DataFrame to_json Column -> [values] orientation and JSON able
#7863  has mentioned that   to_dict(orient='list')  could be a solution.
here is the example code:

```
In [28]: df = pd.DataFrame(np.arange(1, 5).reshape(2,2), index=['r1', 'r2'], columns=['c1','c2'])

In [29]: df
Out[29]: 
    c1  c2
r1   1   2
r2   3   4

In [30]: df.reset_index().to_dict('list')
Out[30]: {'c1': [1, 3], 'c2': [2, 4], 'index': ['r1', 'r2']}
```
howere if the index of dataframe is Timestamp type.  this method would be useless:

```
aa.head(3)
Out[12]: 
            000423.stock  000825.stock  000959.stock  002011.stock  \
settledate                                                           
2017-07-10           0.0           0.0           0.0           0.0   
2017-07-11           0.0           0.0           0.0           0.0   
2017-07-12           0.0           0.0           0.0           0.0   

json.dumps(aa.to_dict(orient='list'))
Traceback (most recent call last):
  File "D:\anaconda\lib\site-packages\IPython\core\interactiveshell.py", line 2885, in run_code
    exec(code_obj, self.user_global_ns, self.user_ns)
  File "<ipython-input-13-8014f98ae600>", line 1, in <module>
    json.dumps(aa.to_dict(orient='list'))
  File "D:\anaconda\lib\json\__init__.py", line 230, in dumps
    return _default_encoder.encode(obj)
  File "D:\anaconda\lib\json\encoder.py", line 199, in encode
    chunks = self.iterencode(o, _one_shot=True)
  File "D:\anaconda\lib\json\encoder.py", line 257, in iterencode
    return _iterencode(o, 0)
  File "D:\anaconda\lib\json\encoder.py", line 180, in default
    raise TypeError(repr(o) + " is not JSON serializable")
TypeError: Timestamp('2017-07-10 00:00:00') is not JSON serializable
```

so, is there a simple way to transfer the dataframe to json  with a format of  "column:[data1,data2]"?