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Python: Datetime to season

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I want to convert a date time series to season, for example for months 3, 4, 5 I want to replace them with 2 (spring); for months 6, 7, 8 I want to replace them with 3 (summer) etc.

So, I have this series

id
1       2011-08-20
2       2011-08-23
3       2011-08-27
4       2011-09-01
5       2011-09-05
6       2011-09-06
7       2011-09-08
8       2011-09-09
Name: timestamp, dtype: datetime64[ns]

and this is the code I have been trying to use, but to no avail.

# Get seasons
spring = range(3, 5)
summer = range(6, 8)
fall = range(9, 11)
# winter = everything else

month = temp2.dt.month
season=[]

for _ in range(len(month)):
    if any(x == spring for x in month):
       season.append(2) # spring 
    elif any(x == summer for x in month):
        season.append(3) # summer
    elif any(x == fall for x in month):
        season.append(4) # fall
    else:
        season.append(1) # winter

and

for _ in range(len(month)):
    if month[_] == 3 or month[_] == 4 or month[_] == 5:
        season.append(2) # spring 
    elif month[_] == 6 or month[_] == 7 or month[_] == 8:
        season.append(3) # summer
    elif month[_] == 9 or month[_] == 10 or month[_] == 11:
        season.append(4) # fall
    else:
        season.append(1) # winter

Neither solution works, specifically in the first implementation I receive an error:

ValueError: The truth value of an array with more than one element is
ambiguous. Use a.any() or a.all()

While in the second is a large list with errors. Any ideas please? Thanks

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1 Answer

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

    >>> [(month%12 + 3)//3 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)//3
    1    3
    2    3
    3    3
    4    4
    5    4
    6    4
    7    4
    8    4
    Name: id, dtype: int64
    
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