每日查询计数，并具有多周的日期限制

2023-12-14

我正在尝试每天查找 # 个活跃用户。

用户在创建后即处于活动状态多于每周 10 个请求，共 4 个连续几周.

IE。 2014 年 10 月 31 日，如果用户在以下时间段内每周发出的请求总数超过 10 个，则该用户处于活动状态：

2014年10月24日至10月30日AND
2014年10月17日-10月23日AND
2014年10月10日-10月16日AND
2014年10月3日-10月9日

我有一张桌子requests:

CREATE TABLE requests (
  id text PRIMARY KEY, -- id of the request
  amount bigint,       -- sum of requests made by accounts_id to recipient_id,
                       -- aggregated on a daily basis based on "date"
  accounts_id text,    -- id of the user
  recipient_id text,   -- id of the recipient
  date timestamp       -- date that the request was made in YYYY-MM-DD
);

样本值：

INSERT INTO requests2
VALUES
    ('1',  19, 'a1', 'b1', '2014-10-05 00:00:00'),
    ('2',  19, 'a2', 'b2', '2014-10-06 00:00:00'),
    ('3',  85, 'a3', 'b3', '2014-10-07 00:00:00'),
    ('4',  11, 'a1', 'b4', '2014-10-13 00:00:00'),
    ('5',  2,  'a2', 'b5', '2014-10-14 00:00:00'),
    ('6',  50, 'a3', 'b5', '2014-10-15 00:00:00'),
    ('7',  787323, 'a1', 'b6', '2014-10-17 00:00:00'),
    ('8',  33, 'a2', 'b8', '2014-10-18 00:00:00'),
    ('9',  14, 'a3', 'b9', '2014-10-19 00:00:00'),
    ('10', 11, 'a4', 'b10', '2014-10-19 00:00:00'),
    ('11', 1628, 'a1', 'b11', '2014-10-25 00:00:00'),
    ('13', 101, 'a2', 'b11', '2014-10-25 00:00:00');

输出示例：

Date       | # Active users
-----------+---------------
10-01-2014 | 600
10-02-2014 | 703
10-03-2014 | 891

以下是我尝试查找特定日期（例如 2014 年 1 月 10 日）的活跃用户数量的方法：

SELECT count(*)
FROM
  (SELECT accounts_id
   FROM requests
   WHERE "date" BETWEEN '2014-10-01'::date - interval '2 weeks' AND '2014-10-01'::date - interval '1 week'
   GROUP BY accounts_id HAVING sum(amount) > 10) week_1
JOIN
  (SELECT accounts_id
   FROM requests
   WHERE "date" BETWEEN '2014-10-01'::date - interval '3 weeks' AND '2014-10-01'::date - interval '2 week'
   GROUP BY accounts_id HAVING sum(amount) > 10) week_2 ON week_1.accounts_id = week_2.accounts_id
JOIN
  (SELECT accounts_id
   FROM requests
   WHERE "date" BETWEEN '2014-10-01'::date - interval '4 weeks' AND '2014-10-01'::date - interval '3 week'
   GROUP BY accounts_id HAVING sum(amount) > 10) week_3 ON week_2.accounts_id = week_3.accounts_id
JOIN
  (SELECT accounts_id
   FROM requests
   WHERE "date" BETWEEN '2014-10-01'::date - interval '5 weeks' AND '2014-10-01'::date - interval '4 week'
   GROUP BY accounts_id HAVING sum(amount) > 10) week_4 ON week_3.accounts_id = week_4.accounts_id

