Pandas - 将两列作为字典转换为新列

Question

我正在尝试使用Pandas将两列转换为一列,该列是两个转换列的字典表示.

df = DataFrame({'Metrics' : [[("P", "P"), ("Q","Q")], ("K", "K"), ("Z", "Z")], 
                'Stage_Name' : ["P", "K", "Z"],  
                'Block_Name' : ["A", "B", "A"]})

基本上我想合并Metrics并Stage_Name:

进入另一个名为的列merged,例如,第一行将是:

{'P': [('P', 'P'), ('Q', 'Q')]}

我知道如何将一行转换为字典表示,但是,我不知道如何在没有for循环的情况下对所有行执行此操作:

something = df.iloc[[0]].set_index('Stage_Name')['Metrics'].to_dict()
print something
Output: {'P': [('P', 'P'), ('Q', 'Q')]}

后来我想基于聚合Block_Name,所以对于合并列,结果将是两个字典加在一起Block_Name:A.

{'P': [('P', 'P'), ('Q', 'Q')], 'Z' : [('Z', 'Z')] }

对于Stage_Name和Metrics,我只是将它附加到列表中,如下所示:

grouped = df.groupby(df['Block_Name'])
df_2 = grouped.aggregate(lambda x: tuple(x))

有人能指出我正确的方向吗？谢谢!

Answer 1

IIUC 正确，然后您apply使用lambda：

In [19]:
df['merged'] = df.apply(lambda row: {row['Stage_Name']:row['Metrics']}, axis=1)
df

Out[19]:
  Block_Name           Metrics Stage_Name                           merged
0          A  [(P, P), (Q, Q)]          P  {'P': [('P', 'P'), ('Q', 'Q')]}
1          B            (K, K)          K                {'K': ('K', 'K')}
2          A            (Z, Z)          Z                {'Z': ('Z', 'Z')}

Answer 2

df['Merged'] = [{key: val} for key, val in zip(df.Stage_Name, df.Metrics)]

>>> df
  Block_Name           Metrics Stage_Name                                Merged
0          A  [(P, P), (Q, Q)]          P  {u'P': [(u'P', u'P'), (u'Q', u'Q')]}
1          B            (K, K)          K                  {u'K': (u'K', u'K')}
2          A            (Z, Z)          Z                  {u'Z': (u'Z', u'Z')}

然后您的代码会产生所需的结果:

grouped = df.groupby(df['Block_Name'])
df_2 = grouped.aggregate(lambda x: tuple(x))[['Metrics', 'Stage_Name']]


>>> df_2
                               Metrics Stage_Name
Block_Name                                       
A           ([(P, P), (Q, Q)], (Z, Z))     (P, Z)
B                            ((K, K),)       (K,)

定时:

%timeit df['Merged'] = [{key: val} for key, val in zip(df.Stage_Name, df.Metrics)]
10000 loops, best of 3: 162 µs per loop

%timeit df['merged'] = df.apply(lambda row: {row['Stage_Name']:row['Metrics']}, axis=1)
1000 loops, best of 3: 332 µs per loop