根据玩过的游戏创建留存 table

Question

情况：

我有一个包含三列的 table：

日期
游戏
user_id

从那个 table 我需要创建一个最终看起来像这样的保留 table：

输出：

+------------+-------------+------+-------------+------+------+-------+
|    date    |    game     | day0 |    day1     | day3 | day7 | day10 |
+------------+-------------+------+-------------+------+------+-------+
| 2019-01-01 | fifa        |  100 | % of day  0 |      |      |       |
| 2019-01-01 | nba         |  100 |             |      |      |       |
| 2019-01-01 | battlefield |  100 |             |      |      |       |
| 2019-01-02 | fifa        |  100 |             |      |      |       |
| 2019-01-02 | battlefield |  100 |             |      |      |       |
| 2019-01-03 | fifa        |  100 |             |      |      |       |
| 2019-01-03 | nba         |  100 |             |      |      |       |
| ...        |             |      |             |      |      |       |
+------------+-------------+------+-------------+------+------+-------+

如果至少有一个人玩过游戏，第 0 天就是 100

day1 是 1 天或 2 天后回来的不同用户的比例

day3 是 3 到 6 天后回来的不同用户的比例。

等等...

如果你绘制数字，它应该给你一个非线性递减曲线。

逻辑：

（见下面的示例数据）

到目前为止我构建的骨架如下所示：

;WITH 
    baseDate AS (
        SELECT
         'target_date' AS [key_name]
        ,CAST('2019-01-01' AS date) AS [key_value]
        ),
durationDate AS (
    SELECT DATEADD(DAY,0,key_value) AS target_date FROM baseDate WHERE [key_name] = 'target_date'
    UNION ALL
    SELECT DATEADD(DAY,1,key_value) AS target_date FROM baseDate WHERE [key_name] = 'target_date'
    UNION ALL
    SELECT DATEADD(DAY,2,key_value) AS target_date FROM baseDate WHERE [key_name] = 'target_date'
    UNION ALL
    SELECT DATEADD(DAY,3,key_value) AS target_date FROM baseDate WHERE [key_name] = 'target_date'
    UNION ALL
    SELECT DATEADD(DAY,4,key_value) AS target_date FROM baseDate WHERE [key_name] = 'target_date'
    UNION ALL
    SELECT DATEADD(DAY,5,key_value) AS target_date FROM baseDate WHERE [key_name] = 'target_date'
    UNION ALL
    SELECT DATEADD(DAY,6,key_value) AS target_date FROM baseDate WHERE [key_name] = 'target_date'
)
    select 
        dd.target_date
        , play.game 
        , play.[count]/play.count *100 as day0
    from durationDate as dd
    left join (
        select t1.date, t1.game, COUNT(distinct t1.user_id) as [count]
        from #t1 t1
        group by t1.date, t1.game
        ) as play on dd.target_date=play.[date]

我可能缺少第 1 天、第 3 天、第 7 天列的另一个基础 table，例如：

durationDateColumn AS (
        SELECT 
              DATEADD(DAY,0,key_value) AS day1
            , DATEADD(DAY,1,key_value) AS day2
            , DATEADD(DAY,2,key_value) AS day3
            , DATEADD(DAY,3,key_value) AS day4
            , DATEADD(DAY,4,key_value) AS day5
            , DATEADD(DAY,5,key_value) AS day6
            , DATEADD(DAY,6,key_value) AS day7
        FROM baseDate
        WHERE [key_name] = 'target_date'
    )

除非有更有效的方法。

示例数据：

drop table if exists #t1
create table #t1 ([date] date, [game] varchar (20),[user_id] bigint)
insert into #t1 values 
('2019-01-01', 'fifa',11),
('2019-01-01', 'fifa',12),
('2019-01-01', 'fifa',13),
('2019-01-01', 'fifa',14),
('2019-01-02', 'fifa',12),
('2019-01-02', 'fifa',13),
('2019-01-02', 'fifa',14),
('2019-01-04', 'fifa',12),
('2019-01-04', 'fifa',13),
('2019-01-08', 'fifa',13),
('2019-01-01', 'nba',11),
('2019-01-01', 'nba',13),
('2019-01-01', 'nba',14),
('2019-01-02', 'nba',13),
('2019-01-02', 'nba',14),
('2019-01-04', 'nba',13),
('2019-01-08', 'nba',13)

Answer 1

加入日期和后续日期数据，计算天数和用户数。使用条件聚合按天数进行透视。

create table #t1 ([date] date, [game] varchar (20),[user_id] bigint);

insert into #t1 values 
('2019-01-01', 'fifa',11),
('2019-01-01', 'fifa',12),
('2019-01-01', 'fifa',13),
('2019-01-01', 'fifa',14),
('2019-01-02', 'fifa',12),
('2019-01-02', 'fifa',13),
('2019-01-02', 'fifa',14),
('2019-01-04', 'fifa',12),
('2019-01-04', 'fifa',13),
('2019-01-08', 'fifa',13),
('2019-01-01', 'nba',11),
('2019-01-01', 'nba',13),
('2019-01-01', 'nba',14),
('2019-01-02', 'nba',13),
('2019-01-02', 'nba',14),
('2019-01-04', 'nba',13),
('2019-01-08', 'nba',13);

select [date], [game] 
  , 100.0 day0
  , cast(100.0 * (0.0 + sum(case ret when 1 then n else 0 end)) / sum(case ret when 0 then n end) as decimal(5,2)) day1
  , cast(100.0 * (0.0 + sum(case ret when 2 then n else 0 end)) / sum(case ret when 0 then n end) as decimal(5,2)) day2
  , cast(100.0 * (0.0 + sum(case ret when 3 then n else 0 end)) / sum(case ret when 0 then n end) as decimal(5,2)) day3
  , cast(100.0 * (0.0 + sum(case ret when 4 then n else 0 end)) / sum(case ret when 0 then n end) as decimal(5,2)) day4
  -- .. more days
from (
    select t1.[date], t1.[game], Datediff(dd, t1.[Date], t2.[Date]) ret, count(distinct t1.[user_id]) n
    from #t1 t1
    left join #t1 t2
        on t1.game = t2.game and t1.user_id = t2.user_id and t1.date <= t2.date
    group by t1.[date], t1.[game], Datediff(dd, t1.[Date], t2.[Date])
) evt
group by [date], [game]
order by [date], [game];

输出

date    game    day0    day1    day2    day3    day4
2019-01-01  fifa    100.0   75.00   0.00    50.00   0.00
2019-01-01  nba     100.0   66.67   0.00    33.33   0.00
2019-01-02  fifa    100.0   0.00    66.67   0.00    0.00
2019-01-02  nba     100.0   0.00    50.00   0.00    0.00
2019-01-04  fifa    100.0   0.00    0.00    0.00    50.00
2019-01-04  nba     100.0   0.00    0.00    0.00    100.00
2019-01-08  fifa    100.0   0.00    0.00    0.00    0.00
2019-01-08  nba     100.0   0.00    0.00    0.00    0.00

根据玩过的游戏创建留存 table

create a retention table base on games played

sql

sql-server

retention