由于这只是获取 1 天号码的查询，因此我需要随着时间的推移每天获取该号码。我认为这个想法是进行连接来获取日期，所以我尝试做这样的事情：

SELECT week_1."Date_series",
       count(*)
FROM
  (SELECT to_char(DAY::date, 'YYYY-MM-DD') AS "Date_series",
          accounts_id
   FROM generate_series('2014-10-01'::date, CURRENT_DATE, '1 day') DAY, requests
   WHERE to_char(DAY::date, 'YYYY-MM-DD')::date BETWEEN requests.date::date - interval '2 weeks' AND requests.date::date - interval '1 week'
   GROUP BY "Date_series",
            accounts_id HAVING sum(amount) > 10) week_1
JOIN
  (SELECT to_char(DAY::date, 'YYYY-MM-DD') AS "Date_series",
          accounts_id
   FROM generate_series('2014-10-01'::date, CURRENT_DATE, '1 day') DAY, requests
   WHERE to_char(DAY::date, 'YYYY-MM-DD')::date BETWEEN requests.date::date - interval '3 weeks' AND requests.date::date - interval '2 week'
   GROUP BY "Date_series",
            accounts_id HAVING sum(amount) > 10) week_2 ON week_1.accounts_id = week_2.accounts_id
AND week_1."Date_series" = week_2."Date_series"
JOIN
  (SELECT to_char(DAY::date, 'YYYY-MM-DD') AS "Date_series",
          accounts_id
   FROM generate_series('2014-10-01'::date, CURRENT_DATE, '1 day') DAY, requests
   WHERE to_char(DAY::date, 'YYYY-MM-DD')::date BETWEEN requests.date::date - interval '4 weeks' AND requests.date::date - interval '3 week'
   GROUP BY "Date_series",
            accounts_id HAVING sum(amount) > 10) week_3 ON week_2.accounts_id = week_3.accounts_id
AND week_2."Date_series" = week_3."Date_series"
JOIN
  (SELECT to_char(DAY::date, 'YYYY-MM-DD') AS "Date_series",
          accounts_id
   FROM generate_series('2014-10-01'::date, CURRENT_DATE, '1 day') DAY, requests
   WHERE to_char(DAY::date, 'YYYY-MM-DD')::date BETWEEN requests.date::date - interval '5 weeks' AND requests.date::date - interval '4 week'
   GROUP BY "Date_series",
            accounts_id HAVING sum(amount) > 10) week_4 ON week_3.accounts_id = week_4.accounts_id
AND week_3."Date_series" = week_4."Date_series"
GROUP BY week_1."Date_series"

然而，我认为我没有得到正确的答案，我也不知道为什么。非常感谢任何提示/指导/指示！ :) :)

附言。我正在使用 Postgres 9.3

这是一个很长的答案，如何使您的查询简短。 :)

Table

在我的表上构建（在您提供不同的表定义之前（odd!) 数据类型：

CREATE TABLE requests (
   id           int
 , accounts_id  int  -- (id of the user)
 , recipient_id int  -- (id of the recipient)
 , date         date -- (date that the request was made in YYYY-MM-DD)
 , amount       int  -- (# of requests by accounts_id for the day)
);

指定日期的活跃用户

“活跃用户”列表某一天:

SELECT accounts_id
FROM  (
   SELECT w.w, r.accounts_id
   FROM  (
      SELECT w
           , day - 6 - 7 * w AS w_start
           , day     - 7 * w AS w_end   
      FROM  (SELECT '2014-10-31'::date - 1 AS day) d  -- effective date here
           , generate_series(0,3) w
      ) w
   JOIN   requests r ON r."date" BETWEEN w_start AND w_end
   GROUP  BY w.w, r.accounts_id
   HAVING sum(r.amount) > 10
   ) sub
GROUP  BY 1
HAVING count(*) = 4;

Step 1

在最里面子查询w（对于“周”）从 4 周的兴趣中建立界限CROSS JOIN给定日期 - 1 的输出为generate_series(0-3).

添加/减去天数date（不是来自时间戳！）只需加/减integer数字。表达方式day - 7 * w从给定日期减去 0-3 乘以 7 天，得出end每周的日期（w_end).
分别减去 6 天（而不是 7 天！）来计算各自的start (w_start).
另外，保留周数w(0-3)用于以后的聚合。

Step 2

In 子查询sub连接来自的行requests到 4 周的集合，其中日期位于开始日期和结束日期之间。GROUP BY周数w和accounts_id.
只有请求总数超过 10 的周才有资格。

Step 3

In the outer SELECT计算每个用户的周数（accounts_id）合格的。必须为 4 才有资格成为“活跃用户”

