid.t5||'.000000' as ts, 
CASE 
     WHEN avg(t.sensorvalue) ISNULL THEN avg(w.sensorvalue) 
     ELSE avg(t.sensorvalue) 
END AS sensorvalue 
, 
CASE 
     WHEN avg(w.sensorvalue) ISNULL THEN avg(t.sensorvalue) 
     ELSE avg(w.sensorvalue) 
END AS sensorvalue2 
FROM 
(SELECT generate_series(min(date_trunc('hour', ts)),max(ts), interval '5 min') AS t5 
FROM device_history_20865735 where  
ts between '2015/05/13 09:00' and '2015/05/14 09:00' ) grid 

LEFT JOIN device_history_20865735 t ON t.ts >= grid.t5 AND t.ts < grid.t5 + interval '5 min' 
LEFT JOIN device_history_493417852 w ON w.ts >= grid.t5 AND w.ts < grid.t5 + interval '5 min' 
GROUP BY grid.t5 ORDER BY grid.t5

但随后计算平均值我要做的另一选择在此之上每列数devide （又名传感器），如果他们只是两个，那就OK，但如果是因为可能有多个传感器每行有NULL值...

SQL是使用Postgres 9.4从应用程序（使用Python）语法派生的，所以有一个简单的方法为了实现我所需要的，我觉得我正在走上一条相当复杂的路线......？

编辑＃2：有了您的输入我已经产生这个SQL代码，又似乎相当复杂，但开到你的想法和审查，如果它是可靠的，可维护：

SELECT ts, sensortotal, sensorcount, 
CASE 
    WHEN sensorcount = 0 THEN -1000 
    ELSE sensortotal/sensorcount 
END AS sensorAvg 

FROM (
    WITH grid as (
      SELECT t5 
      FROM (SELECT generate_series(min(date_trunc('hour', ts)), max(ts), interval '5 min') as t5 
       FROM device_history_20865735 
       ) d 
      WHERE t5 between '2015-05-13 09:00' and '2015-05-14 09:00' 
     ) 
    SELECT d1.t5 || '.000000' as ts 
      , Coalesce(avg(d1.sensorvalue), 0) + Coalesce(avg(d2.sensorvalue),0) as sensorTotal 
      , (CASE 
        WHEN avg(d1.sensorvalue) ISNULL THEN 0 
        ELSE 1 
      END + CASE 
      WHEN avg(d2.sensorvalue) ISNULL THEN 0 
      ELSE 1 
      END) as sensorCount 

    FROM (SELECT grid.t5, avg(t.sensorvalue) as sensorvalue 
      FROM grid LEFT JOIN 
       device_history_20865735 t 
       ON t.ts >= grid.t5 AND t.ts <grid.t5 + interval '5 min' 
      GROUP BY grid.t5 
     ) d1 LEFT JOIN 
     (SELECT grid.t5, avg(t.sensorvalue) as sensorvalue 
      FROM grid LEFT JOIN 
       device_history_493417852 t 
       ON t.ts >= grid.t5 AND t.ts <grid.t5 + interval '5 min' 
     GROUP BY grid.t5 
     ) d2 on d1.t5 = d2.t5 
    GROUP BY d1.t5 
    ORDER BY d1.t5 
) tmp;

谢谢！

来源

2015-05-14 Kostas

我不知道如何计算平均值，但你可以做'（coalesce（sum（t.sensorvalue），0）+ coalesce（sum（w.sensorvalue），0））/ count（t.sensorvalue）+ count（（w.sensorvalue））''。这可以很容易地扩展到任何数量的传感器。 – dnoeth

谢谢@dnoeth！我需要在网格的每一行计算它，例如每5分钟，而不是整个列... – Kostas

为了得到精确的平均值，则需要分别计算每一个之前联接：

WITH grid as (
     SELECT t5 
     FROM (SELECT generate_series(min(date_trunc('hour', ts)), max(ts), interval '5 min') as t5 
      FROM device_history_20865735 
      ) d 
     WHERE t5 between '2015-05-13 09:00' and '2015-05-14 09:00' 
    ) 
SELECT d1.t5 || '.000000' as ts, 
     avg(d1.sensorvalue) as sensorvalue1 
     , avg(d2.sensorvalue) as sensorvalue2 
FROM (SELECT grid.t5, avg(t.sensorvalue) as sensorvalue 
     FROM grid LEFT JOIN 
      device_history_20865735 t 
      ON t.ts >= grid.t5 AND t.ts <grid.t5 + interval '5 min' 
     GROUP BY grid.t5 
    ) d1 LEFT JOIN 
    (SELECT grid.t5, avg(t.sensorvalue) as sensorvalue 
     FROM grid LEFT JOIN 
      device_history_493417852 t 
      ON t.ts >= grid.t5 AND t.ts <grid.t5 + interval '5 min' 
    GROUP BY grid.t5 
    ) d2 on d1.t5 = d2.t5 
GROUP BY d1.t5 
ORDER BY d1.t5;

来源

2015-05-14 10:35:46

谢谢@戈登！ - 虽然我得到语法错误...错误：在“GROUP”处或附近的语法错误行21：GROUP BY d1.t5 ^ – Kostas

在LEFT JOIN中d1和d2之间没有关系，ON条件丢失 –

我已经设法让它运行，结果与上面的SQL完全相同......在平均值问题上，虽然有更“优雅”的解决方案吗？ :) – Kostas

这听起来像你想是这样的：

(coalesce(value1,0) + coalesce(value2,0) + coalesce(value3,0))/
(value1 IS NOT NULL::int + value2 IS NOT NULL::int + value3 IS NOT NULL::int) 
AS average

基本上，只是做你想为每一行做数学。唯一“棘手”的部分是如何“计数”非空值 - 我使用了一个强制转换，但还有其他选项，如：

CASE WHEN value1 IS NULL THEN 0 ELSE 1 END

来源

2015-07-23 00:41:01

SQL平均值和零点

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