Python 将长度可变的元组列表转换为数据帧

我有这个：

[['COMPANY:', [('U S News & World Report Inc', 63)]],
 ['ORGANIZATION:',
  [('Ashoka', 0),
   ('Innovators For The Public', 91),
   ('Us Environmental  Protection Agency', 55)]]]

我希望它成为一个熊猫数据帧，如：

NAME           ORGS                            PERCENT
Company        US News & World Report          63
Organization   Ashoka                          0
Organization   US Environmental Protection     55

结果:

           NAME                                 ORGS  PERCENT
0       Company          U S News & World Report Inc       63
1  Organization                               Ashoka        0
2  Organization            Innovators For The Public       91
3  Organization  Us Environmental  Protection Agency       55

结果:

           NAME                                 ORGS  PERCENT
0       Company          U S News & World Report Inc       63
1  Organization                               Ashoka        0
2  Organization            Innovators For The Public       91
3  Organization  Us Environmental  Protection Agency       55

结果:

           NAME                                 ORGS  PERCENT
0       Company          U S News & World Report Inc       63
1  Organization                               Ashoka        0
2  Organization            Innovators For The Public       91
3  Organization  Us Environmental  Protection Agency       55

结果:

           NAME                                 ORGS  PERCENT
0       Company          U S News & World Report Inc       63
1  Organization                               Ashoka        0
2  Organization            Innovators For The Public       91
3  Organization  Us Environmental  Protection Agency       55

---

这是一种您应该研究的

from_dict

read方法，适用于这种情况，只需将列表转换为字典即可：

L=[['COMPANY:', [('U S News & World Report Inc', 63)]],
 ['ORGANIZATION:',
  [('Ashoka', 0),
   ('Innovators For The Public', 91),
   ('Us Environmental  Protection Agency', 55)]]]

In [160]:
df=pd.DataFrame.from_dict(dict(L), orient="index").stack().reset_index(level=0)
df['Name']=df[0].apply(lambda x: x[0])
df['Val']=df[0].apply(lambda x: x[1])
df['Type']=df.level_0.str.slice(stop=-1)
df.__delitem__(0)
df.__delitem__('level_0')

In [161]:
print df
                                  Name  Val          Type
0                               Ashoka    0  ORGANIZATION
1            Innovators For The Public   91  ORGANIZATION
2  Us Environmental  Protection Agency   55  ORGANIZATION
0          U S News & World Report Inc   63       COMPANY

---

这是一种您应该研究的

from_dict

read方法，适用于这种情况，只需将列表转换为字典即可：

L=[['COMPANY:', [('U S News & World Report Inc', 63)]],
 ['ORGANIZATION:',
  [('Ashoka', 0),
   ('Innovators For The Public', 91),
   ('Us Environmental  Protection Agency', 55)]]]

In [160]:
df=pd.DataFrame.from_dict(dict(L), orient="index").stack().reset_index(level=0)
df['Name']=df[0].apply(lambda x: x[0])
df['Val']=df[0].apply(lambda x: x[1])
df['Type']=df.level_0.str.slice(stop=-1)
df.__delitem__(0)
df.__delitem__('level_0')

In [161]:
print df
                                  Name  Val          Type
0                               Ashoka    0  ORGANIZATION
1            Innovators For The Public   91  ORGANIZATION
2  Us Environmental  Protection Agency   55  ORGANIZATION
0          U S News & World Report Inc   63       COMPANY

---

这是一种您应该研究的

from_dict

read方法，适用于这种情况，只需将列表转换为字典即可：

L=[['COMPANY:', [('U S News & World Report Inc', 63)]],
 ['ORGANIZATION:',
  [('Ashoka', 0),
   ('Innovators For The Public', 91),
   ('Us Environmental  Protection Agency', 55)]]]

In [160]:
df=pd.DataFrame.from_dict(dict(L), orient="index").stack().reset_index(level=0)
df['Name']=df[0].apply(lambda x: x[0])
df['Val']=df[0].apply(lambda x: x[1])
df['Type']=df.level_0.str.slice(stop=-1)
df.__delitem__(0)
df.__delitem__('level_0')

In [161]:
print df
                                  Name  Val          Type
0                               Ashoka    0  ORGANIZATION
1            Innovators For The Public   91  ORGANIZATION
2  Us Environmental  Protection Agency   55  ORGANIZATION
0          U S News & World Report Inc   63       COMPANY

---

这是一种您应该研究的

from_dict

read方法，适用于这种情况，只需将列表转换为字典即可：

L=[['COMPANY:', [('U S News & World Report Inc', 63)]],
 ['ORGANIZATION:',
  [('Ashoka', 0),
   ('Innovators For The Public', 91),
   ('Us Environmental  Protection Agency', 55)]]]

In [160]:
df=pd.DataFrame.from_dict(dict(L), orient="index").stack().reset_index(level=0)
df['Name']=df[0].apply(lambda x: x[0])
df['Val']=df[0].apply(lambda x: x[1])
df['Type']=df.level_0.str.slice(stop=-1)
df.__delitem__(0)
df.__delitem__('level_0')

In [161]:
print df
                                  Name  Val          Type
0                               Ashoka    0  ORGANIZATION
1            Innovators For The Public   91  ORGANIZATION
2  Us Environmental  Protection Agency   55  ORGANIZATION
0          U S News & World Report Inc   63       COMPANY

---

嗯，我喜欢有一个具体的方法来解决这个问题，但是我不喜欢ti比使用for循环更复杂。不过，我将进一步研究这种方法。谢谢嗯，我喜欢有一个具体的方法来解决这个问题，但是我不喜欢ti比使用for循环更复杂。不过，我将进一步研究这种方法。谢谢嗯，我喜欢有一个具体的方法来解决这个问题，但是我不喜欢ti比使用for循环更复杂。不过，我将进一步研究这种方法。谢谢嗯，我喜欢有一个具体的方法来解决这个问题，但是我不喜欢ti比使用for循环更复杂。不过，我将进一步研究这种方法。谢谢