如何将两个或多个相关但不同的特性组合为一个功能，并为该功能分配新值

Question

fuel_type= {'fuel_Petrol':1,'fuel_Diesel':2,'fuel_Electric':3,'fuel_LPG':4,'fuel_CNG':5}
df['fueltype']= df['fuel_Petrol','fuel_Diesel','fuel_Electric','fuel_LPG','fuel_CNG'].apply(lamda x: fuel_type(x))

我得到了这个错误：

 File "<ipython-input-31-b238d0a16f2d>", line 2
    df['fueltype']= df['fuel_Petrol','fuel_Diesel','fuel_Electric','fuel_LPG','fuel_CNG'].apply(lamda x: fuel_type(x))
                                                                                                      ^ SyntaxError: invalid syntax

Answer 1

如果您有一个值列，您希望使用字典对其进行编码，则应该使用：

fuel_type= {
    'fuel_Petrol':1,'fuel_Diesel':2,
    'fuel_Electric':3,'fuel_LPG':4,'fuel_CNG':5}

df = pd.DataFrame({
    'fuel_type': ['fuel_Petrol','fuel_Diesel',
                  'fuel_Electric','fuel_LPG','fuel_CNG']
})

df['fueltype'] = df['fuel_type'].apply(lambda  x: fuel_type[x])

print (df)

输出：

       fuel_type  fueltype
0    fuel_Petrol         1
1    fuel_Diesel         2
2  fuel_Electric         3
3       fuel_LPG         4
4       fuel_CNG         5

但是，当您有太多不同的值要编码时，这并不是推荐的方法。相反，使用下面的方式进行编码，这不需要我们手动创建字典。

df["fuel_type"] = df["fuel_type"].astype('category')
df["fueltype"] = df["fuel_type"].cat.codes
print (df)

输出：

       fuel_type  fueltype
0    fuel_Petrol         4
1    fuel_Diesel         1
2  fuel_Electric         2
3       fuel_LPG         3
4       fuel_CNG         0

sklearn还具有执行标签编码的功能。