Pandas Dataframe可按列对dict分组

Sem*_*aki 1 python dictionary dataframe pandas

我有一个这样的数据框:

Subject_id    Subject    Score    
Subject_1        Math        5                 
Subject_1    Language        4                 
Subject_1       Music        8
Subject_2        Math        8                 
Subject_2    Language        3                 
Subject_2       Music        9
Run Code Online (Sandbox Code Playgroud)

我想将其转换为字典,按subject_id分组

{'Subject_1': {'Math': 5,
               'Language': 4,
               'Music': 8},
{'Subject_2': {'Math': 8,
               'Language': 3,
               'Music': 9}
}
Run Code Online (Sandbox Code Playgroud)

如果我只有一个主题,那么我可以这样做:

my_dict['Subject_1'] = dict(zip(df['Subject'],df['Score']))
Run Code Online (Sandbox Code Playgroud)

但是由于我有几个主题,所以密钥列表会重复,因此我不能直接使用zip。

数据框有一种.to_dict('index')方法,但是创建字典时我需要能够按特定列分组。

我该如何实现?

谢谢。

jez*_*ael 5

使用groupby自定义lambda函数和最后转换输出Series to_dict

d = (df.groupby('Subject_id')
       .apply(lambda x: dict(zip(x['Subject'],x['Score'])))
       .to_dict())

print (d)
{'Subject_2': {'Math': 8, 'Music': 9, 'Language': 3}, 
 'Subject_1': {'Math': 5, 'Music': 8, 'Language': 4}}
Run Code Online (Sandbox Code Playgroud)

详情:

print (df.groupby('Subject_id').apply(lambda x: dict(zip(x['Subject'],x['Score']))))

Subject_id
Subject_1    {'Math': 5, 'Music': 8, 'Language': 4}
Subject_2    {'Math': 8, 'Music': 9, 'Language': 3}
dtype: object
Run Code Online (Sandbox Code Playgroud)


Zer*_*ero 5

使用to_dictpivot

In [29]: df.pivot('Subject_id', 'Subject', 'Score').to_dict('index')
Out[29]:
{'Subject_1': {'Language': 4L, 'Math': 5L, 'Music': 8L},
 'Subject_2': {'Language': 3L, 'Math': 8L, 'Music': 9L}}
Run Code Online (Sandbox Code Playgroud)

或者,

In [25]: df.set_index(['Subject_id', 'Subject']).unstack()['Score'].to_dict('index')
Out[25]:
{'Subject_1': {'Language': 4L, 'Math': 5L, 'Music': 8L},
 'Subject_2': {'Language': 3L, 'Math': 8L, 'Music': 9L}}
Run Code Online (Sandbox Code Playgroud)