基于复杂字典为数据框创建新列

Question

我有一个复杂的字典来重新映射值。我如何在 python 中实现这一点？

cols_to_check = ["ColA","ColB","ColC"]
dic_string = "{(Type 1 | Type 2) : Type dual,
       (Type 3 | Type 4) : Type many,
        ELSE: Not listed
       }"
df = pd.DataFrame(
{
        'ID': ['AB01', 'AB02', 'AB03', 'AB04', 'AB05','AB06','AB07','AB08'],
        'ColA': ["Type 1","Undef",np.nan,"Undef",
                 "Type 1", "","", "Undef"],
        'ColB': ["N","Type 2","","",
                 "Y", np.nan,"", "N"],
        'ColC': [np.nan,"Undef","Type 3",np.nan,"Undef",
                 "Undef", "","Type 2"]
})

如果它是一个简单的字典并且其中没有提到 ELSE，我可以做到。假设我可以将“dic_string”转换为以下内容：

dic = {"Type 1" : "Type dual","Type 2" : "Type dual",
   "Type 3" : "Type many", "Type 4" : "Type many",
    "ELSE": "Not listed"
   }

如何使用新列“结果”制作像这样的最终结果。如何在不硬编码 dic 内容的情况下实现这一点？

Answer 1

使用np.select:

dic = {('Type 1', 'Type 2'): 'Type dual',
       ('Type 3', 'Type 4'): 'Type many',}
default = 'Not listed'

condlist = [df[cols_to_check].isin(k).any(axis=1) for k in dic]
choicelist = dic.values()

df['Result'] = np.select(condlist, choicelist, default)

输出：

     ID    ColA    ColB    ColC      Result
0  AB01  Type 1       N     NaN   Type dual
1  AB02   Undef  Type 2   Undef   Type dual
2  AB03     NaN          Type 3   Type many
3  AB04   Undef             NaN  Not listed
4  AB05  Type 1       Y   Undef   Type dual
5  AB06             NaN   Undef  Not listed
6  AB07                          Not listed
7  AB08   Undef       N  Type 2   Type dual