熊猫将输出转换为逗号单独的字符串，而不是列表

Question

我有一个看起来像这样的DataFrame：

df = pd.DataFrame({'ID':[1,1,2,2,3,4],'Name':['John Doe','Jane Doe','John Smith','Jane Smith','Jack Hill','Jill Hill']})

    ID  Name
0   1   John Doe
1   1   Jane Doe
2   2   John Smith
3   2   Jane Smith
4   3   Jack Hill
5   4   Jill Hill

然后，我按ID添加了另一列分组，并取名称中的唯一值：

df['Multi Name'] = df.groupby('ID')['Name'].transform('unique')

    ID  Name    Multi Name
0   1   John Doe    [John Doe, Jane Doe]
1   1   Jane Doe    [John Doe, Jane Doe]
2   2   John Smith  [John Smith, Jane Smith]
3   2   Jane Smith  [John Smith, Jane Smith]
4   3   Jack Hill   [Jack Hill]
5   4   Jill Hill   [Jill Hill]

如何从多个名称中删除括号？

我试过：

df['Multi Name'] = df['Multi Name'].str.strip('[]')


ID  Name    Multi Name
0   1   John Doe    NaN
1   1   Jane Doe    NaN
2   2   John Smith  NaN
3   2   Jane Smith  NaN
4   3   Jack Hill   NaN
5   4   Jill Hill   NaN

期望产出：

    ID  Name    Multi Name
0   1   John Doe    John Doe, Jane Doe
1   1   Jane Doe    John Doe, Jane Doe
2   2   John Smith  John Smith, Jane Smith
3   2   Jane Smith  John Smith, Jane Smith
4   3   Jack Hill   Jack Hill
5   4   Jill Hill   Jill Hill

Answer 1

transform

df.join(df.groupby('ID').Name.transform('unique').rename('Multi Name'))

   ID        Name                Multi Name
0   1    John Doe      [John Doe, Jane Doe]
1   1    Jane Doe      [John Doe, Jane Doe]
2   2  John Smith  [John Smith, Jane Smith]
3   2  Jane Smith  [John Smith, Jane Smith]
4   3   Jack Hill               [Jack Hill]
5   4   Jill Hill               [Jill Hill]

df.join(df.groupby('ID').Name.transform('unique').str.join(', ').rename('Multi Name'))

   ID        Name              Multi Name
0   1    John Doe      John Doe, Jane Doe
1   1    Jane Doe      John Doe, Jane Doe
2   2  John Smith  John Smith, Jane Smith
3   2  Jane Smith  John Smith, Jane Smith
4   3   Jack Hill               Jack Hill
5   4   Jill Hill               Jill Hill

map

df.join(df.ID.map(df.groupby('ID').Name.unique().str.join(', ')).rename('Multi Name'))

   ID        Name              Multi Name
0   1    John Doe      John Doe, Jane Doe
1   1    Jane Doe      John Doe, Jane Doe
2   2  John Smith  John Smith, Jane Smith
3   2  Jane Smith  John Smith, Jane Smith
4   3   Jack Hill               Jack Hill
5   4   Jill Hill               Jill Hill

itertools.groupby

from itertools import groupby

d = {
    k: ', '.join(x[1] for x in v)
    for k, v in groupby(sorted(set(zip(df.ID, df.Name))), key=lambda x: x[0])
}

df.join(df.ID.map(d).rename('Multi Name'))

   ID        Name              Multi Name
0   1    John Doe      Jane Doe, John Doe
1   1    Jane Doe      Jane Doe, John Doe
2   2  John Smith  Jane Smith, John Smith
3   2  Jane Smith  Jane Smith, John Smith
4   3   Jack Hill               Jack Hill
5   4   Jill Hill               Jill Hill

Answer 2

在这里，unique似乎是函数的错误选择。我推荐一个使用str.join的定制lambda函数

df['Multi Name'] = df.groupby('ID')['Name'].transform(lambda x: ', '.join(set(x)))

​

df
   ID        Name              Multi Name
0   1    John Doe      John Doe, Jane Doe
1   1    Jane Doe      John Doe, Jane Doe
2   2  John Smith  Jane Smith, John Smith
3   2  Jane Smith  Jane Smith, John Smith
4   3   Jack Hill               Jack Hill
5   4   Jill Hill               Jill Hill

Answer 3

使用map和join

df['Multi Name'] = df.groupby('ID')['Name'].transform('unique').map(', '.join)

输出：

   ID        Name              Multi Name
0   1    John Doe      John Doe, Jane Doe
1   1    Jane Doe      John Doe, Jane Doe
2   2  John Smith  John Smith, Jane Smith
3   2  Jane Smith  John Smith, Jane Smith
4   3   Jack Hill               Jack Hill
5   4   Jill Hill               Jill Hill