Python 我可以使用pandas loc方法选择多个列并用NaN替换结果行值吗

我有这样一个数据帧：

students = {'ID': [2, 3, 5, 7, 11, 13], 
        'Name':['John','Jane','Sam','James','Stacy','Mary'],
        'Gender':['M','F','F','M','F','F'],
        'school_name':['College2','College2','College10','College2','College2','College2'],
        'grade':['9th','10th','9th','9th','8th','5th'],
        'math_score':[90,89,88,89,89,90],
        'art_score':[90,89,89,78,90,94]}
        
        students_df = pd.DataFrame(students)

我是否可以使用学生df上的loc方法来选择College2九年级的所有数学和艺术分数，并将其替换为NaN？ 是否有一种干净的方法可以做到这一点，而不会将过程分成两部分：一部分用于子集，另一部分用于替换

我试图选择这种方式：

students_df.loc[(students_df['school_name'] == 'College2') & (students_df['grade'] == "9th"),['grade','school_name','math_score','art_score']]

students_df['math_score'] = np.where((students_df['school_name']=='College2') & (students_df['grade']=='9th'), np.NaN, students_df['math_score'])

我这样替换：

students_df.loc[(students_df['school_name'] == 'College2') & (students_df['grade'] == "9th"),['grade','school_name','math_score','art_score']]

students_df['math_score'] = np.where((students_df['school_name']=='College2') & (students_df['grade']=='9th'), np.NaN, students_df['math_score'])

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我可以使用loc和np.NaN以更干净、更高效的方式实现同样的目标吗？

首先选择列替换缺少的值，然后设置

NaN

：

students_df.loc[(students_df['school_name'] == 'College2') & (students_df['grade'] == "9th"),['math_score','art_score']] = np.nan
print (students_df)
   ID   Name Gender school_name grade  math_score  art_score
0   2   John      M    College2   9th         NaN        NaN
1   3   Jane      F    College2  10th        89.0       89.0
2   5    Sam      F   College10   9th        88.0       89.0
3   7  James      M    College2   9th         NaN        NaN
4  11  Stacy      F    College2   8th        89.0       90.0
5  13   Mary      F    College2   5th        90.0       94.0

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谢谢你，耶斯雷尔。你太棒了！！