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185. 部门工资前三高的所有员工

English Version

题目描述

表: Employee

+--------------+---------+
| Column Name  | Type    |
+--------------+---------+
| id           | int     |
| name         | varchar |
| salary       | int     |
| departmentId | int     |
+--------------+---------+
id 是该表的主键列(具有唯一值的列)。
departmentId 是 Department 表中 ID 的外键（reference 列）。
该表的每一行都表示员工的ID、姓名和工资。它还包含了他们部门的ID。

 

表: Department

+-------------+---------+
| Column Name | Type    |
+-------------+---------+
| id          | int     |
| name        | varchar |
+-------------+---------+
id 是该表的主键列(具有唯一值的列)。
该表的每一行表示部门ID和部门名。

 

公司的主管们感兴趣的是公司每个部门中谁赚的钱最多。一个部门的 高收入者 是指一个员工的工资在该部门的 不同 工资中 排名前三 。

编写解决方案，找出每个部门中 收入高的员工 。

以 任意顺序 返回结果表。

返回结果格式如下所示。

 

示例 1:

输入: 
Employee 表:
+----+-------+--------+--------------+
| id | name  | salary | departmentId |
+----+-------+--------+--------------+
| 1  | Joe   | 85000  | 1            |
| 2  | Henry | 80000  | 2            |
| 3  | Sam   | 60000  | 2            |
| 4  | Max   | 90000  | 1            |
| 5  | Janet | 69000  | 1            |
| 6  | Randy | 85000  | 1            |
| 7  | Will  | 70000  | 1            |
+----+-------+--------+--------------+
Department  表:
+----+-------+
| id | name  |
+----+-------+
| 1  | IT    |
| 2  | Sales |
+----+-------+
输出: 
+------------+----------+--------+
| Department | Employee | Salary |
+------------+----------+--------+
| IT         | Max      | 90000  |
| IT         | Joe      | 85000  |
| IT         | Randy    | 85000  |
| IT         | Will     | 70000  |
| Sales      | Henry    | 80000  |
| Sales      | Sam      | 60000  |
+------------+----------+--------+
解释:
在IT部门:
- Max的工资最高
- 兰迪和乔都赚取第二高的独特的薪水
- 威尔的薪水是第三高的

在销售部:
- 亨利的工资最高
- 山姆的薪水第二高
- 没有第三高的工资，因为只有两名员工

解法

方法一

import pandas as pd


def top_three_salaries(
    employee: pd.DataFrame, department: pd.DataFrame
) -> pd.DataFrame:
    salary_cutoff = (
        employee.drop_duplicates(["salary", "departmentId"])
        .groupby("departmentId")["salary"]
        .nlargest(3)
        .groupby("departmentId")
        .min()
    )
    employee["Department"] = department.set_index("id")["name"][
        employee["departmentId"]
    ].values
    employee["cutoff"] = salary_cutoff[employee["departmentId"]].values
    return employee[employee["salary"] >= employee["cutoff"]].rename(
        columns={"name": "Employee", "salary": "Salary"}
    )[["Department", "Employee", "Salary"]]
SELECT
    Department.NAME AS Department,
    Employee.NAME AS Employee,
    Salary
FROM
    Employee,
    Department
WHERE
    Employee.DepartmentId = Department.Id
    AND (
        SELECT
            COUNT(DISTINCT e2.Salary)
        FROM Employee AS e2
        WHERE e2.Salary > Employee.Salary AND Employee.DepartmentId = e2.DepartmentId
    ) < 3;

方法二

# Write your MySQL query statement below
WITH
    T AS (
        SELECT
            *,
            DENSE_RANK() OVER (
                PARTITION BY departmentId
                ORDER BY salary DESC
            ) AS rk
        FROM Employee
    )
SELECT d.name AS Department, t.name AS Employee, salary AS Salary
FROM
    T AS t
    JOIN Department AS d ON t.departmentId = d.id
WHERE rk <= 3;