我每天都在尝试查找＃个活跃用户。
用户在每周超过 10次请求<连续4周时处于有效状态
。即。在2014年10月31日，如果用户每周总共发出超过10个请求，则该用户处于活动状态：
我有requests的表格：
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
);
示例值：
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');
示例输出：
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的活跃用户数量（例如10-01-2014）：
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天数的查询，因此我需要每天获得此数字。我认为这个想法是做一个联接以获取日期，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"
但是，我认为我没有得到正确答案，我不知道为什么。任何提示/指导/指针非常感谢！ :)：）
PS。我使用的是Postgres 9.3
答案 0 :(得分：6)
如何简化您的查询，这是一个很长的答案。 :)
在我的表上构建（在您使用不同的（ odd！）数据类型提供表定义之前：
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)
);
＆＃34;活跃用户列表＆＃34; 给定的一天：
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;
在最里面的子查询w （对于＆＃34;周＆＃34;）从给定日期的CROSS JOIN构建感兴趣的4周的界限 - 1，输出为generate_series(0-3)。
要在date（不是时间戳！）中添加/减去天数，只需添加/减去integer个数字。表达式day - 7 * w从给定日期开始7天减去0-3次，到达每周的结束日期（w_end）。
从每个中减去另外6天（不是7！）以计算相应的开始（w_start）。
此外，请保留周数w（0-3）以用于以后的聚合。
在子查询sub 中，将requests行添加到4周的集合中，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
只有超过10个请求的周数才符合条件。
在外部SELECT 计算每个用户（accounts_id）合格的周数。必须为4才能符合＆＃34;活跃用户＆＃34;
这是 炸药 包含在一个简单的SQL函数中以简化一般用途，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;
该功能需要任何一天（_now），＆＃34;今天＆＃34;默认情况下，结果中的天数（_days），默认为3。拨打：
SELECT * FROM f_active_users('2014-10-31', 5);
或者没有参数来使用默认值：
SELECT * FROM f_active_users();
该方法与第一个查询不同。
SQL Fiddle包含表定义的查询和变体。
仅在感兴趣的时段内，每r次CTE (accounts_id, date)预聚合金额，以获得更好的效果。该表只扫描一次，建议的索引（见打击）将在这里开始。
在内部子查询d中生成必要的天数列表：27 + _days行，其中_days是输出中所需的行数，有效期为28天或更长。
在此过程中，计算在步骤3中用于聚合的星期几（dow）。i%7与每周时间间隔一致，查询适用于任何时间间隔但是。
在内部子查询a中生成一个唯一的用户列表（accounts_id），它存在于CTE r中并通过一些初步的表面测试（足够的行跨越足够的时间并有足够的总请求）。
从d和a生成一个笛卡尔积，其中CROSS JOIN为每个相关用户的每一天都有一行。 LEFT JOIN到r附加请求数量（如果有）。没有WHERE条件，我们希望结果中的每一天，即使根本没有活跃用户。
使用Window functions with a custom frame.计算同一步骤中过去一周（w_amount）的总金额示例：
现在切断过去6天;这是可选，可能有助于也可能没有帮助。测试一下：WHERE date >= _now - (21 + _days)
在类似的窗口函数中计算满足最小金额（w_ct）的周数，此时由dow分区，另外在框架中过去4周只有相同的工作日（携带相应过去一周的总和）。
表达式count(w_amount > 10 OR NULL)仅计算超过10个请求的行。详细解释：
在SELECT外的date组中，计算通过所有4周（count(w_ct = 4 OR NULL)）的用户。在日期中添加1，以便按照1，ORDER和LIMIT补偿所需的天数。
两个查询的完美索引是：
CREATE INDEX foo ON requests (date, accounts_id, amount);
由于新的移动聚合支持，性能应该很好，但是即将推出的Postgres 9.4 会更好（更好）：
Moving-aggregate support in the Postgres Wiki.
Moving aggregates in the 9.4 manual
除此之外：不要拨打timestamp列＆＃34;日期＆＃34;，它是timestamp，而不是date。更好的是，永远不要使用date或timestamp等基本类型名称作为标识符。如初。