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Cómo convertir el campo de una columna de objeto dtype en particular en una columna de marco de datos en pandas

Puede crear un nuevo marco de datos desde la columna page_view_count y join columna _id . Último sort_index :

df1 = pd.DataFrame([x for x in df['page_view_count']]).join(df['_id'])
df1 = df1.sort_index(1)
print df1
                            _id  comments  detailed-rating  detailed-rating2  \
0      568a8c25cac4991645c287ac       NaN                2                 1   
1      568cd22e9e82dfc166d7dff1         1                1                 1   
2      568e5a38b4a797c664143dda       NaN                1                 1   
3      568e5a561ae56e09656bfb99       NaN                1               NaN   
4      568df45a177e30c6487d3600         1                1                 1   

   main-rating  thank-you  
0            2        NaN  
1            1          1  
2            1        NaN  
3            1        NaN  
4            1          1  

EDITAR:

Todavía hay problemas con NaN con join.Solution es reemplazar NaN para vaciar dictionary por fillna y luego crea Dataframe :

import pandas as pd
import numpy as np


df = pd.DataFrame([[1,  {'name':'Jack','email':'abc'} ], 
                     [2, np.nan],
                     [3,  {'name':'Ram','email':'xyz'}  ],
                     ], columns=['_id','page_view_count'])

print df[df['page_view_count'].isnull()].index
#Int64Index([1], dtype='int64')

print pd.Series([{}], index=df[df['page_view_count'].isnull()].index , name='page_view_count')
#1    {}
#Name: page_view_count, dtype: object

df['page_view_count'] = df['page_view_count'].fillna(pd.Series([{}], index=df[df['page_view_count'].isnull()].index , name='page_view_count'))
print df
#   _id                       page_view_count
#0    1  {u'name': u'Jack', u'email': u'abc'}
#1    2                                    {}
#2    3   {u'name': u'Ram', u'email': u'xyz'}


df1 = pd.DataFrame([x for x in df['page_view_count']]).join(df['_id'], how='right')
df1 = df1.sort_index(1)
print df1
#   _id email  name
#0    1   abc  Jack
#1    2   NaN   NaN
#2    3   xyz   Ram