基于dataframe中给定NaN值的现有列添加新列

Question

我想基于dataframe的现有列添加列。框架包含5列。我需要用数值替换category列。基于此，我需要添加'Class'列，并根据前面提到的条件指定值0或1

Desired result:
        File         Task  Category  Class
0    g0pA_taska.txt    a      0       0
1    g0pA_taskb.txt    b      3       1
2    g0pA_taskc.txt    c      2       1
3    g0pA_taskd.txt    d      1       1
4    g0pA_taske.txt    e      0       0
...
...
99   orig_taske.txt    e     -1      -1
 plagiarism_df.replace({'Category' : {'non':0,'heavy':1,'light':2,'cut':3,'orig':-1}})
    plagiarism_df.loc[plagiarism_df['Category']==0, 'Class'] = 0
    plagiarism_df.loc[plagiarism_df['Category']==1, 'Class'] = 1
    plagiarism_df.loc[plagiarism_df['Category']==2, 'Class'] = 1
    plagiarism_df.loc[plagiarism_df['Category']==3, 'Class'] = 1
    plagiarism_df.loc[plagiarism_df['Category']==-1,'Class'] = 1 

Answer 1

您没有修改数据帧，replace返回一个新的数据帧，您必须分配它： plagiarism_df = plagiarism_df.replace({'Category': { 'non': 0, 'heavy': 1, 'light': 2, 'cut': 3, 'orig': -1 }})

或者使用que paraminplace = True修改DataFrame对象，如下所示：

plagiarism_df.replace({'Category':{ 'non': 0, 'heavy': 1, 'light': 2, 'cut': 3, 'orig': -1}}, inplace=True)

Answer 2

或者，您可以使用map函数，然后应用lambda来获得所需的结果： plagiarism_df['Category'] = plagiarism_df['Category'].map({ 'non': 0, 'heavy': 1, 'light': 2, 'cut': 3, 'orig': -1})

plagiarism_df['Class'] = plagiarism_df['Category'].apply(lambda x: 1 if x in [1,2,3,-1] else 0)