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181. 超过经理收入的员工

题目描述

表：Employee 

+-------------+---------+
| Column Name | Type    |
+-------------+---------+
| id          | int     |
| name        | varchar |
| salary      | int     |
| managerId   | int     |
+-------------+---------+
id 是该表的主键（具有唯一值的列）。
该表的每一行都表示雇员的ID、姓名、工资和经理的ID。

 

编写解决方案，找出收入比经理高的员工。

以 任意顺序 返回结果表。

结果格式如下所示。

 

示例 1:

输入: 
Employee 表:
+----+-------+--------+-----------+
| id | name  | salary | managerId |
+----+-------+--------+-----------+
| 1  | Joe   | 70000  | 3         |
| 2  | Henry | 80000  | 4         |
| 3  | Sam   | 60000  | Null      |
| 4  | Max   | 90000  | Null      |
+----+-------+--------+-----------+
输出: 
+----------+
| Employee |
+----------+
| Joe      |
+----------+
解释: Joe 是唯一挣得比经理多的雇员。

解法

方法一

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import pandas as pd


def find_employees(employee: pd.DataFrame) -> pd.DataFrame:
    df = employee.merge(right=employee, how="left", left_on="managerId", right_on="id")
    emp = df[df["salary_x"] > df["salary_y"]]["name_x"]

    return pd.DataFrame({"Employee": emp})

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SELECT Name AS Employee
FROM Employee AS Curr
WHERE
    Salary > (
        SELECT Salary
        FROM Employee
        WHERE Id = Curr.ManagerId
    );

方法二

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# Write your MySQL query statement below
SELECT
    e1.name AS Employee
FROM
    Employee AS e1
    JOIN Employee AS e2 ON e1.managerId = e2.id
WHERE e1.salary > e2.salary;

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