如何根据另一个 table 的 GROUP BY 函数计算 table 中的每日平均频率？

Question

我想知道每天哪个电影院更频繁，我需要这些值作为相对频率。

我的表是：

表 1:

client_id	cinema_room
1	A
2	B
3	C
4	C
5	A
6	C
7	B
8	A
9	B
10	B
11	B
12	A
13	C
14	A
15	A
16	B

Table 2:

day	client_id
01/01/2022	1
01/01/2022	1
01/02/2022	1
01/02/2022	2
01/02/2022	4
01/03/2022	8
01/04/2022	14
01/04/2022	15
01/04/2022	16

所以我需要这样的结果：

day	cinema_room	avg_freq
01/01/2022	A	1
01/01/2022	B	0
01/01/2022	C	0
01/02/2022	A	0.33
01/02/2022	B	0.33
01/02/2022	C	0.33
01/03/2022	A	1
01/03/2022	B	0
01/03/2022	C	0
01/04/2022	A	0.66
01/04/2022	B	0.33
01/04/2022	C	0

我现在得到的是：

SELECT day , cinema_room, COUNT(t2.client_id) as t2_tot
FROM table1 t1
    LEFT JOIN table2 t2 ON t1.client_id = t2.client_id
GROUP BY day, cinema_room
ORDER BY day

其不足的结果是：

NOTE 1: I'm counting, not even averaging.

NOTE 2: The first three rows are wrong af.

day	cinema_room	count_freq
None	A	0
None	B	0
None	C	0
01/01/2022	A	3
01/01/2022	B	0
01/01/2022	C	0
01/02/2022	A	1
01/02/2022	B	1
01/02/2022	C	1
01/03/2022	A	1
01/03/2022	B	0
01/03/2022	C	0
01/04/2022	A	2
01/04/2022	B	1
01/04/2022	C	0

Answer 1

您可以尝试在一个子查询中使用 CROSS JOIN，这将是一个日历 table，然后在另一个子查询中聚合 table1 & table2 每个 get 计数cinema_room 每天。

SELECT t1.day , t1.cinema_room, SUM(IFNULL(t2.cnt,0)) / (SELECT COUNT(*) FROM table2 tt2 WHERE t1.day = tt2.day) as t2_tot
FROM (
  SELECT DISTINCT day,cinema_room
 FROM table2 t2 CROSS JOIN table1
) t1
LEFT JOIN (
 SELECT day,cinema_room,COUNT(*) cnt
  FROM table1 t1 INNER JOIN table2 t2
  ON t1.client_id = t2.client_id
  GROUP BY day,cinema_room
) t2 ON t1.day = t2.day AND t1.cinema_room = t2.cinema_room
GROUP BY t1.day , t1.cinema_room
ORDER BY t1.day

sqlfiddle

如何根据另一个 table 的 GROUP BY 函数计算 table 中的每日平均频率？

How to calculate daily average frequency in a table in function of GROUP BY from another table?

mysql

sql

average

date