如何获取 24 小时数据作为 Postgresql 中的单个列？

Question

我有一个如下所示的表net_score （仅提供了一个示例）

            date            |       ip        | up_score 
----------------------------+-----------------+----------
 2022-09-09 07:30:04.485979 | 12.22.19.0      |       51
 2022-09-09 07:30:04.485979 | 10.22.39.1      |       95
 2022-09-09 07:30:04.485979 | 14.260.13.1     |      100
 2022-09-09 07:30:04.485979 | 252.229.219.43  |       97
 2022-09-09 07:30:04.485979 | 10.551.343.10   |       97
 2022-09-09 08:30:04.485979 | 12.22.19.0      |       11
 2022-09-09 08:30:04.485979 | 10.22.39.1      |       54
 2022-09-09 08:30:04.485979 | 14.260.13.1     |       89
 2022-09-09 08:30:04.485979 | 252.229.219.43  |       37
 2022-09-09 08:30:04.485979 | 10.551.343.10   |       11
 2022-09-09 09:30:04.485979 | 12.22.19.0      |       54
 2022-09-09 09:30:04.485979 | 10.22.39.1      |       15
 2022-09-09 09:30:04.485979 | 14.260.13.1     |       90
 2022-09-09 09:30:04.485979 | 252.229.219.43  |       17
 2022-09-09 09:30:04.485979 | 10.551.343.10   |       50

如您所见，我有一个date 、 ip和up_score列，并且每小时计算一次数据。

我想要的是一种获取24小时每小时数据的方法，每个ip有24 个单独的列，如果数据不存在，只需为该列输入-1 。

我可以通过执行以下操作来获取每小时数据

select date_trunc('hour', date) as hourly, ip, up_score 
from net_score 
where date between '2022-09-09 05:30:00' and '2022-09-10 05:30:00' 
group by ip, hourly, up_score 
order by ip, hourly

但我预期的查询应该产生一个表 output 如下所示（并用-1填充缺失的小时值）

       ip       | hour_0  |  hour_1  |  hour_2  |  ..
--------------------------+--------- +----------+------
12.22.19.0      |   51    |    11    |     54   |  ..
10.22.39.1      |   95    |    54    |     15   |  ..
14.260.13.1     |   100   |    89    |     90   |  ..
252.229.219.43  |   97    |    37    |     17   |  ..
10.551.343.10   |   97    |    11    |     50   |  ..

我怎样才能做到这一点？

笔记：

这样做的原因是，通过使用我知道的查询获取 output 将返回很多行，并且对于添加的每个新ip ， Z78E6221F6393D1356681DB398F24ZZ行将再增加 6 行。 随着我处理越来越多的ips ，获取这个结果并处理它（我在我的代码中使用python ）将需要更长的时间。

但是，如果我按照我提到的方式（24 列方式）尝试它，我将只处理每个ip的额外行，并且列数继续仅为 2 4 。 因此，我正在考虑这样做。 如果我的方法错误或可以改进，请纠正我。

Answer 1

您可以通过以下方式使用PIVOT技术：

select 
    date_trunc('day', date) as daily, ip, 
    coalesce(sum(up_score) filter (where extract('hour' from date) = 6 ), 0) up_score_6,
    sum(up_score) filter (where extract('hour' from date) = 7 ) up_score_7,
    sum(up_score) filter (where extract('hour' from date) = 8 ) up_score_8,
    sum(up_score) filter (where extract('hour' from date) = 9 ) up_score_9
from net_score 
where date between '2022-09-09 05:30:00' and '2022-09-10 05:30:00' 
group by ip, daily
order by ip, daily

当然，您需要为每个 24 小时扩展我的示例查询

在线 SQL 编辑器

+=====================+================+============+============+============+============+
| daily               | ip             | up_score_6 | up_score_7 | up_score_8 | up_score_9 |
+=====================+================+============+============+============+============+
| 2022-09-09 00:00:00 | 10.22.39.1     | 0          | 95         | 54         | 15         |
+---------------------+----------------+------------+------------+------------+------------+
| 2022-09-09 00:00:00 | 10.551.343.10  | 0          | 97         | 11         | 50         |
+---------------------+----------------+------------+------------+------------+------------+
| 2022-09-09 00:00:00 | 12.22.19.0     | 0          | 51         | 11         | 54         |
+---------------------+----------------+------------+------------+------------+------------+
| 2022-09-09 00:00:00 | 14.260.13.1    | 0          | 100        | 89         | 90         |
+---------------------+----------------+------------+------------+------------+------------+
| 2022-09-09 00:00:00 | 252.229.219.43 | 0          | 97         | 37         | 17         |
+---------------------+----------------+------------+------------+------------+------------+

当使用旧的 PostgreSQL 版本时，可以使用 CASE 语句：

select 
    date_trunc('day', date) as daily, ip, 
    sum(case when extract('hour' from date) = 6 then up_score else 0 end) up_score_6,
    sum(case when extract('hour' from date) = 7 then up_score else 0 end) up_score_7,
    sum(case when extract('hour' from date) = 8 then up_score else 0 end) up_score_8,
    sum(case when extract('hour' from date) = 9 then up_score else 0 end) up_score_9
from net_score 
where date between '2022-09-09 05:30:00' and '2022-09-10 05:30:00' 
group by ip, daily
order by ip, daily