好了，我有两个框架：

df = pd.DataFrame({'A':['German Shepherd','Border Collie','Golden Retriever','Beagle','Daschund']})
df = df.T
df.columns = df.iloc[0]
df = df.drop(df.index[0])

A   German Shepherd     Border Collie   Golden Retriever    Beagle  Daschund

df2 = pd.DataFrame({'ID':['A','A','A','B','C','C','C','C','C'],
                   'Breed':['German Shepherd','Beagle','Dashung','Border Collie',
                           'German Shepherd','Border Collie','Golden Retriever','Beagle','Daschund']})

ID  Breed
0   A   German Shepherd
1   A   Beagle
2   A   Dashung
3   B   Border Collie
4   C   German Shepherd
5   C   Border Collie
6   C   Golden Retriever
7   C   Beagle
8   C   Daschund

我想在df 2中找到狗品种的ID，然后更新df，如果它存在于该ID中：

dogs_grouped = df2.groupby('ID')
missing_dogs = []
vals = [np.nan for i in df.columns]
for group_name, df_group in dogs_grouped:
    print(f'Cluster: {group_name}')
    cluster_dogs = sorted(list(set(df_group['Breed'].to_list())))
    cluster_dogs = [i for i in cluster_dogs if i in all_dogs]
    weird_dogs = [i for i in cluster_dogs if i not in all_dogs]
    missing_dogs.append(weird_dogs)
    df = df.append(pd.Series(vals, index=df.columns, name=group_name))
    df.loc[group_name][cluster_dogs] = 1
df = df.fillna(0)

我的代码可以工作，但对于大型数据集来说非常慢。我有一个50万行的数据集，我正在迭代，创建一个4000 x 30，000的矩阵需要几个小时。

A   German Shepherd     Border Collie   Golden Retriever    Beagle      Daschund
A        1                    0               0                1           0
B        0                    1               0                0           0
C        1                    1               1                1           1

必须有一个更pythonic/Pandas的方式来处理这个问题？