CODEWITHSUNDEEP
pythonpandasdata-manipulation
Greg Rosen
Greg Rosen

Reputation: 197

Is it possible to translate pandas datetime to weekday/weekend and season?

I have a pandas dataframe with one if its columns in datetime format (year-month-day). Is there any way to create a new column that says weekend or weekday, and a column that says what season it is? Even better (but this one seems much more complex so not totally necessary) would also be whether it's a federal holiday (United States) or not.

For example, an instance at date 2019-10-23 is a Wednesday, so I'd want to create a column called day_type that fills 'weekday', and a column called season that fills 'fall'.

Upvotes: 2

Views: 2328

Answers (2)

mcrrnz
mcrrnz

Reputation: 595

It's easy with strftime:

import pandas as pd 

df = pd.DataFrame({'string_date': ['2019-10-23', '2019-10-24', '2019-10-23']})
df['date'] = pd.to_datetime(df['string_date'], format='%Y-%m-%d', errors='ignore')
df['day_of_week'] = df['date'].dt.strftime('%A')

Result1 :

  string_date       date    day_of_week
0  2019-10-23   2019-10-23   Wednesday
1  2019-10-24   2019-10-24    Thursday
2  2019-10-23   2019-10-23   Wednesday

For adding the season you can use this formula : Link

df['season'] = (df['date'].dt.month%12 + 3)//3

Result2:

      string_date       date day_of_week  season
0  2019-10-23 2019-10-23   Wednesday       4
1  2019-10-24 2019-10-24    Thursday       4
2  2019-10-23 2019-10-23   Wednesday       4

Strftime reference: http://strftime.org/

Upvotes: 6

irahorecka
irahorecka

Reputation: 1807

I wrote a function that takes in a string argument in the form 'yyyy-mm-dd' and returns a tuple of season and day-of-week. Please modify the function accordingly to your needs.

from datetime import date, datetime

def date_week(date_str):
    """
    this method returns season and day of week tuple from str
    arg in the format 'yyyy-mm-dd'
    """

    datetime_obj = datetime.strptime(date_str, '%Y-%m-%d')
    weekdays = {0:'monday',
               1:'tuesday',
               2:'wednesday',
               3:'thursday',
               4:'friday',
               5:'saturday',
               6:'sunday'}
    day_of_week = weekdays[datetime_obj.weekday()]  # returns day of week


    Y = 2000 # dummy leap year to allow input X-02-29 (leap day)
    seasons = [('winter', (date(Y,  1,  1),  date(Y,  3, 20))),
               ('spring', (date(Y,  3, 21),  date(Y,  6, 20))),
               ('summer', (date(Y,  6, 21),  date(Y,  9, 22))),
               ('autumn', (date(Y,  9, 23),  date(Y, 12, 20))),
               ('winter', (date(Y, 12, 21),  date(Y, 12, 31)))]
    if isinstance(datetime_obj, datetime):
        datetime_obj = datetime_obj.date()
    datetime_obj = datetime_obj.replace(year=Y)
    season = next(season for season, (start, end) in seasons
                if start <= datetime_obj <= end)  # returns season

    return day_of_week, season

If we call date_week("2019-10-23"), it will return ('wednesday', 'autumn')

Upvotes: 3

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