Question

我有时态数据库，我想获得时间序列数据的每小时平均值。

示例数据：

1     -1.64 2007-09-29 00:01:09
2     -1.76 2007-09-29 00:03:09
3     -1.83 2007-09-29 00:05:09
4     -1.86 2007-09-29 00:07:09
5     -1.94 2007-09-29 00:09:09
6     -1.87 2007-09-29 00:11:09
7     -1.87 2007-09-29 00:13:09
8     -1.80 2007-09-29 00:15:09
9     -1.64 2007-09-29 00:17:09
10    -1.60 2007-09-29 00:19:09
11    -1.90 2007-09-29 00:21:09
12    -2.08 2007-09-29 00:23:09
13    -1.94 2007-09-29 00:25:09
14    -2.12 2007-09-29 00:27:09
15    -1.87 2007-09-29 00:29:09
16    -2.18 2007-09-29 00:31:09
17    -1.98 2007-09-29 00:33:09
18    -1.73 2007-09-29 00:35:09
19    -1.84 2007-09-29 00:37:09
20    -2.04 2007-09-29 00:39:09
21    -1.86 2007-09-29 00:41:09
22    -1.94 2007-09-29 00:43:09
23    -1.77 2007-09-29 00:45:09
24    -1.78 2007-09-29 00:47:09
25    -1.50 2007-09-29 00:49:09
26    -1.46 2007-09-29 00:51:09
27    -1.72 2007-09-29 00:53:09
28    -1.67 2007-09-29 00:55:09
29    -1.56 2007-09-29 00:57:09
30    -1.69 2007-09-29 00:59:09
31    -1.97 2007-09-29 01:01:09
32    -1.79 2007-09-29 01:03:09
33    -1.79 2007-09-29 01:05:09
34    -1.84 2007-09-29 01:07:09
35    -1.91 2007-09-29 01:09:09
36    -1.87 2007-09-29 01:11:09
37    -1.98 2007-09-29 01:13:09
38    -1.83 2007-09-29 01:15:09
39    -1.88 2007-09-29 01:17:09
40    -1.88 2007-09-29 01:19:09
41    -1.78 2007-09-29 01:21:09
42    -1.78 2007-09-29 01:23:09
43    -1.66 2007-09-29 01:25:09
44    -1.70 2007-09-29 01:27:09
45    -1.46 2007-09-29 01:29:09
46    -1.36 2007-09-29 01:31:09
47    -1.40 2007-09-29 01:33:09
48    -1.34 2007-09-29 01:35:09
49    -1.34 2007-09-29 01:37:09
50    -1.30 2007-09-29 01:39:09
51    -1.36 2007-09-29 01:41:09
52    -1.40 2007-09-29 01:43:09
53    -1.43 2007-09-29 01:45:09
54    -1.38 2007-09-29 01:47:09
55    -1.40 2007-09-29 01:49:09
56    -1.42 2007-09-29 01:51:09
57    -1.47 2007-09-29 01:53:09
58    -1.66 2007-09-29 01:55:09
59    -1.84 2007-09-29 01:57:09
60    -1.92 2007-09-29 01:59:09
61    -1.88 2007-09-29 02:01:09
62    -2.11 2007-09-29 02:03:09
63    -1.91 2007-09-29 02:05:09
64    -2.04 2007-09-29 02:07:09
65    -1.94 2007-09-29 02:09:09
66    -1.92 2007-09-29 02:11:09
67    -1.80 2007-09-29 02:13:09
68    -1.74 2007-09-29 02:15:09
69    -1.74 2007-09-29 02:17:09
70    -1.76 2007-09-29 02:19:09
71    -1.74 2007-09-29 02:21:09
72    -1.80 2007-09-29 02:23:09
73    -1.80 2007-09-29 02:25:09
74    -1.80 2007-09-29 02:27:09
75    -1.82 2007-09-29 02:29:09
76    -1.90 2007-09-29 02:31:09
77    -1.93 2007-09-29 02:33:09
78    -2.06 2007-09-29 02:35:09
79    -2.08 2007-09-29 02:37:09
80    -1.95 2007-09-29 02:39:09
81    -1.98 2007-09-29 02:41:09
82    -2.32 2007-09-29 02:43:09
83    -1.86 2007-09-29 02:45:09
84    -1.97 2007-09-29 02:47:09
85    -1.64 2007-09-29 02:49:09
86    -2.00 2007-09-29 02:51:09
87    -1.48 2007-09-29 02:53:09
88    -1.74 2007-09-29 02:55:09
89    -1.85 2007-09-29 02:57:09
90    -1.82 2007-09-29 02:59:09
91    -1.82 2007-09-29 03:01:09
92    -1.92 2007-09-29 03:03:09
93    -1.80 2007-09-29 03:05:09
94    -1.54 2007-09-29 03:07:09
95    -1.36 2007-09-29 03:09:09
96    -1.50 2007-09-29 03:11:09
97    -1.59 2007-09-29 03:13:09
98    -1.60 2007-09-29 03:15:09
99    -1.58 2007-09-29 03:17:09
100   -1.81 2007-09-29 03:19:09
101   -2.16 2007-09-29 03:21:09
102   -1.97 2007-09-29 03:23:09
103   -1.94 2007-09-29 03:25:09
104   -2.29 2007-09-29 03:27:09
105   -2.46 2007-09-29 03:29:09
106   -2.42 2007-09-29 03:31:09
107   -2.34 2007-09-29 03:33:09
108   -2.38 2007-09-29 03:35:09
109   -2.44 2007-09-29 03:37:09
110   -2.28 2007-09-29 03:39:09
111   -2.24 2007-09-29 03:41:09
112   -2.26 2007-09-29 03:43:09

