对比MySQL和Pandas统计直播间各时间段上线人数

一道面试题如下：

简单理解就是求每个直播间，所有的最短时间段的上线人数。简单图解下：

有了上图，我们应该就秒懂了这题的解题思路，先将所有的时间点排序，按顺序分配得到每个时间分区，绿色为进入起始时间点，红色为结束时间点，我们进入起始点时+1，进入结束时间点时-1，这样就可以得到每个区间的在线人数了。当然假如user3也从​​10:21:10​​进去，则这个时间点位置+2，后续的时间片段内在线人数就是累加后的值。

理解了思路，我们就可以开始编码了：

pandas处理代码

最终完整处理代码为：

import pandas as pd

df = pd.DataFrame([
    ["a", "user1", "2022-02-01 10:12:13", "2022-02-01 10:30:23"],
    ["a", "user2", "2022-02-01 10:21:10", "2022-02-01 11:02:06"],
    ["b", "user1", "2022-02-01 10:12:13", "2022-02-01 10:30:23"],
    ["b", "user2", "2022-02-01 10:21:10", "2022-02-01 11:02:06"],
    ["b", "user3", "2022-02-01 10:19:10", "2022-02-01 11:05:06"],
], columns=["room_id", "user_id", "start_time", "end_time"])
df


def func(df):
    times = pd.concat([df.start_time.value_counts(),
                       -df.end_time.value_counts()]).sort_index()
    times.index.name = "start_time"
    r = times.cumsum().to_frame("user_cnt").reset_index()
    r.insert(1, "end_time", r.start_time.shift(-1))
    r.insert(0, "room_id", df.room_id.iat[0])
    return r.query("user_cnt>0")


df.groupby("room_id", as_index=False).apply(func).reset_index(drop=True)

结果：

注意：使用如下代码即可设置一个单元格可以显示全部输出（）

from IPython.core.interactiveshell import InteractiveShell InteractiveShell.ast_node_interactivity = "all"

默认是输出结尾表达式：

​​from IPython.core.interactiveshell import InteractiveShell print(InteractiveShell.ast_node_interactivity.default_value) print(InteractiveShell.ast_node_interactivity.values) ​​​​last_expr ['all', 'last', 'last_expr', 'none', 'last_expr_or_assign'] ​​

MySQL处理代码

首先我们创建表并插入数据：

CREATE TABLE `t1` (
  `room_id` VARCHAR(20),
  `user_id` VARCHAR(20),
  `start_time` DATETIME,
  `end_time` DATETIME
);

INSERT  INTO `t1`(`room_id`,`user_id`,`start_time`,`end_time`) VALUES ('a','user1','2022-02-01 10:12:13','2022-02-01 10:30:23');
INSERT  INTO `t1`(`room_id`,`user_id`,`start_time`,`end_time`) VALUES ('a','user2','2022-02-01 10:21:10','2022-02-01 11:02:06');
INSERT  INTO `t1`(`room_id`,`user_id`,`start_time`,`end_time`) VALUES ('b','user1','2022-02-01 10:12:13','2022-02-01 10:30:23');
INSERT  INTO `t1`(`room_id`,`user_id`,`start_time`,`end_time`) VALUES ('b','user2','2022-02-01 10:21:10','2022-02-01 11:02:06');
INSERT  INTO `t1`(`room_id`,`user_id`,`start_time`,`end_time`) VALUES ('b','user3','2022-02-01 10:19:10','2022-02-01 11:05:06');

然后编写SQL：

SELECT
  room_id, start_time, end_time, user_cnt
FROM(
  SELECT
    room_id, start_time,
    lead(start_time) over(PARTITION BY room_id ORDER BY start_time) end_time,
    SUM(i) over(PARTITION BY room_id ORDER BY start_time) user_cnt
  FROM(
    SELECT room_id, start_time, COUNT(1) AS i
    FROM t1 GROUP BY room_id, start_time
    UNION ALL 
    SELECT room_id, end_time, -COUNT(1) AS i
    FROM t1 GROUP BY room_id, end_time
  ) a
) b
WHERE user_cnt>0;

OK，问题解决。若读者有更优的解决方案欢迎分享！！！