Python 生成错误结果的DataFrame.groupby.ank

这里有两张桌子-

Employees
+----+-------+--------+--------------+
| id | name  | salary | departmentId |
+----+-------+--------+--------------+
| 1  | Joe   | 70000  | 1            |
| 2  | Jim   | 90000  | 1            |
| 3  | Henry | 80000  | 2            |
| 4  | Sam   | 60000  | 2            |
| 5  | Max   | 90000  | 1            |
+----+-------+--------+--------------+

Department table:
+----+-------+
| id | name  |
+----+-------+
| 1  | IT    |
| 2  | Sales |
+----+-------+

Required Output

Output: 
+------------+----------+--------+
| Department | Employee | Salary |
+------------+----------+--------+
| IT         | Jim      | 90000  |
| Sales      | Henry    | 80000  |
| IT         | Max      | 90000  |
+------------+----------+--------+

问题是找到每个部门的最高工资.

我的Python代码是这样的-

import pandas as pd

def department_highest_salary(employee: pd.DataFrame, department: pd.DataFrame) -> pd.DataFrame:
    df = employee.merge(department, how = "inner", left_on ="departmentId", right_on = "id")
    df["rank"] = employee.groupby("departmentId")["salary"].rank(method = "dense", ascending=False)
    return( df.query("rank == 1")[["name_y","name_x","salary"]].rename(columns = {"name_y":"Department", "name_x":"Employee", "salary":"Salary"}))

这是错误的，因为groupby.rank返回错误的结果-

   id_x name_x  salary  departmentId  id_y name_y  rank
0     1    Joe   70000             1     1     IT   2.0
1     2    Jim   90000             1     1     IT   1.0
2     5    Max   90000             1     1     IT   1.0
3     3  Henry   80000             2     2  Sales   2.0
4     4    Sam   60000             2     2  Sales   1.0

为什么Henry的排名是2，不应该是1吗？我到底做错了什么？