176. 第二高的薪水
176. 第二高的薪水
2023年8月12日19:03:40
中等
SQL Schema
Pandas Schema
Employee 表：
+-------------+------+
| Column Name | Type |
+-------------+------+
| id | int |
| salary | int |
+-------------+------+
在 SQL 中，id 是这个表的主键。
表的每一行包含员工的工资信息。
查询并返回 Employee 表中第二高的薪水 。如果不存在第二高的薪水，查询应该返回 null(Pandas 则返回 None) 。
查询结果如下例所示。
示例 1：
输入：
Employee 表：
+----+--------+
| id | salary |
+----+--------+
| 1 | 100 |
| 2 | 200 |
| 3 | 300 |
+----+--------+
输出：
+---------------------+
| SecondHighestSalary |
+---------------------+
| 200 |
+---------------------+
示例 2：
输入：
Employee 表：
+----+--------+
| id | salary |
+----+--------+
| 1 | 100 |
+----+--------+
输出：
+---------------------+
| SecondHighestSalary |
+---------------------+
| null |
+---------------------+
这错在哪里了
import pandas as pd
def second_highest_salary(employee: pd.DataFrame) -> pd.DataFrame:
employee['SecondHighestSalary'] = employee.drop_duplicates(subset=['salary']).sort_values(by="salary",ascending=False)[["salary"]]
if(len(employee)>=2):
return employee[['SecondHighestSalary']].iloc[1:2,[0]]
else:
return pd.DataFrame({"SecondHighestSalary":['null']})
我的错
import pandas as pd
def second_highest_salary(employee: pd.DataFrame) -> pd.DataFrame:
employee['SecondHighestSalary'] = employee.drop_duplicates(subset=['salary']).sort_values(by="salary",ascending=False)[["salary"]]
if(len(employee)>=2):
return employee[['SecondHighestSalary']].iloc[1:2,[0]]
else:
return pd.DataFrame({"SecondHighestSalary":[np.NaN]})
答案
import pandas as pd
def second_highest_salary(employee: pd.DataFrame) -> pd.DataFrame:
employee['SecondHighestSalary'] = employee.drop_duplicates(subset=['salary'])[["salary"]]
if(len(employee)>=2):
return employee[['SecondHighestSalary']].sort_values(by="SecondHighestSalary",ascending=False).iloc[1:2,[0]]
else:
return pd.DataFrame({"SecondHighestSalary":[np.NaN]})
