Question

表结构包括：

User_ID , User_Type , Time_Period (integer)
---------------------------------
12345   ,      1     ,    201501
12346   ,      1     ,    201501
12347   ,      2     ,    201501
12345   ,      1     ,    201502
12346   ,      2     ,    201502

随着时间的推移，

唯一的User_ID可以完全保留User_Type
可以在任何Time_Period
User_ID可以迁移到每个Time_Period中的其他User_Types。

我需要编写代码来了解用户类型超过12个周期的用户的稳定性，例如，70％的不同User_ID在201501到201512期间保留在User_Type 1中

输出应该是User_Type的列表，其中列出了在12个时间段内保持相同User_Type的不同User_ID的计数，以及同一时间段内不同User_ID总数的第二列

User_Type , Count Distinct Same User_IDs , Count Distinct Total User_IDs
---------------------------------------------------------------------
1         ,              146,023         ,       201,501
2         ,              46,124          ,       147,234
3         ,              27,500          ,       87,954

这是我第一次发帖，所以如果您需要更多细节并提前致谢，请告诉我

编辑 - 由于每个时间段，User_ID可以多次出现在表格中，但每个时段只能出现一次。

Answer 1

如果每个用户在每个月都有，并且您想要找到所有12个月内存在的用户比例并保持＆＃34; 1＆＃34;，那么一种方法是：

select avg(minut = 1 and maxut = 1)
from (select user_id, min(user_type) as minut, max(user_type) as maxut
      from t
      where left(time_period, 4) = '2015'
      group by user_id
      having count(*) = 12 
     ) t;

Answer 2

这将显示用户的行为每个用户在其中一行中计数一次一行描述了用户的阶段 - 每个月都存在的用户类型和用户类型如果您增加数据样本，则更容易理解此报告。

select      count(*)    as users
           ,`1`,`2`,`3`,`4`,`5`,`6`,`7`,`8`,`9`,`10`,`11`,`12`

from       (select      user_id

                       ,min(case when time_period = 201501 then user_type end) as `1`
                       ,min(case when time_period = 201502 then user_type end) as `2`
                       ,min(case when time_period = 201503 then user_type end) as `3`
                       ,min(case when time_period = 201504 then user_type end) as `4`
                       ,min(case when time_period = 201505 then user_type end) as `5`
                       ,min(case when time_period = 201506 then user_type end) as `6`
                       ,min(case when time_period = 201507 then user_type end) as `7`
                       ,min(case when time_period = 201508 then user_type end) as `8`
                       ,min(case when time_period = 201509 then user_type end) as `9`
                       ,min(case when time_period = 201510 then user_type end) as `10`
                       ,min(case when time_period = 201511 then user_type end) as `11`
                       ,min(case when time_period = 201512 then user_type end) as `12`

            from        mytable

            group by    user_id
            ) t

group by    `1`,`2`,`3`,`4`,`5`,`6`,`7`,`8`,`9`,`10`,`11`,`12`          

order by    users desc

+-------+---+--------+--------+--------+--------+--------+--------+--------+--------+--------+--------+--------+
| users | 1 | 2      | 3      | 4      | 5      | 6      | 7      | 8      | 9      | 10     | 11     | 12     |
+-------+---+--------+--------+--------+--------+--------+--------+--------+--------+--------+--------+--------+
| 1     | 1 | 2      | (null) | (null) | (null) | (null) | (null) | (null) | (null) | (null) | (null) | (null) |
+-------+---+--------+--------+--------+--------+--------+--------+--------+--------+--------+--------+--------+
| 1     | 2 | (null) | (null) | (null) | (null) | (null) | (null) | (null) | (null) | (null) | (null) | (null) |
+-------+---+--------+--------+--------+--------+--------+--------+--------+--------+--------+--------+--------+
| 1     | 1 | 1      | (null) | (null) | (null) | (null) | (null) | (null) | (null) | (null) | (null) | (null) |
+-------+---+--------+--------+--------+--------+--------+--------+--------+--------+--------+--------+--------+

用户类型的SQL不同用户ID随时间的变化

2 个答案: