统计活跃间隔对用户分级结果

id	uid	artical_id	in_time	out_time	sign_cin
1	109	9001	2021-08-31 10:00:00	2021-08-31 10:00:09	0
2	109	9002	2021-11-04 11:00:55	2021-11-04 11:00:59	0
3	108	9001	2021-09-01 10:00:01	2021-09-01 10:01:50	0
4	108	9001	2021-11-03 10:00:01	2021-11-03 10:01:50	0
5	104	9001	2021-11-02 10:00:28	2021-11-02 10:00:50	0
6	104	9003	2021-09-03 11:00:45	2021-09-03 11:00:55	0
7	105	9003	2021-11-03 11:00:53	2021-11-03 11:00:59	0
8	102	9001	2021-10-30 10:00:00	2021-10-30 10:00:09	0
9	103	9001	2021-10-21 10:00:00	2021-10-21 10:00:09	0
10	101	0	2021-10-01 10:00:00	2021-10-01 10:00:42	1

（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：统计活跃间隔对用户分级后，各活跃等级用户占比，结果保留两位小数，且按占比降序排序。

注：

用户等级标准简化为：忠实用户(近7天活跃过且非新晋用户)、新晋用户(近7天新增)、沉睡用户(近7天未活跃但更早前活跃过)、流失用户(近30天未活跃但更早前活跃过)。
假设今天就是数据中所有日期的最大值。
近7天表示包含当天T的近7天，即闭区间[T-6, T]。

输出示例：

示例数据的输出结果如下

user_grade	ratio
忠实用户	0.43
新晋用户	0.29
沉睡用户	0.14
流失用户	0.14

解释：

今天日期为2021.11.04，根据用户分级标准，用户行为日志表tb_user_log中忠实用户有：109、108、104；新晋用户有105、102；沉睡用户有103；流失用户有101；共7个用户，因此他们的比例分别为0.43、0.29、0.14、0.14。

示例1

输入:

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
  (109, 9001, '2021-08-31 10:00:00', '2021-08-31 10:00:09', 0),
  (109, 9002, '2021-11-04 11:00:55', '2021-11-04 11:00:59', 0),
  (108, 9001, '2021-09-01 10:00:01', '2021-09-01 10:01:50', 0),
  (108, 9001, '2021-11-03 10:00:01', '2021-11-03 10:01:50', 0),
  (104, 9001, '2021-11-02 10:00:28', '2021-11-02 10:00:50', 0),
  (104, 9003, '2021-09-03 11:00:45', '2021-09-03 11:00:55', 0),
  (105, 9003, '2021-11-03 11:00:53', '2021-11-03 11:00:59', 0),
  (102, 9001, '2021-10-30 10:00:00', '2021-10-30 10:00:09', 0),
  (103, 9001, '2021-10-21 10:00:00', '2021-10-21 10:00:09', 0),
  (101, 0, '2021-10-01 10:00:00', '2021-10-01 10:00:42', 1);

输出:

忠实用户|0.43
新晋用户|0.29
沉睡用户|0.14
流失用户|0.14

原站题解

上次编辑到这里，代码来自缓存点击恢复默认模板

Mysql 解法, 执行用时: 36ms, 内存消耗: 6396KB, 提交时间: 2021-12-31



select user_grade,
round(count(uid)/(select count(distinct uid) from tb_user_log),2) ratio


from
-- --------------------------------------------------
(select uid, 
(case
when datediff(td,last_active)<=6 
and  datediff(td,first_reg)>6   then '忠实用户'
when datediff(td,last_active)<=6
and  datediff(td,first_reg)<=6  then '新晋用户'
when datediff(td,last_active)>6
and  datediff(td,last_active)<=29 then '沉睡用户'
when datediff(td,last_active)>29  then '流失用户'
else '未分类' end) user_grade
from
-- --------------------------------------------------
(select uid,
max(out_time) last_active,
min(in_time) first_reg,
(select max(out_time) from tb_user_log) td
from tb_user_log
group by uid) a) b
-- --------------------------------------------------
-- --------------------------------------------------
group by user_grade
order by ratio desc

