Question

数据框内的case语句，我在这里做什么错了？

df1['Response'] = [4 if x =='Closed with monetary relief' 
               3 if x =='Closed with non-monetary relief'
               2 if x =='Closed with explanation' 
               1 if x =='Closed' 
               0 if x =='Untimely response' for x in df1['Response']]

我看到的错误：

3，如果x =='通过非货币救济关闭' ^ SyntaxError：语法无效

Answer 1

我认为最好按字典使用Series.map：

d = {'Closed with monetary relief' : 4,
     'Closed with non-monetary relief':3.
     'Closed with explanation':2,
     'Closed':1,
     'Untimely response':0}

df1['Response'] = df1['Response'].map(d)

如果某些不匹配的值丢失了值，则可以将其替换为原始值：

df1['Response'] = df1['Response'].map(d).fillna(df1['Response'])

或其他一些值，例如-1，然后还将值转换为整数，因为至少有一个NaN创建了float列：

df1['Response'] = df1['Response'].map(d).fillna(-1).astype(int)

Answer 2

尝试这种格式，它应该可以工作：

df1.Response.apply(lambda x: 4 if x =='Closed with monetary relief' else
               3 if x =='Closed with non-monetary relief' else
               2 if x =='Closed with explanation' else
               1 if x =='Closed' else
               0 if x =='Untimely response' else None )

根据条件更改数据框的值

2 个答案: