CODEWITHSUNDEEP
pythonpandasdatedatetime
Emac
Emac

Reputation: 1167

Pandas return the next Sunday for every row

In Pandas for Python, I have a data set that has a column of datetimes in it. I need to create a new column that has the date of the following Sunday for each row.

I've tried various methods trying to use iterrows and then figure out the day of the week, and add a day until the day is 7, but it hasn't worked and I'm not even sure how I'd return the date instead of just the day number then. I also don't feel like iterrows would be the best way to do it either.

What is the best way to return a column of the following Sunday from a date column?

Upvotes: 6

Views: 4853

Answers (4)

ifly6
ifly6

Reputation: 5331

Use the Pandas date offsets, e.g.:

>>> pd.to_datetime('2019-04-09') + pd.offsets.Week(n=0, weekday=6)
Timestamp('2019-04-14 00:00:00')

For example, this changes the provided datetime over a week. This is vectorised, so you can run it against a series like so:

temp['sunday_dates'] = temp['our_dates'] + pd.offsets.Week(n=0, weekday=6)

    our_dates  random_data sunday_dates
0  2010-12-31         4012   2011-01-02
1  2007-12-31         3862   2008-01-06
2  2006-12-31         3831   2007-01-07
3  2011-12-31         3811   2012-01-01

N.b. Pass n=0 to keep a day, which is already on a Sunday, on that day. Pass n=1 if you want to force it to the next Sunday. The Week(weekday=INT) parameter is 0 indexed on Monday and takes values from 0 to 6 (inclusive). Thus, passing 0 yields all Mondays, 1 yields all Tuesdays, etc. Using this, you can make everything any day of the week you would like.

N.b. If you want to go to the last Sunday, just swap + to - to go back.

N.b. (Such note, much bene) The specific documentation on time series functionality can be found here: https://pandas.pydata.org/pandas-docs/stable/user_guide/timeseries.html

Upvotes: 9

Pedro Lobito
Pedro Lobito

Reputation: 98921

The accepted answer is the way to go, but you can also use Series.apply() and pandas.Timedelta() for this, i.e.:

df["ns"] = df["d"].apply(lambda d: d + pd.Timedelta(days=(6 if d.weekday() == 6 else 6-d.weekday())))

    d                             ns
0   2019-04-09 21:22:10.886702    2019-04-14 21:22:10.886702

Demo

Upvotes: 0

Dmitry
Dmitry

Reputation: 21

I suggest to use calendar library 

 import calendar
 import datetime as dt

 #today date 
 now = datetime.datetime.now()
 print (now.year, now.month, now.day, now.hour, now.minute, now.second)

 # diffrence in days between current date  and  Sunday
 difday =  7 - calendar.weekday(now.year, now.month, now.day)

 # Afterwards next Sunday from today
 nextsunday  = datetime.date(now.year, now.month , now.day + difday)

 print(nextsunday)

Write this function and use

Upvotes: 0

d4tm4x
d4tm4x

Reputation: 558

The function

import datetime
def datetime_to_next_sunday(original_datetime):
    return original_datetime + datetime.timedelta(days=6-original_datetime.weekday())

returns the datetime.datetime shifted to the next sunday. Having

import pandas as pd
df = pd.DataFrame({'A': ['Foo', 'Bar'],
                   'datetime': [datetime.datetime.now(),
                                datetime.datetime.now() + datetime.timedelta(days=1)]})

the following line should to the job:

df['dt_following_sunday'] = df[['datetime']].applymap(datetime_to_next_sunday)

Upvotes: 1

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