1050. 合作过至少三次的演员和导演

1050. 合作过至少三次的演员和导演

2023年8月14日10:35:10

1050. 合作过至少三次的演员和导演

简单

SQL Schema

---

Pandas Schema

---

ActorDirector 表：

+-------------+---------+
| Column Name | Type    |
+-------------+---------+
| actor_id    | int     |
| director_id | int     |
| timestamp   | int     |
+-------------+---------+
timestamp 是这张表的主键(具有唯一值的列).

编写解决方案找出合作过至少三次的演员和导演的 id 对 (actor_id, director_id)

示例 1：

输入：
ActorDirector 表：
+-------------+-------------+-------------+
| actor_id    | director_id | timestamp   |
+-------------+-------------+-------------+
| 1           | 1           | 0           |
| 1           | 1           | 1           |
| 1           | 1           | 2           |
| 1           | 2           | 3           |
| 1           | 2           | 4           |
| 2           | 1           | 5           |
| 2           | 1           | 6           |
+-------------+-------------+-------------+
输出：
+-------------+-------------+
| actor_id    | director_id |
+-------------+-------------+
| 1           | 1           |
+-------------+-------------+
解释：
唯一的 id 对是 (1, 1)，他们恰好合作了 3 次。

通过次数

55.6K

提交次数

72.6K

通过率

76.5%

答案

import pandas as pd

def actors_and_directors(actor_director: pd.DataFrame) -> pd.DataFrame:

    # cnts = actor_director.groupby(['actor_id', 'director_id']).size().reset_index(name='counts')
    # return cnts[cnts['counts'] >= 3][['actor_id', 'director_id']]

    actor_director = actor_director.groupby(["actor_id","director_id"],as_index=False).agg({"timestamp": lambda x:x.count()>=3})
    return actor_director[actor_director["timestamp"]==True][["actor_id","director_id"]]