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문제
Table: Employee
+-------------+---------+
| Column Name | Type |
+-------------+---------+
| empId | int |
| name | varchar |
| supervisor | int |
| salary | int |
+-------------+---------+
empId is the column with unique values for this table.
Each row of this table indicates the name and the ID of an employee in addition to their salary and the id of their manager.
Table: Bonus
+-------------+------+
| Column Name | Type |
+-------------+------+
| empId | int |
| bonus | int |
+-------------+------+empId is the column of unique values for this table.
empId is a foreign key (reference column) to empId from the Employee table.
Each row of this table contains the id of an employee and their respective bonus.
Write a solution to report the name and bonus amount of each employee with a bonus less than 1000.
Return the result table in any order.
https://leetcode.com/problems/employee-bonus/description/?lang=pythondata
예시
Input:
Employee table:
+-------+--------+------------+--------+
| empId | name | supervisor | salary |
+-------+--------+------------+--------+
| 3 | Brad | null | 4000 |
| 1 | John | 3 | 1000 |
| 2 | Dan | 3 | 2000 |
| 4 | Thomas | 3 | 4000 |
+-------+--------+------------+--------+
Bonus table:
+-------+-------+
| empId | bonus |
+-------+-------+
| 2 | 500 |
| 4 | 2000 |
+-------+-------+
Output:
+------+-------+
| name | bonus |
+------+-------+
| Brad | null |
| John | null |
| Dan | 500 |
+------+-------+문제 풀이
import pandas as pd
def employee_bonus(employee: pd.DataFrame, bonus: pd.DataFrame) -> pd.DataFrame:
# empId 기준으로 merge
merged = pd.merge(employee, bonus, on = 'empId', how='left')
# bonus 1,000이하인 행 필터링(null은 0원으로 null도 포함)
filtered = merged.loc[(merged['bonus'] < 1000) | merged['bonus'].isna(), :]
result = filtered.loc[:,['name','bonus']]
return result파이썬을 독학하시는 분들에게 도움이 되길 바라며,
혹 더 좋은 방법이 있거나 오류가 있다면 편하게 말씀 부탁드립니다.
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