Question

考虑以下数据框

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]

我想获得一个专栏，跟踪name中每个col组合中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]中包含的所有唯一字符串。

即，预期输出为

[one two, three four]

事实上，对于A组，您可以看到[]，[one two]和[one two]中包含的唯一字符串集是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

我可以使用Pandas : how to get the unique number of values in cells when cells contain lists?获取相应数量的唯一值：

nunique

但将unique替换为http-proxy-host = 192.168.1.21 http-proxy-port = 3690 http-proxy-username = [username] http-proxy-password = [password]失败。

有什么想法吗？谢谢！

Answer 1

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')

期望的输出：

  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]

Answer 2

这是解决方案

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

熊猫：如何获取包含值列表的列的唯一值？

2 个答案: