Python Pandas: detecting frequency of time series

Question

Assume I have loaded time series data from SQL or CSV (not created in Python), the index would be:

DatetimeIndex(['2015-03-02 00:00:00', '2015-03-02 01:00:00',
               '2015-03-02 02:00:00', '2015-03-02 03:00:00',
               '2015-03-02 04:00:00', '2015-03-02 05:00:00',
               '2015-03-02 06:00:00', '2015-03-02 07:00:00',
               '2015-03-02 08:00:00', '2015-03-02 09:00:00', 
               ...
               '2015-07-19 14:00:00', '2015-07-19 15:00:00',
               '2015-07-19 16:00:00', '2015-07-19 17:00:00',
               '2015-07-19 18:00:00', '2015-07-19 19:00:00',
               '2015-07-19 20:00:00', '2015-07-19 21:00:00',
               '2015-07-19 22:00:00', '2015-07-19 23:00:00'],
              dtype='datetime64[ns]', name=u'hour', length=3360, freq=None, tz=None)

As you can see, the freq is None. I am wondering how can I detect the frequency of this series and set the freq as its frequency. If possible, I would like this to work in the case of data which isn't continuous (there are plenty of breaks in the series).

I was trying to find the mode of all the differences between two timestamps, but I am not sure how to transfer it into a format that is readable by Series

Answer 1

It is worth mentioning that if data is continuous, you can use pandas.DateTimeIndex.inferred_freq property:

dt_ix = pd.date_range('2015-03-02 00:00:00', '2015-07-19 23:00:00', freq='H')
dt_ix._set_freq(None)
dt_ix.inferred_freq
Out[2]: 'H'

or pandas.infer_freq method:

pd.infer_freq(dt_ix)
Out[3]: 'H'

If not continuous pandas.infer_freq will return None. Similarly to what has been proposed yet, another alternative is using pandas.Series.diff method:

split_ix = dt_ix.drop(pd.date_range('2015-05-01 00:00:00','2015-05-30 00:00:00', freq='1H'))
split_ix.to_series().diff().min()
Out[4]: Timedelta('0 days 01:00:00')