184. 部门工资最高的员工

题目描述

表： Employee

+--------------+---------+
| 列名          | 类型    |
+--------------+---------+
| id           | int     |
| name         | varchar |
| salary       | int     |
| departmentId | int     |
+--------------+---------+
在 SQL 中，id是此表的主键。
departmentId 是 Department 表中 id 的外键（在 Pandas 中称为 join key）。
此表的每一行都表示员工的 id、姓名和工资。它还包含他们所在部门的 id。

 

表： Department

+-------------+---------+
| 列名         | 类型    |
+-------------+---------+
| id          | int     |
| name        | varchar |
+-------------+---------+
在 SQL 中，id 是此表的主键列。
此表的每一行都表示一个部门的 id 及其名称。

 

查找出每个部门中薪资最高的员工。
按 任意顺序 返回结果表。
查询结果格式如下例所示。

 

示例 1:

输入：
Employee 表:
+----+-------+--------+--------------+
| id | name  | salary | departmentId |
+----+-------+--------+--------------+
| 1  | Joe   | 70000  | 1            |
| 2  | Jim   | 90000  | 1            |
| 3  | Henry | 80000  | 2            |
| 4  | Sam   | 60000  | 2            |
| 5  | Max   | 90000  | 1            |
+----+-------+--------+--------------+
Department 表:
+----+-------+
| id | name  |
+----+-------+
| 1  | IT    |
| 2  | Sales |
+----+-------+
输出：
+------------+----------+--------+
| Department | Employee | Salary |
+------------+----------+--------+
| IT         | Jim      | 90000  |
| Sales      | Henry    | 80000  |
| IT         | Max      | 90000  |
+------------+----------+--------+
解释：Max 和 Jim 在 IT 部门的工资都是最高的，Henry 在销售部的工资最高。

解法

方法一：等值连接 + 子查询

我们可以使用等值连接，将 Employee 表和 Department 表连接起来，连接条件为 Employee.departmentId = Department.id，然后使用子查询来找到每个部门的最高工资，最后使用 WHERE 子句来筛选出每个部门中薪资最高的员工。

MySQLPython3

# Write your MySQL query statement below
SELECT d.name AS department, e.name AS employee, salary
FROM
    Employee AS e
    JOIN Department AS d ON e.departmentId = d.id
WHERE
    (d.id, salary) IN (
        SELECT departmentId, MAX(salary)
        FROM Employee
        GROUP BY 1
    );

import pandas as pd


def department_highest_salary(
    employee: pd.DataFrame, department: pd.DataFrame
) -> pd.DataFrame:
    # Merge the two tables on departmentId and department id
    merged = employee.merge(department, left_on='departmentId', right_on='id')

    # Find the maximum salary for each department
    max_salaries = merged.groupby('departmentId')['salary'].transform('max')

    # Filter employees who have the highest salary in their department
    top_earners = merged[merged['salary'] == max_salaries]

    # Select required columns and rename them
    result = top_earners[['name_y', 'name_x', 'salary']].copy()
    result.columns = ['Department', 'Employee', 'Salary']

    return result

方法二：等值连接 + 窗口函数

我们可以使用等值连接，将 Employee 表和 Department 表连接起来，连接条件为 Employee.departmentId = Department.id，然后使用窗口函数 rank()，它可以为每个部门的每个员工分配一个排名，然后我们可以选择排名为 \(1\) 的行即可。

MySQL

# Write your MySQL query statement below
WITH
    T AS (
        SELECT
            d.name AS department,
            e.name AS employee,
            salary,
            RANK() OVER (
                PARTITION BY d.name
                ORDER BY salary DESC
            ) AS rk
        FROM
            Employee AS e
            JOIN Department AS d ON e.departmentId = d.id
    )
SELECT department, employee, salary
FROM T
WHERE rk = 1;

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