Question

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;