Mysql 解法, 执行用时: 36ms, 内存消耗: 6480KB, 提交时间: 2022-01-22

select user_grade,round(count(uid)/(select count(distinct uid) from tb_user_log),2) q
from  
(
select uid,(case when datediff((select max(in_time) from tb_user_log),max(in_time)) <=6 
            and  datediff((select max(in_time) from tb_user_log),min(in_time)) >6 then '忠实用户'
            when datediff((select max(in_time) from tb_user_log),max(in_time)) <=6 
            and  datediff((select max(in_time) from tb_user_log),min(in_time)) <=6 then '新晋用户'
             when datediff((select max(in_time) from tb_user_log),max(in_time)) >6 
             and  datediff((select max(in_time) from tb_user_log),min(in_time)) <=29 then '沉睡用户'
             else '流失用户' end ) user_grade
from tb_user_log
group by uid
) f1
group by user_grade
order by q desc

Mysql 解法, 执行用时: 36ms, 内存消耗: 6504KB, 提交时间: 2021-12-27

WITH today AS(
SELECT MAX(DATE(in_time)) FROM tb_user_log #今天的日期
)

SELECT 
  user_grade,
  ROUND(COUNT(uid) / (SELECT COUNT(DISTINCT uid) FROM tb_user_log), 2) ratio
FROM (
  SELECT 
    uid,
    CASE 
      WHEN MAX(DATE(in_time)) BETWEEN DATE_SUB((SELECT * FROM today), INTERVAL 6 DAY)
                              AND
                              (SELECT * FROM today) #最后活跃在近七天(T-6)
           AND
           MIN(DATE(in_time)) < DATE_SUB((SELECT * FROM today), INTERVAL 6 DAY) #非新晋用户：注册日期在近七天前
           THEN '忠实用户'
      WHEN MIN(DATE(in_time)) BETWEEN DATE_SUB((SELECT * FROM today), INTERVAL 6 DAY) 
                                      AND #新晋用户：首次登陆日期在近七天（T-6）之内
                                      (SELECT * FROM today) THEN '新晋用户'
      WHEN MAX(DATE(in_time)) BETWEEN DATE_SUB((SELECT * FROM today), INTERVAL 29 DAY) 
                              AND #沉睡用户：最后活跃时间在7天之前（T-7），30天之内（T-29）
                              DATE_SUB((SELECT * FROM today), INTERVAL 7 DAY) THEN '沉睡用户'
      ELSE '流失用户'
    END user_grade
  FROM tb_user_log
  GROUP BY 1
) t
GROUP BY 1
ORDER BY 2 DESC

Mysql 解法, 执行用时: 36ms, 内存消耗: 6512KB, 提交时间: 2021-12-14

WITH today AS(
SELECT MAX(DATE(in_time)) FROM tb_user_log #今天的日期
)
SELECT 
  user_grade,
  ROUND(COUNT(uid) / (SELECT COUNT(DISTINCT uid) FROM tb_user_log), 2) ratio
FROM (
  SELECT 
    uid,
CASE WHEN MAX(DATE(in_time)) BETWEEN DATE_SUB((SELECT * FROM today), INTERVAL 6 DAY)
AND (SELECT * FROM today) #最后活跃在近七天(T-6)
AND MIN(DATE(in_time)) < DATE_SUB((SELECT * FROM today), INTERVAL 6 DAY) #非新晋用户：注册日期在近七天前 
    THEN '忠实用户'
WHEN MIN(DATE(in_time)) BETWEEN DATE_SUB((SELECT * FROM today), INTERVAL 6 DAY) 
AND (SELECT * FROM today) THEN '新晋用户'
WHEN MAX(DATE(in_time)) BETWEEN DATE_SUB((SELECT * FROM today), INTERVAL 29 DAY) 
AND #沉睡用户：最后活跃时间在7天之前（T-7），30天之内（T-29）
DATE_SUB((SELECT * FROM today), INTERVAL 7 DAY) THEN '沉睡用户'
ELSE '流失用户'
END user_grade
  FROM tb_user_log
  GROUP BY 1
) t
GROUP BY 1
ORDER BY 2 DESC

Mysql 解法, 执行用时: 36ms, 内存消耗: 6512KB, 提交时间: 2021-12-03

select user_grade,
       round(count(uid)/(select count(distinct uid) from tb_user_log),2) ratio
from ( SELECT uid,
              case
              when (max(date(in_time)) BETWEEN '2021-10-29' and '2021-11-04')
                   and (min(date(in_time)) <'2021-10-29' ) then '忠实用户'
              when min(date(in_time)) BETWEEN '2021-10-29'
                   and '2021-11-04' then '新晋用户'
              when (max(date(in_time)) <= '2021-10-28')
                    and (max(date(in_time)) >= '2021-10-06') then '沉睡用户'
              else '流失用户' end as user_grade
              FROM tb_user_log
              GROUP BY uid ) a
GROUP BY user_grade
ORDER BY ratio desc

上一题

下一题

详情