Pandas efficient groupby season for every year

Question

I have a multi-year time series an want the bounds between which 95% of my data lie. I want to look at this by season of the year (\'DJF\', \'MAM\', \'JJA\', \'SON\').

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Answer 1

The fastest so far is a combination of creating a low-frequency timeseries with which to do the season lookup and @Garrett's method of using a numpy.array index lookup rather than a dict.

season_lookup = np.array([
    None,
    'DJF', 'DJF',
    'MAM', 'MAM', 'MAM',
    'JJA', 'JJA', 'JJA',
    'SON', 'SON', 'SON',
    'DJF'])
SEASON_HALO = pd.datetools.relativedelta(months=4)
start_with_halo = df.index.min() - SEASON_HALO
end_with_halo = df.index.max() + SEASON_HALO
seasonal_idx = pd.DatetimeIndex(start=start_with_halo, end=end_with_halo, freq='QS-DEC')
seasonal_ts = pd.DataFrame(index=seasonal_idx)
seasonal_ts[SEAS] = season_lookup[seasonal_ts.index.month]
seasonal_minutely_ts = seasonal_ts.resample(df.index.freq, fill_method='ffill')
df_via_resample = df.join(seasonal_minutely_ts)
gp_up_sample = df_via_resample.groupby(SEAS)
gp_up_sample.quantile(FRAC_2_TAIL)

with 10 years of minute data, on my machine: this is about:

• 2% faster than low frequency dict lookup then up-sample
• 7% faster than the normal frequency np.array lookup
• >400% improvement on my original method

YMMV

Answer 2

In case it helps, I would suggest replacing the following list comprehension and dict lookup that you identified as slow:

month_to_season_dct = {
    1: 'DJF', 2: 'DJF',
    3: 'MAM', 4: 'MAM', 5: 'MAM',
    6: 'JJA', 7: 'JJA', 8: 'JJA',
    9: 'SON', 10: 'SON', 11: 'SON',
    12: 'DJF'
}
grp_ary = [month_to_season_dct.get(t_stamp.month) for t_stamp in df.index]

with the following, which uses a numpy array as a lookup table.

month_to_season_lu = np.array([
    None,
    'DJF', 'DJF',
    'MAM', 'MAM', 'MAM',
    'JJA', 'JJA', 'JJA',
    'SON', 'SON', 'SON',
    'DJF'
])
grp_ary = month_to_season_lu[df.index.month]

Here's a timeit comparison of the two approaches on ~3 years of minutely data:

In [16]: timeit [month_to_season_dct.get(t_stamp.month) for t_stamp in df.index]
1 loops, best of 3: 12.3 s per loop

In [17]: timeit month_to_season_lu[df.index.month]
1 loops, best of 3: 549 ms per loop