将时间戳截短为任意间隔 [英] Truncate timestamp to arbitrary intervals

问题描述

我想在时间轴上比在数据库中具有更低的分辨率来绘制 my_values 。我使用以下SQL来获取每个时间间隔的平均值（例如，一个小时）：

I want to plot my_values with some lower resolution on the time axis than I have in the database. I use something like the following SQL to get the average values per time interval (e.g., an hour):

SELECT DATE_TRUNC('hour', my_timestamps) AS my_time_lowres
     , AVG(my_values) 
FROM my_table 
GROUP BY my_time_lowres

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使用 date_trunc（）可以在一定程度上减少时间戳的分辨率（根据文档）：

---

Using date_trunc() it is possible to reduce the resolution of the timestamp to a certain degree (according to the docs):

select date_trunc('hour', timestamp '2001-02-16 20:38:40') 
-- the output is: 2001-02-16 20:00:00

这样，我可以为以下间隔大小（和一些较大/较小的大小）：

This way, I can do this for the following interval sizes (and some larger/smaller sizes):

...
second
minute
hour
day
week
...

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有没有一种方法来实现这个其他时间int例如3小时6小时？

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Is there a way to achieve this for other time intervals as well, e.g., 3 hours, 6 hours?

推荐答案

考虑这个演示将时间戳降低到15分钟并且导致重复的结果：

Consider this demo to bring timestamps down to a resolution of 15 minutes and aggregate resulting dupes:

WITH tbl(id, ts) AS ( VALUES
    (1::int, '2012-10-04 00:00:00'::timestamp)
   ,(2, '2012-10-04 18:23:01')
   ,(3, '2012-10-04 18:30:00')
   ,(4, '2012-10-04 18:52:33')
   ,(5, '2012-10-04 18:55:01')
   ,(6, '2012-10-04 18:59:59')
   ,(7, '2012-10-05 11:01:01')
   )
SELECT to_timestamp((extract(epoch FROM ts)::bigint / 900)*900)::timestamp
                                                            AS lower_bound
     , to_timestamp(avg(extract(epoch FROM ts)))::timestamp AS avg_ts
     , count(*) AS ct
FROM   tbl
GROUP  BY 1
ORDER  BY 1;

结果：

     lower_bound     |       avg_ts        | ct
---------------------+---------------------+----
 2012-10-04 00:00:00 | 2012-10-04 00:00:00 |  1
 2012-10-04 18:15:00 | 2012-10-04 18:23:01 |  1
 2012-10-04 18:30:00 | 2012-10-04 18:30:00 |  1
 2012-10-04 18:45:00 | 2012-10-04 18:55:51 |  3
 2012-10-05 11:00:00 | 2012-10-05 11:01:01 |  1

诀窍是提取一个像@迈克尔已经发布的unix纪元。整数除法将所有分辨率的数据块合并在一起，因为小数位数被截断。

The trick is to extract a unix epoch like @Michael already posted. Integer division lumps them together in buckets of the chosen resolution, because fractional digits are truncated.

除以900，因为15分钟= 900秒。

I divide by 900, because 15 minutes = 900 seconds.

乘以相同的数字，得到最终的 lower_bound 。
使用 将unix纪元转换回时间戳code> to_timestamp（）

Multiply by the same number to get the resulting lower_bound. Convert the unix epoch back to a timestamp with to_timestamp().

这对于没有分数十进制数字。对于更多的多功能性，使用经常被忽视的功能 width_bucket（） ，就像我在这个最近的密切相关的答案。更多解释，链接和sqlfiddle演示。

This works great for intervals that can be represented without fractional digits in the decimal system. For even more versatility use the often overlooked function width_bucket() like I demonstrate in this recent, closely related answer. More explanation, links and an sqlfiddle demo over there.

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