应根据以下时间间隔执行聚合： HH =（HH-1）：41 -HH：40 示例：13 =观察期12:41至13:40

Answer 1

应该像这样工作：

SELECT  date_trunc('hour', ts + interval '20 min') AS h
       ,avg(val) as avg_val
FROM    t
GROUP   BY 1
ORDER   BY 1;

我添加20分钟，然后用date_trunc()将时间缩小，然后按它聚合请注意，边框时间08:40最终以9点的平均值结束。

常规每小时网格

...包括基表中没有任何行的小时数：

SELECT *
FROM   generate_series('2007-09-28 22:00'  -- first hour
                      ,'2007-09-29 05:00'  -- last hour
                      ,interval '1h') AS h
LEFT   JOIN (
    SELECT  date_trunc('hour', ts + interval '20 min') AS h
           ,avg(val) as avg_val
    FROM    t
    GROUP   BY 1
    ORDER   BY 1
    ) x USING (h);

使用generate_series()和LEFT JOIN。

-> Updated the sqlfiddle.

Answer 2

我的系统很慢，所以我不能给你一个sql小提琴。请尝试以下方法。假设你只需要获得每小时的平均值。

SELECT datecol, DATEPART(hour,timescolumn) as hourcol, AVG(valuecol)
FROM Yourtable
GROUP BY hourcol, dateCol;

修改

归功于@Erwin。我使用他的sqlfiddle来运行我的查询。有几个语法错误要做。它只是按小时计算。

SQLFIDDLE DEMO

Postegres查询：

SELECT DATE(ts) AS DT, DATE_PART('hour', ts) as hourcol, avg(val) FROM t GROUP BY hourcol, DT ORDER BY hourcol ;

结果：

DT HOURCOL AVG September, 29 2007 00:00:00+0000 0 -1.814666666667 September, 29 2007 00:00:00+0000 1 -1.638 September, 29 2007 00:00:00+0000 2 -1.879333333333 September, 29 2007 00:00:00+0000 3 -1.986363636364

PostgreSQL中的小时平均聚合HH =（HH-1）：41 - HH：40格式

2 个答案:

常规每小时网格