Python: Datetime to season

It's, also, possible to use dictionary mapping.

1. Create a dictionary that maps a month to a season:

   In [27]: seasons = [1, 1, 2, 2, 2, 3, 3, 3, 4, 4, 4, 1]
   
   In [28]: month_to_season = dict(zip(range(1,13), seasons))
   
   In [29]: month_to_season 
   Out[29]: {1: 1, 2: 1, 3: 2, 4: 2, 5: 2, 6: 3, 7: 3, 8: 3, 9: 4, 10: 4, 11: 4, 12: 1}
   

2. Use it to convert the months to seasons

   In [30]: df.id.dt.month.map(month_to_season) 
   Out[30]: 
   1    3
   2    3
   3    3
   4    4
   5    4
   6    4
   7    4
   8    4
   Name: id, dtype: int64
   

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Performance: This is fairly fast

In [35]: %timeit df.id.dt.month.map(month_to_season) 
1000 loops, best of 3: 422 µs per loop

You can use a simple mathematical formula to compress a month to a season, e.g.:

>>> [month%12 // 3 + 1 for month in range(1, 13)]
[1, 1, 2, 2, 2, 3, 3, 3, 4, 4, 4, 1]

So for your use-case using vector operations (credit @DSM):

>>> temp2.dt.month%12 // 3 + 1
1    3
2    3
3    3
4    4
5    4
6    4
7    4
8    4
Name: id, dtype: int64