每天的日活数及新用户占比

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；

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
  (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);

输出:

2021-10-31|2|1.00
2021-11-01|3|0.33
2021-11-02|3|0.67
2021-11-03|5|0.40

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: 6436KB, 提交时间: 2022-02-09

select dt, count(*) as dau, round(sum(new)/count(*), 2) as uv_new_ratio
from (
select uid, dt, case when dt = first_dt then 1 else 0 end as new
from 
(select uid, date(in_time) as dt
from tb_user_log
UNION
select uid, date(out_time) as dt
from tb_user_log) t1 
left join
(select uid, min(date(in_time)) as first_dt
from tb_user_log
group by uid) t2
using(uid)
) t
group by dt
order by dt

Mysql 解法, 执行用时: 36ms, 内存消耗: 6500KB, 提交时间: 2022-01-23



select dt

,count(distinct t1.uid) as dau
,round(count(distinct case when dt=new_date then t1.uid else null end )/count(distinct t1.uid),2) as uv_new_ratio
FROM
(
select uid
,date(in_time) as dt
from tb_user_log
    
union 
    
select uid
,date(out_time) as dt
from tb_user_log

group by uid
,dt
)t1

left join
(
select uid
,date(min(in_time)) as new_date

from tb_user_log
group by uid
)t2 on t1.uid=t2.uid 

group by dt

order by dt

Mysql 解法, 执行用时: 37ms, 内存消耗: 6372KB, 提交时间: 2022-01-22

select new_t.dt,dau_t.dau,round(new_t.new/dau_t.dau,2) as uv_new_ratio
from
(select a.dt,if(b.day is null,0,b.new) as new
from
(select distinct date(in_time) as dt from tb_user_log) as a
left join 
(select date(new_day) as day,count(uid) as new
 FROM
 (select uid,min(in_time) as new_day from tb_user_log group by uid) as t
 group by day) as b
 on a.dt=b.day) as new_t
 JOIN
 (select p.day,if(p.dau>q.dau,p.dau,q.dau) as dau
from
(select date(in_time) as day,count(distinct uid) as dau
from tb_user_log
group by day) as p
JOIN
(select date(out_time) as day,count(distinct uid) as dau
from tb_user_log
group by day) as q
on p.day=q.day) as dau_t
on new_t.dt=dau_t.day
order by new_t.dt

Mysql 解法, 执行用时: 37ms, 内存消耗: 6392KB, 提交时间: 2022-01-06

select
	dt,
	count(uid)as dau,
	round(sum(if(times=1,1,0))/count(uid),2)as nv_new_ratio
from
(
select 
	*,
	count(*) over (partition by uid order by dt)as times
from
(
select 
	uid,
	date(in_time) as dt
from tb_user_log
union
select
	uid,
	date(out_time) as dt
from tb_user_log
) t1
) t2
group by dt
order by dt

Mysql 解法, 执行用时: 37ms, 内存消耗: 6392KB, 提交时间: 2021-12-27

with t1 as (
    select uid,min(date(in_time)) min_date
    from tb_user_log
    group by uid),
t2 as(
    select uid,date(in_time) dt
    from tb_user_log
    union
    select uid,date(out_time) dt
    from tb_user_log)

select dt,count(*),round(count(min_date)/count(*),2)
from(
    SELECT t2.uid,dt,min_date
    from t2
    left join t1 on t2.uid=t1.uid and t2.dt=t1.min_date) t3
group by dt
order by dt

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