SQL166 每天的日活数及新用户占比

题目描述

用户行为日志表tb_user_log

id	uid	artical_id	in_time	out_time	sign_cin
1	101	9001	2021-10-31 10:00:00	2021-10-31 10:00:09	0
2	102	9001	2021-10-31 10:00:00	2021-10-31 10:00:09	0
3	101	0	2021-11-01 10:00:00	2021-11-01 10:00:42	1
4	102	9001	2021-11-01 10:00:00	2021-11-01 10:00:09	0
5	108	9001	2021-11-01 10:00:01	2021-11-01 10:00:50	0
6	108	9001	2021-11-02 10:00:01	2021-11-02 10:00:50	0
7	104	9001	2021-11-02 10:00:28	2021-11-02 10:00:50	0
8	106	9001	2021-11-02 10:00:28	2021-11-02 10:00:50	0
9	108	9001	2021-11-03 10:00:01	2021-11-03 10:00:50	0
10	109	9002	2021-11-03 11:00:55	2021-11-03 11:00:59	0
11	104	9003	2021-11-03 11:00:45	2021-11-03 11:00:55	0
12	105	9003	2021-11-03 11:00:53	2021-11-03 11:00:59	0
13	106	9003	2021-11-03 11:00:45	2021-11-03 11:00:55	0

（uid-用户ID, artical_id-文章ID, in_time-进入时间, out_time-离开时间, sign_in-是否签到）

问题：统计每天的日活数及新用户占比

注：

• 新用户占比=当天的新用户数÷当天活跃用户数（日活数）。
• 如果in_time-进入时间和out_time-离开时间跨天了，在两天里都记为该用户活跃过。
• 新用户占比保留2位小数，结果按日期升序排序。

输出示例：

示例数据的输出结果如下：

dt	dau	uv_new_ratio
2021-10-30	2	1.00
2021-11-01	3	0.33
2021-11-02	3	0.67
2021-11-03	5	0.40

解释：

2021年10月31日有2个用户活跃，都为新用户，新用户占比1.00；

2021年11月1日有3个用户活跃，其中1个新用户，新用户占比0.33；

SQL Schema

DROP TABLE IF EXISTS tb_user_log;
CREATE TABLE tb_user_log
(
    id         INT PRIMARY KEY AUTO_INCREMENT COMMENT '自增ID',
    uid        INT NOT NULL COMMENT '用户ID',
    artical_id INT NOT NULL COMMENT '视频ID',
    in_time    datetime COMMENT '进入时间',
    out_time   datetime COMMENT '离开时间',
    sign_in    TINYINT DEFAULT 0 COMMENT '是否签到'
) CHARACTER SET utf8
  COLLATE utf8_bin;

INSERT INTO tb_user_log(uid, artical_id, in_time, out_time, sign_in)
VALUES (101, 9001, '2021-10-31 10:00:00', '2021-10-31 10:00:09', 0),
       (102, 9001, '2021-10-31 10:00:00', '2021-10-31 10:00:09', 0),
       (101, 0, '2021-11-01 10:00:00', '2021-11-01 10:00:42', 1),
       (102, 9001, '2021-11-01 10:00:00', '2021-11-01 10:00:09', 0),
       (108, 9001, '2021-11-01 10:00:01', '2021-11-01 10:01:50', 0),
       (108, 9001, '2021-11-02 10:00:01', '2021-11-02 10:01:50', 0),
       (104, 9001, '2021-11-02 10:00:28', '2021-11-02 10:00:50', 0),
       (106, 9001, '2021-11-02 10:00:28', '2021-11-02 10:00:50', 0),
       (108, 9001, '2021-11-03 10:00:01', '2021-11-03 10:01:50', 0),
       (109, 9002, '2021-11-03 11:00:55', '2021-11-03 11:00:59', 0),
       (104, 9003, '2021-11-03 11:00:45', '2021-11-03 11:00:55', 0),
       (105, 9003, '2021-11-03 11:00:53', '2021-11-03 11:00:59', 0),
       (106, 9003, '2021-11-03 11:00:45', '2021-11-03 11:00:55', 0);

答案

SELECT dt, COUNT(uid) AS `dau`, ROUND(AVG(IF(dt = first_time, 1, 0)), 2) AS `uv_new_ratio`
FROM (SELECT uid, dt, MIN(dt) OVER (PARTITION BY uid ORDER BY dt) AS `first_time`
      FROM (SELECT uid, DATE_FORMAT(in_time, '%Y-%m-%d') AS `dt`
            FROM tb_user_log
            UNION
            SELECT uid, DATE_FORMAT(out_time, '%Y-%m-%d') AS `dt`
            FROM tb_user_log) a) b
GROUP BY dt
ORDER BY dt;