Datetime objects with pandas mean function

Question

I am new to programming so I apologize in advance if this question does not make any sens. I noticed that when I try to calculate the mean value of a pandas data frame with a date time object formatted like this: datetime.datetime(2014, 7, 10), it can not calculate the mean value of it however it seems to be able to calculate the minimum and maximum value of that same data frame with out a problem.

d={'one' : Series([1, 2, 3], index=['a', 'b', 'c']), 'two' :Series([datetime.datetime(2014, 7, 9) , datetime.datetime(2014, 7, 10) , datetime.datetime(2014, 7, 11) ], index=['a', 'b', 'c'])}
df=pd.DataFrame(d)

df
Out[18]: 
      one        two    
   a    1 2014-07-09
   b    2 2014-07-10
   c    3 2014-07-11

df.min()
Out[19]: 
   one             1
   two    2014-07-09
dtype: object

df.mean()
Out[20]: 
   one    2
dtype: float64

I did notice that the min and the max function converted all the columns to objects, where as the mean function only outputs floats. Could anyone explain to me why the mean function can only handle floats? Is there another way I to get the mean values of a data frame with a date time object? I can work around it by using epoch time (as integer), but it would be very convenient if there was a direct way. I use Python 2.7

I am grateful for any hints.

Answer 1

To simplify Alex's answer (I would have added this as a comment but I don't have sufficient reputation):

import datetime
import pandas as pd

d={'one': pd.Series([1, 2, 3], index=['a', 'b', 'c']), 
   'two': pd.Series([datetime.datetime(2014, 7, 9), 
           datetime.datetime(2014, 7, 10), 
           datetime.datetime(2014, 7, 11) ], 
           index=['a', 'b', 'c'])}
df = pd.DataFrame(d)

Which looks like:

   one   two
a   1   2014-07-09
b   2   2014-07-10
c   3   2014-07-11

Then calculate the mean of column "two" by:

(df.two - df.two.min()).mean() + df.two.min()

So, subtract the min of the timeseries, calculate the mean (or median) of the resulting timedeltas, and add back the min.