python - pandas dataframe - 根据唯一用户对艺术家进行分组

Question

为了避免同一用户的重复，我想使用 pandas groupby 函数整齐地组织一个 {k:artist1,artist2,artist3,etc} 的嵌套字典。这是示例数据(我的直觉告诉我链接一个聚合函数？)

...像df.groupby('users')？

    users                                       artist
0   00001411dc427966b17297bf4d69e7e193135d89    the most serene republic
1   00001411dc427966b17297bf4d69e7e193135d89    stars
2   00001411dc427966b17297bf4d69e7e193135d89    broken social scene
3   00001411dc427966b17297bf4d69e7e193135d89    have heart
4   00001411dc427966b17297bf4d69e7e193135d89    luminous orange
5   00001411dc427966b17297bf4d69e7e193135d89    boris
6   00001411dc427966b17297bf4d69e7e193135d89    arctic monkeys
7   00001411dc427966b17297bf4d69e7e193135d89    bright eyes
8   00001411dc427966b17297bf4d69e7e193135d89    coaltar of the deepers
9   00001411dc427966b17297bf4d69e7e193135d89    polar bear club
10  00001411dc427966b17297bf4d69e7e193135d89    the libertines
11  00001411dc427966b17297bf4d69e7e193135d89    death from above 1979
12  00001411dc427966b17297bf4d69e7e193135d89    owl city
13  00001411dc427966b17297bf4d69e7e193135d89    coldplay
14  00001411dc427966b17297bf4d69e7e193135d89    okkervil river
15  00001411dc427966b17297bf4d69e7e193135d89    jim sturgess
16  00001411dc427966b17297bf4d69e7e193135d89    deerhoof
17  00001411dc427966b17297bf4d69e7e193135d89    fear before the march of flames
18  00001411dc427966b17297bf4d69e7e193135d89    breathe carolina
19  00001411dc427966b17297bf4d69e7e193135d89    mstrkrft

Answer 1

我相信您正在此处寻找 groupby + agg。

df.groupby('users').artist.apply(list).to_dict()

{'00001411dc427966b17297bf4d69e7e193135d89': ['the most serene republic',
  'stars',
  'broken social scene',
  'have heart',
  'luminous orange',
  'boris',
  ...
]
}