리트코드 : 607. Sales Person
문제
Table: SalesPerson
+-----------------+---------+
| Column Name | Type |
+-----------------+---------+
| sales_id | int |
| name | varchar |
| salary | int |
| commission_rate | int |
| hire_date | date |
+-----------------+---------+
sales_id is the primary key (column with unique values) for this table.
Each row of this table indicates the name and the ID of a salesperson alongside their salary, commission rate, and hire date.
Table: Company
+-------------+---------+
| Column Name | Type |
+-------------+---------+
| com_id | int |
| name | varchar |
| city | varchar |
+-------------+---------+
com_id is the primary key (column with unique values) for this table.
Each row of this table indicates the name and the ID of a company and the city in which the company is located.
Table: Orders
+-------------+------+
| Column Name | Type |
+-------------+------+
| order_id | int |
| order_date | date |
| com_id | int |
| sales_id | int |
| amount | int |
+-------------+------+
order_id is the primary key (column with unique values) for this table.
com_id is a foreign key (reference column) to com_id from the Company table.
sales_id is a foreign key (reference column) to sales_id from the SalesPerson table.
Each row of this table contains information about one order. This includes the ID of the company, the ID of the salesperson, the date of the order, and the amount paid.
Write a solution to find the names of all the salespersons who did not have any orders related to the company with the name "RED".
Return the result table in any order.
The result format is in the following example.
Example 1:
Input:
SalesPerson table:
+----------+------+--------+-----------------+------------+
| sales_id | name | salary | commission_rate | hire_date |
+----------+------+--------+-----------------+------------+
| 1 | John | 100000 | 6 | 4/1/2006 |
| 2 | Amy | 12000 | 5 | 5/1/2010 |
| 3 | Mark | 65000 | 12 | 12/25/2008 |
| 4 | Pam | 25000 | 25 | 1/1/2005 |
| 5 | Alex | 5000 | 10 | 2/3/2007 |
+----------+------+--------+-----------------+------------+
Company table:
+--------+--------+----------+
| com_id | name | city |
+--------+--------+----------+
| 1 | RED | Boston |
| 2 | ORANGE | New York |
| 3 | YELLOW | Boston |
| 4 | GREEN | Austin |
+--------+--------+----------+
Orders table:
+----------+------------+--------+----------+--------+
| order_id | order_date | com_id | sales_id | amount |
+----------+------------+--------+----------+--------+
| 1 | 1/1/2014 | 3 | 4 | 10000 |
| 2 | 2/1/2014 | 4 | 5 | 5000 |
| 3 | 3/1/2014 | 1 | 1 | 50000 |
| 4 | 4/1/2014 | 1 | 4 | 25000 |
+----------+------------+--------+----------+--------+
Output:
+------+
| name |
+------+
| Amy |
| Mark |
| Alex |
+------+
Explanation:
According to orders 3 and 4 in the Orders table, it is easy to tell that only salesperson John and Pam have sales to company RED, so we report all the other names in the table salesperson.
문제 풀이
MySQL
SELECT NAME
FROM SALESPERSON
WHERE SALES_ID NOT IN (
SELECT SALES_ID
FROM ORDERS
WHERE COM_ID = (SELECT COM_ID FROM COMPANY WHERE NAME = 'RED'))
- RED 회사 넘버를 찾은 다음 ORDER TABLE에서 COM_ID가 RED와 관련있는 것들을 모두 뽑는다.
- SALES_ID가 이 서브쿼리에 없으면 된다. NOT IN 사용해서 작성하기.
Pandas
import pandas as pd
def sales_person(sales_person: pd.DataFrame, company: pd.DataFrame, orders: pd.DataFrame) -> pd.DataFrame:
red = company[company['name'] == 'RED'][['com_id']]
red_sales = orders[orders['com_id'].isin(red['com_id'])][['sales_id']]
return sales_person[~sales_person['sales_id'].isin(red_sales['sales_id'])][['name']]
- isin을 사용해서 데이터프레임에 조건 걸어주기.
- SQL에서 두 번의 서브쿼리를 사용하듯이 두 개의 임시 데이터프레임을 만들고 isin으로 조건을 걸어준다.
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