使用嵌套字典根据条件映射数据帧中的值

首先，为嵌套字典中元组的单独值重塑嵌套字典：

from collections import defaultdict

d = defaultdict(dict)

for k, v in dict_map.items():
    for k1, v1 in v.items():
        if isinstance(v1, tuple):
            for x in v1:
                d[k1][x] = k
        else:
            d[k1][v1] = k

print (d)
defaultdict(<class 'dict'>, {'Drug': {'A': 'Anti', 'B': 'Anti', 
                                      'C': 'Anti', 'D': 'Undef'},
                             'Name': {'Type X': 'Undef', 'Vitamin C': 'Vit ',
                                      'Placebo': 'PPL'}})

df = pd.DataFrame(
    {
            'ID': ['AB01', 'AB02', 'AB03', 'AB04', 'AB05','AB06'],
            'Drug': ["A","B","A",np.nan,
                     "D","D"],
            'Name': ['Placebo', 'Vitamin C', np.nan, 'Placebo', '', 'Type X']
    }
    )

然后按字典映射，优先级是按列表中列的顺序排列的：cols_to_map

cols_to_map = ["Drug", "Name"]

df['Result'] = np.nan
for col in cols_to_map:
    df['Result'] = df['Result'].combine_first(df[col].map(d[col]))

print (df)
     ID Drug       Name Result
0  AB01    A    Placebo   Anti
1  AB02    B  Vitamin C   Anti
2  AB03    A        NaN   Anti
3  AB04  NaN    Placebo    PPL
4  AB05    D             Undef
5  AB06    D     Type X  Undef

cols_to_map = [ "Name","Drug"]

df['Result'] = np.nan
for col in cols_to_map:
    df['Result'] = df['Result'].combine_first(df[col].map(d[col]))

print (df)
     ID Drug       Name Result
0  AB01    A    Placebo    PPL
1  AB02    B  Vitamin C   Vit 
2  AB03    A        NaN   Anti
3  AB04  NaN    Placebo    PPL
4  AB05    D             Undef
5  AB06    D     Type X  Undef

编辑：

df['Result1'] = df['Drug'].map(d['Drug'])
df['Result2'] = df['Name'].map(d['Name'])
print (df)
     ID Drug       Name Result1 Result2
0  AB01    A    Placebo    Anti     PPL
1  AB02    B  Vitamin C    Anti    Vit 
2  AB03    A        NaN    Anti     NaN
3  AB04  NaN    Placebo     NaN     PPL
4  AB05    D              Undef     NaN
5  AB06    D     Type X   Undef   Undef

由于 dict 和预期结果之间的关系非常复杂，我将使用一个函数来应用于您的 DataFrame。这使我们免于操纵字典：

def get_result(row):
    result = np.nan
    for k,v in dict_map.items():
        if row['Name'] in v.values():
            result = k
        if row['Name'] and type(row['Drug']) == str and 'Drug' in v.keys() and row['Drug'] in v['Drug']:
            return k
    return result


df['Result'] = df.apply(lambda row: get_result(row), axis=1)
print(df)

输出：

     ID Drug       Name Result
0  AB01    A    Placebo   Anti
1  AB02    B  Vitamin C   Anti
2  AB03    A        NaN   Anti
3  AB04  NaN    Placebo    PPL
4  AB05    D               NaN
5  AB06    D     Type X  Undef

在更新您的问题后，我将函数更改为通用函数。不过，我不太确定它是否会涵盖您的所有情况，因为您的输出不会随着新列而发生太大变化：

col_to_maps = ["Drug", "Name", "Batch"]

def get_result(row, dict_map):
    result = np.nan
    for k,v in dict_map.items():
        for i,col in enumerate(col_to_maps[:-1]):    
            if type(v)==dict:
                if str(row[col]) and \
                 all(str(row[other_col])
                     and (not(str(other_col) in v.keys()) and str(col) in v.keys() and str(row[col]) in v[col]
                          or str(other_col) in v.keys() and str(row[other_col]) in v[other_col]
                          )
                     for other_col in col_to_maps[i+1:]
                    ):
                    return k 
            elif str(row[col]) in v:
                result = k
    return result


df['Result'] = df.apply(lambda row: get_result(row, dict_map), axis=1)
print(df)

输出：

     ID Drug       Name Batch          Result
0  AB01    A    Placebo   ABC            Anti
1  AB02    B  Vitamin C   NaN            Anti
2  AB03    A        NaN   NaN            Anti
3  AB04  NaN    Placebo   XYZ  Placebo Effect
4  AB05    D              NaN             NaN
5  AB06    D     Type X   NaN           Undef