将数据框转换为元组列表字典 [英] Convert dataframe to dictionary of list of tuples

问题描述

我有一个如下所示的数据框

I have a dataframe that looks like the following

    user                             item  
0  b80344d063b5ccb3212f76538f3d9e43d87dca9e          The Cove - Jack Johnson   
1  b80344d063b5ccb3212f76538f3d9e43d87dca9e  Entre Dos Aguas - Paco De Lucia   
2  b80344d063b5ccb3212f76538f3d9e43d87dca9e            Stronger - Kanye West   
3  b80344d063b5ccb3212f76538f3d9e43d87dca9e    Constellations - Jack Johnson   
4  b80344d063b5ccb3212f76538f3d9e43d87dca9e      Learn To Fly - Foo Fighters   

rating  
0       1  
1       2  
2       1  
3       1  
4       1  

并想实现如下结构:

dict-> list of tuples
user-> (item, rating)

b80344d063b5ccb3212f76538f3d9e43d87dca9e -> list((The Cove - Jack 
Johnson, 1), ... , )

我能做到:

item_set = dict((user, set(items)) for user, items in 
data.groupby('user')['item'])

但这只会让我半途而废.如何从 groupby 中获取相应的评分"值?

But that only gets me halfways. How do I get the corresponding "rating" value from the groupby?

推荐答案

设置user为索引，使用df.apply转换成元组，使用分组索引df.groupby(level=0) 并使用 dfGroupBy.agg 获取列表并使用 df.to_dict 转换为字典:

Set user as index, convert to tuple using df.apply, groupby index using df.groupby(level=0) and get a list using dfGroupBy.agg and convert to dictionary using df.to_dict:

In [1417]: df
Out[1417]: 
                                       user                             item  
0  b80344d063b5ccb3212f76538f3d9e43d87dca9e          The Cove - Jack Johnson   
1  b80344d063b5ccb3212f76538f3d9e43d87dca9e  Entre Dos Aguas - Paco De Lucia   
2  b80344d063b5ccb3212f76538f3d9e43d87dca9e            Stronger - Kanye West   
3  b80344d063b5ccb3212f76538f3d9e43d87dca9e    Constellations - Jack Johnson   
4  b80344d063b5ccb3212f76538f3d9e43d87dca9e      Learn To Fly - Foo Fighters   

   rating  
0       1  
1       2  
2       2  
3       2  
4       2  

In [1418]: df.set_index('user').apply(tuple, 1)
             .groupby(level=0).agg(lambda x: list(x.values))
             .to_dict()
Out[1418]: 
{'b80344d063b5ccb3212f76538f3d9e43d87dca9e': [('The Cove - Jack Johnson', 1),
  ('Entre Dos Aguas - Paco De Lucia', 2),
  ('Stronger - Kanye West', 2),
  ('Constellations - Jack Johnson', 2),
  ('Learn To Fly - Foo Fighters', 2)]}

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