Converting Pandas DatetimeIndex to a numeric format

I found another solution:

df['date'] = df['date'].astype('datetime64').astype(int).astype(float)

Convert to Timedelta and extract the total seconds from dt.total_seconds:

df

        date
0 2013-01-01
1 2013-01-02
2 2013-01-03
3 2013-01-04
4 2013-01-05
5 2013-01-06
6 2013-01-07
7 2013-01-08
8 2013-01-09
9 2013-01-10

pd.to_timedelta(df.date).dt.total_seconds()

0    1.356998e+09
1    1.357085e+09
2    1.357171e+09
3    1.357258e+09
4    1.357344e+09
5    1.357430e+09
6    1.357517e+09
7    1.357603e+09
8    1.357690e+09
9    1.357776e+09
Name: date, dtype: float64

Or, maybe, the data would be more useful presented as an int type:

pd.to_timedelta(df.date).dt.total_seconds().astype(int)

0    1356998400
1    1357084800
2    1357171200
3    1357257600
4    1357344000
5    1357430400
6    1357516800
7    1357603200
8    1357689600
9    1357776000
Name: date, dtype: int64

I believe this offers another solution, here assuming a dataframe with a DatetimeIndex.

pd.to_numeric(df.index, downcast='float')
# although normally I would prefer an integer, and to coerce errors to NaN
pd.to_numeric(df.index, errors = 'coerce',downcast='integer')

Use astype float i.e if you have a dataframe like

df = pd.DataFrame({'date': ['1998-03-01 00:00:01', '2001-04-01 00:00:01','1998-06-01 00:00:01','2001-08-01 00:00:01','2001-05-03 00:00:01','1994-03-01 00:00:01'] })
df['date'] = pd.to_datetime(df['date'])
df['x'] = list('abcdef')
df = df.set_index('date')

Then

df.index.values.astype(float)

array([  8.88710401e+17,   9.86083201e+17,   8.96659201e+17,
     9.96624001e+17,   9.88848001e+17,   7.62480001e+17])

pd.to_datetime(df.index.values.astype(float))

DatetimeIndex(['1998-03-01 00:00:01', '2001-04-01 00:00:01',
           '1998-06-01 00:00:01', '2001-08-01 00:00:01',
           '2001-05-03 00:00:01', '1994-03-01 00:00:01'],
          dtype='datetime64[ns]', freq=None)