lambda中的语法错误，带有if表达式

Question

basic_salary['dem_education_level_numeric']=basic_salary['dem_education_level'].apply(lambda x : 3 if (x == 'high'))

我收到以下错误：

SyntaxError: invalid syntax (<ipython-input-20-80adac538241>, line 1)
  File "<ipython-input-20-80adac538241>", line 1
    basic_salary['dem_education_level_numeric']=basic_salary['dem_education_level'].apply(lambda x : 3 if (x == 'high'))
                                                                                                                       ^
SyntaxError: invalid syntax

请帮忙

Answer 1

如果您要在if语句中添加lambda，则还必须定义else。每个lambda语句都必须返回一些内容，即使它是None。如果lambda不是None，则返回x，将您的"high"语句更改为此

lambda x : 3 if x == 'high' else None # paralysis isn't necessary 

Answer 2

Lambda语句需要返回一些内容，如果没有else子句，则返回可能不明确。改为：

basic_salary['dem_education_level_numeric']=basic_salary['dem_education_level'].apply(lambda x : 3 if x == 'high' else 0)