DataFrame.apply(axis=1) to return a series with dictionary as value
``` python
# Your code here
df = pd.util.testing.makeTimeDataFrame(5)
df2 = df.reset_index()
ff = lambda row: {'some key': row['A']}

# this is what I want, but my dataframe is unfortunately like df2
df.apply(ff, axis = 1)
2000-01-03      {u'some key': 1.70177486613}
2000-01-04    {u'some key': -0.833481157829}
2000-01-05    {u'some key': -0.117445006362}
2000-01-06     {u'some key': -1.12638444617}
2000-01-07    {u'some key': -0.981868953754}
Freq: B, dtype: object

# not correct
df2.apply(ff, axis = 1)
   index   A   B   C   D
0    NaN NaN NaN NaN NaN
1    NaN NaN NaN NaN NaN
2    NaN NaN NaN NaN NaN
3    NaN NaN NaN NaN NaN
4    NaN NaN NaN NaN NaN
>>> 
```
#### Expected Output

returns me a pandas Series with dictionary as value
#### Output of `pd.show_versions()`
## INSTALLED VERSIONS

commit: None
python: 2.7.12.final.0
python-bits: 64
OS: Windows
OS-release: 7
machine: AMD64
processor: Intel64 Family 6 Model 44 Stepping 2, GenuineIntel
byteorder: little
LC_ALL: None
LANG: en_GB

pandas: 0.18.1
nose: 1.3.7
pip: 8.1.0
setuptools: 0.6c11
Cython: 0.21
numpy: 1.10.1
scipy: 0.16.0
statsmodels: 0.6.1
xarray: None
IPython: 5.1.0
sphinx: 1.3.1
patsy: 0.4.0
dateutil: 2.4.2
pytz: 2015.7
blosc: None
bottleneck: None
tables: 3.1.1
numexpr: 2.4.4
matplotlib: 1.4.3
openpyxl: 1.8.5
xlrd: 0.9.3
xlwt: 0.7.5
xlsxwriter: 0.5.7
lxml: 3.5.0
bs4: 4.4.1
html5lib: None
httplib2: None
apiclient: None
sqlalchemy: 0.9.7
pymysql: None
psycopg2: None
jinja2: 2.8
boto: 2.32.1
pandas_datareader: None
<details>
# Paste the output here

</details>
