您可以首先生成一个将类别Map到产品的反转字典，例如a -> [good, neutral]。然后使用此反转字典将df中的值拆分为,、explode和map。然后使用groupby和set将它们重新聚集到展开列表产品上，最后将它们与,连接：

from collections import defaultdict
from itertools import chain

# form the dictionary
reversed_dict = defaultdict(list)
[reversed_dict[cat].append(prod) for prod, categs in mydict.items()
                                 for cat in categs]

# apply over the df
df["quality"] = (df.products
                   .str.split(",")
                   .explode()
                   .map(reversed_dict)
                   .groupby(level=0)
                   .agg(lambda s: ", ".join(set(chain.from_iterable(s)))))

得到

>>> df

  products        quality
0    a,b,c  good, neutral
1      a,c  good, neutral
2      b,d   bad, neutral
3    a,b,c  good, neutral

我们试试看

help = pd.Series(mydict).explode().reset_index().groupby(0)['index'].agg(','.join)

df['quality'] = df.products.replace(help,regex=True).str.split(',').map(set).str.join(',')
Out[150]: 
0    good,neutral
1    good,neutral
2     bad,neutral
3    good,neutral
Name: products, dtype: object

你应该这样定义mydict：

mydict = {'a': ['good', 'neutral'],
          'b': ['neutral'],
          'c': ['neutral'],
          'd': ['bad']}

然后：

def func(row):
    categories = []
    for item in row['products'].split(','):
        categories = categories + mydict[item]
    return ','.join(sorted(list(set(categories))))

    
df['quality'] = df.apply(lambda row: func(row), axis=1)

退货：

products    quality
0   a,b,c       good,neutral
1   a,c         good,neutral
2   b,d         bad,neutral
3   a,b,c       good,neutral

下面是另一种方法：

d = {'a': ['good', 'neutral'],
          'b': ['neutral'],
          'c': ['neutral'],
          'd': ['bad']}

df['quality'] = df['products'].str.split(',').explode().map(d).explode().groupby(level=0).unique().str.join(',')

或（第一部分将当前词典转换为新格式）

s = df['products'].str.split(',').explode()

d = {i:[] for i in set(s)}

for k,v in mydict.items():
    for i in v:
        d.get(i).append(k)

s.map(d).map(set).groupby(level=0).agg(lambda x: set.union(*x)).str.join(',')