Question

550. Game Play Analysis IV

中文文档

Description

Table: Activity

+--------------+---------+
| Column Name  | Type  |
+--------------+---------+
| player_id  | int   |
| device_id  | int   |
| event_date   | date  |
| games_played | int   |
+--------------+---------+
(player_id, event_date) is the primary key (combination of columns with unique values) of this table.
This table shows the activity of players of some games.
Each row is a record of a player who logged in and played a number of games (possibly 0) before logging out on someday using some device.

 

Write a solution to report the fraction of players that logged in again on the day after the day they first logged in, rounded to 2 decimal places. In other words, you need to count the number of players that logged in for at least two consecutive days starting from their first login date, then divide that number by the total number of players.

The result format is in the following example.

 

Example 1:

Input: 
Activity table:
+-----------+-----------+------------+--------------+
| player_id | device_id | event_date | games_played |
+-----------+-----------+------------+--------------+
| 1     | 2     | 2016-03-01 | 5      |
| 1     | 2     | 2016-03-02 | 6      |
| 2     | 3     | 2017-06-25 | 1      |
| 3     | 1     | 2016-03-02 | 0      |
| 3     | 4     | 2018-07-03 | 5      |
+-----------+-----------+------------+--------------+
Output: 
+-----------+
| fraction  |
+-----------+
| 0.33    |
+-----------+
Explanation: 
Only the player with id 1 logged back in after the first day he had logged in so the answer is 1/3 = 0.33

Solutions

Solution 1: Grouping and Minimum Value + Left Join

We can first find the first login date of each player, and then perform a left join with the original table, with the join condition being that the player ID is the same and the date difference is $-1$, which means the player logged in on the second day. Then, we only need to calculate the ratio of non-null players among the players who logged in on the second day.

import pandas as pd


def gameplay_analysis(activity: pd.DataFrame) -> pd.DataFrame:
  activity["first"] = activity.groupby("player_id").event_date.transform(min)
  activity_2nd_day = activity[
    activity["first"] + pd.DateOffset(1) == activity["event_date"]
  ]

  return pd.DataFrame(
    {"fraction": [round(len(activity_2nd_day) / activity.player_id.nunique(), 2)]}
  )

# Write your MySQL query statement below
SELECT ROUND(AVG(b.event_date IS NOT NULL), 2) AS fraction
FROM
  (
    SELECT player_id, MIN(event_date) AS event_date
    FROM Activity
    GROUP BY 1
  ) AS a
  LEFT JOIN Activity AS b
    ON a.player_id = b.player_id AND DATEDIFF(a.event_date, b.event_date) = -1;

Solution 2: Window Function

We can use the LEAD window function to get the next login date of each player. If the next login date is one day after the current login date, it means that the player logged in on the second day, and we use a field $st$ to record this information. Then, we use the RANK window function to rank the player IDs in ascending order by date, and get the login ranking of each player. Finally, we only need to calculate the ratio of non-null $st$ values among the players with a ranking of $1$.

# Write your MySQL query statement below
WITH
  T AS (
    SELECT
      player_id,
      DATEDIFF(
        LEAD(event_date) OVER (
          PARTITION BY player_id
          ORDER BY event_date
        ),
        event_date
      ) = 1 AS st,
      RANK() OVER (
        PARTITION BY player_id
        ORDER BY event_date
      ) AS rk
    FROM Activity
  )
SELECT ROUND(COUNT(IF(st = 1, player_id, NULL)) / COUNT(DISTINCT player_id), 2) AS fraction
FROM T
WHERE rk = 1;