每日活跃用户数

This is dynamite.
封装在一个简单的 SQL 函数中以简化一般使用，但查询也可以单独使用：

CREATE FUNCTION f_active_users (_now date = now()::date, _days int = 3)
  RETURNS TABLE (day date, users int) AS
$func$
WITH r AS (
   SELECT accounts_id, date, sum(amount)::int AS amount
   FROM   requests
   WHERE  date BETWEEN _now - (27 + _days) AND _now - 1
   GROUP  BY accounts_id, date
   )
SELECT date + 1, count(w_ct = 4 OR NULL)::int
FROM  (
   SELECT accounts_id, date
        , count(w_amount > 10 OR NULL)
                         OVER (PARTITION BY accounts_id, dow ORDER BY date DESC
                         ROWS BETWEEN CURRENT ROW AND 3 FOLLOWING) AS w_ct
   FROM  (
      SELECT accounts_id, date, dow   
           , sum(amount) OVER (PARTITION BY accounts_id ORDER BY date DESC
                         ROWS BETWEEN CURRENT ROW AND 6 FOLLOWING) AS w_amount
      FROM  (SELECT _now - i AS date, i%7 AS dow
             FROM   generate_series(1, 27 + _days) i) d -- period of interest
      CROSS  JOIN (
             SELECT accounts_id FROM r
             GROUP  BY 1
             HAVING count(*) > 3 AND sum(amount) > 39  -- enough rows & requests
             AND    max(date) > min(date) + 15) a      -- can cover 4 weeks
      LEFT   JOIN r USING (accounts_id, date)
      ) sub1
   WHERE date > _now - (22 + _days)  -- cut off 6 trailing days now - useful?
   ) sub2
GROUP  BY date
ORDER  BY date DESC
LIMIT  _days
$func$ LANGUAGE sql STABLE;

该函数需要任意一天（_now)，默认为“今天”，以及天数 (_days) 结果中，默认为 3。称呼：

SELECT * FROM f_active_users('2014-10-31', 5);

或者不带参数使用默认值：

SELECT * FROM f_active_users();

方法是与第一个查询不同.

SQL小提琴包含表定义的查询和变体。

Step 0

在 CTE 中r每个预聚合金额(accounts_id, date)仅适用于感兴趣的时期，以获得更好的性能。该表仅被扫描once，建议的索引（见下文）将在此处生效。

Step 1

在内部子查询中d生成必要的天数列表：27 + _days行，其中_days是输出中所需的行数，有效为 28 天或更长。
在此过程中，计算一周中的哪一天（dow) 用于步骤 3 中的聚合。i%7与每周间隔一致，查询适用于any不过，间隔。

在内部子查询中a生成唯一的用户列表（accounts_id) 存在于 CTE 中r并通过一些初步的表面测试（足够的行跨越足够的时间和足够的总请求）。

Step 2

生成笛卡尔积d and a with a CROSS JOIN具有每个相关用户的每个相关日期一行. LEFT JOIN to r附加请求数量（如果有）。不WHERE在这种情况下，我们希望每天都有结果，即使根本没有活跃用户。

计算过去一周的总金额（w_amount）在同一步骤中使用窗口函数具有定制框架.例子：

如何在窗口函数中使用环形数据结构

Step 3

现在切断最后6天；这是optional并且可能会或可能不会帮助性能。测试一下：WHERE date >= _now - (21 + _days)

计算满足最低金额的周数（w_ct）在类似的窗口函数中，这次划分为dow另外，框架中仅包含过去 4 周的相同工作日（包含过去一周的总和）。表达方式count(w_amount > 10 OR NULL)仅计算请求数超过 10 的行。详细解释：

在同一 SELECT sql 查询中从 SUM() 计算百分比

Step 4

在外层SELECT通过...分组date并对过去 4 周的用户进行计数（count(w_ct = 4 OR NULL)）。日期加 1 以补偿偏差 1，ORDER and LIMIT到要求的天数。

业绩与展望

这两个查询的完美索引是：

CREATE INDEX foo ON requests (date, accounts_id, amount);

性能应该不错，但是使用即将推出的 Postgres 会更好9.4，由于新的移动聚合支持：

移动聚合支持在 Postgres Wiki 中。
9.4 手册中的移动聚合

旁白：不要打电话timestamp“日期”列，它是timestamp, not a date。更好的是，永远不要使用基本类型名称，例如date or timestamp作为标识符。曾经。

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