4

我有一张这样的桌子。

ID (integer)
event_name(varchar(20))
event_date(timestamp)

下面给出了一些示例数据。

ID         event_date                          event_name
101        2013-04-24 18:33:37.694818          event_A
102        2013-04-24 20:34:37.000000          event_B
103        2013-04-24 20:40:37.000000          event_A
104        2013-04-25 01:00:00.694818          event_A
105        2013-04-25 12:00:15.694818          event_A
106        2013-04-26 00:56:10.800000          event_A
107        2013-04-27 12:00:15.694818          event_A
108        2013-04-27 12:00:15.694818          event_B

我需要生成基于窗口的报告。这里窗口代表一组行。例如：如果我选择窗口大小为2，我需要连续显示两天的每个事件的总计数，即同一天和前一天。如果我选择窗口大小 3 ，我需要连续三天生成每个事件的计数。

所以如果选择 2 天窗口，结果应该如下所示。

Date                                       Count_eventA                 Count_eventB
2013-04-27 (this counts sum of 27th, 26th)       2                           1 
2013-04-26 (this counts sum of 26th, 25th)       3                           0
2013-04-25 (this counts sum of 25th, 24th)       4                           1
2013-04-24 (this counts sum of 24th      )       2                           1

我已经阅读了 postgres 中的窗口函数。有人可以指导我如何为此报告编写 sql 查询！

4

1 回答 1

6

您想将count聚合用作窗口函数，例如count(id) over (partition by event_date rows 3 preceeding)……但是由于数据的性质，它变得非常复杂。您正在存储时间戳，而不仅仅是日期，并且您希望按天而不是按先前事件的数量进行分组。最重要的是，您需要对结果进行交叉制表。

如果 PostgreSQLRANGE在窗口函数中支持，这将比现在简单得多。事实上，你必须以艰难的方式去做。

然后，您可以通过窗口对其进行过滤以获取每个事件每天的滞后计数...除了您的事件天不连续，不幸的是 PostgreSQL 窗口函数仅支持ROWS，不支持RANGE，因此您必须加入生成的系列首先是日期。

WITH
/* First, get a listing of event counts by day */
event_days(event_name, event_day, event_day_count) AS (
        SELECT event_name, date_trunc('day', event_date), count(id)
        FROM Table1
        GROUP BY event_name, date_trunc('day', event_date)
        ORDER BY date_trunc('day', event_date), event_name
),
/* 
 * Then fill in zeros for any days within the range that didn't have any events.
 * If PostgreSQL supported RANGE windows, not just ROWS, we could get rid of this/
 */
event_days_contiguous(event_name, event_day, event_day_count) AS (
        SELECT event_names.event_name, gen_day, COALESCE(event_days.event_day_count,0)
        FROM generate_series( (SELECT min(event_day)::date FROM event_days), (SELECT max(event_day)::date FROM event_days), INTERVAL '1' DAY ) gen_day
        CROSS JOIN (SELECT DISTINCT event_name FROM event_days) event_names(event_name)
        LEFT OUTER JOIN event_days ON (gen_day = event_days.event_day AND event_names.event_name = event_days.event_name)
),
/*
 * Get the lagged counts by using the sum() function over a row window...
 */
lagged_days(event_name, event_day_first, event_day_last, event_days_count) AS (
        SELECT event_name, event_day, first_value(event_day) OVER w, sum(event_day_count) OVER w
        FROM event_days_contiguous
        WINDOW w AS (PARTITION BY event_name ORDER BY event_day ROWS 1 PRECEDING)
)
/* Now do a manual pivot. For arbitrary column counts use an external tool
 * or check out the 'crosstab' function in the 'tablefunc' contrib module 
 */
SELECT d1.event_day_first, d1.event_days_count AS "Event_A", d2.event_days_count AS "Event_B"
FROM lagged_days d1
INNER JOIN lagged_days d2 ON (d1.event_day_first = d2.event_day_first AND d1.event_name = 'event_A' AND d2.event_name = 'event_B')
ORDER BY d1.event_day_first;

输出样本数据：

    event_day_first     | Event_A | Event_B 
------------------------+---------+---------
 2013-04-24 00:00:00+08 |       2 |       1
 2013-04-25 00:00:00+08 |       4 |       1
 2013-04-26 00:00:00+08 |       3 |       0
 2013-04-27 00:00:00+08 |       2 |       1
(4 rows)

通过将三个 CTE 子句组合成一个嵌套查询，您可以潜在地使查询更快但更丑陋，使用FROM (SELECT...)并将它们包装在视图中而不是 CTE 以供外部查询使用。这将允许 Pg 将谓词“下推”到查询中，从而大大减少您在查询数据子集时必须处理的数据。

SQLFiddle 目前似乎无法正常工作，但这是我使用的演示设置：

CREATE TABLE Table1 
(id integer primary key, "event_date" timestamp not null, "event_name" text);

INSERT INTO Table1
("id", "event_date", "event_name")
VALUES
(101, '2013-04-24 18:33:37', 'event_A'),
(102, '2013-04-24 20:34:37', 'event_B'),
(103, '2013-04-24 20:40:37', 'event_A'),
(104, '2013-04-25 01:00:00', 'event_A'),
(105, '2013-04-25 12:00:15', 'event_A'),
(106, '2013-04-26 00:56:10', 'event_A'),
(107, '2013-04-27 12:00:15', 'event_A'),
(108, '2013-04-27 12:00:15', 'event_B');

我将最后一个条目的 ID 从 107 更改为 108，因为我怀疑这只是您手动编辑中的错误。

以下是如何将其表达为视图：

CREATE VIEW lagged_days AS
SELECT event_name, event_day AS event_day_first, sum(event_day_count) OVER w AS event_days_count 
FROM (
        SELECT event_names.event_name, gen_day, COALESCE(event_days.event_day_count,0)
        FROM generate_series( (SELECT min(event_date)::date FROM Table1), (SELECT max(event_date)::date FROM Table1), INTERVAL '1' DAY ) gen_day
        CROSS JOIN (SELECT DISTINCT event_name FROM Table1) event_names(event_name)
        LEFT OUTER JOIN (
                SELECT event_name, date_trunc('day', event_date), count(id)
                FROM Table1
                GROUP BY event_name, date_trunc('day', event_date)
                ORDER BY date_trunc('day', event_date), event_name
        ) event_days(event_name, event_day, event_day_count)
        ON (gen_day = event_days.event_day AND event_names.event_name = event_days.event_name)
) event_days_contiguous(event_name, event_day, event_day_count)
WINDOW w AS (PARTITION BY event_name ORDER BY event_day ROWS 1 PRECEDING);

然后，您可以在要编写的任何交叉表查询中使用该视图。它将与先前的手动交叉表查询一起使用：

SELECT d1.event_day_first, d1.event_days_count AS "Event_A", d2.event_days_count AS "Event_B"
FROM lagged_days d1
INNER JOIN lagged_days d2 ON (d1.event_day_first = d2.event_day_first AND d1.event_name = 'event_A' AND d2.event_name = 'event_B')
ORDER BY d1.event_day_first;

...或者crosstab从tablefunc扩展中使用，我会让你学习。

为了一笑，这是explain上面基于视图的查询：http ://explain.depesz.com/s/nvUq

于 2013-06-10T05:12:01.053 回答