LeetCode（数据库）- 游戏玩法分析 V

Posted 2021-08-28 程序员牧码

题目链接：点击打开链接

题目大意：略。

解题思路：略。

AC 代码

-- 解决方案(1)
WITH t1 AS(SELECT *, ROW_NUMBER() OVER(PARTITION BY player_id ORDER BY event_date) rw FROM Activity),
t2 AS(SELECT event_date, COUNT(*) installs FROM t1 WHERE rw = 1 GROUP BY event_date),
t3 AS(SELECT st2.event_date, COUNT(*) installs FROM t1 st1 JOIN t1 st2 ON st1.player_id = st2.player_id AND st1.rw = 1 AND st2.rw = 2 AND DATEDIFF(st2.event_date, st1.event_date) = 1
GROUP BY st2.event_date)

SELECT t2.event_date install_dt, t2.installs, FORMAT(IF(t3.event_date IS NULL, 0, t3.installs) / t2.installs, 2) + 0 Day1_retention
FROM t2 LEFT JOIN t3 ON DATEDIFF(t3.event_date, t2.event_date) = 1
ORDER BY t2.event_date

-- 解决方案(2)
SELECT
    first_day as install_dt,
    count(DISTINCT player_id) as installs,
    ROUND(
        (SUM(if(datediff(event_date, first_day) = 1, 1, 0))) / (count(DISTINCT player_id)), 2
    ) as Day1_retention
FROM(
    SELECT
        player_id,
        event_date,
        MIN(event_date) over(partition by player_id) as first_day
    FROM
        Activity
) a 
GROUP BY 
    first_day