熊猫：如何获取包含值列表的列的唯一值？

Question

考虑下面的数据帧

df = pd.DataFrame({'name' : [['one two','three four'], ['one'],[], [],['one two'],['three']], 
        'col' : ['A','B','A','B','A','B']})  
df.sort_values(by='col',inplace=True) 

df 
Out[62]: 
    col     name 
0 A [one two, three four] 
2 A      [] 
4 A    [one two] 
1 B     [one] 
3 B      [] 
5 B    [three] 

我想获得一个跟踪列入name为col每个组合的所有唯一字符串的列。

也就是说，预期产量

df 
Out[62]: 
    col     name unique_list 
0 A [one two, three four] [one two, three four] 
2 A      [] [one two, three four] 
4 A    [one two] [one two, three four] 
1 B     [one] [one, three] 
3 B      [] [one, three] 
5 B    [three] [one, three] 

事实上，说为一组，你可以看到，唯一的一组字符串包含在[one two, three four]，[]和[one two]是[one two]

我能获得相应使用的唯一值数量Pandas : how to get the unique number of values in cells when cells contain lists?：

df['count_unique']=df.groupby('col')['name'].transform(lambda x: list(pd.Series(x.apply(pd.Series).stack().reset_index(drop=True, level=1).nunique()))) 


df 
Out[65]: 
    col     name count_unique 
0 A [one two, three four]   2 
2 A      []   2 
4 A    [one two]   2 
1 B     [one]   2 
3 B      []   2 
5 B    [three]   2 

，但替换nunique与unique以上失败。

任何想法？ 谢谢！

Answer 1

下面是解

df['unique_list'] = df.col.map(df.groupby('col')['name'].sum().apply(np.unique)) 
    df 

Answer 2

尝试：

uniq_df = df.groupby('col')['name'].apply(lambda x: list(set(reduce(lambda y,z: y+z,x)))).reset_index() 
uniq_df.columns = ['col','uniq_list'] 
pd.merge(df,uniq_df, on='col', how='left') 

所需的输出：

col     name    uniq_list 
0 A [one two, three four] [one two, three four] 
1 A      [] [one two, three four] 
2 A    [one two] [one two, three four] 
3 B     [one]   [three, one] 
4 B      []   [three, one] 
5 B    [three]   [three, one]