CODEWITHSUNDEEP
pythonpython-3.xpandas
Amelio Vazquez-Reina
Amelio Vazquez-Reina

Reputation: 96294

Getting the date of the last day of this [week/month/quarter/year]

Is there any way to get the date (a datetime, pd.Timestamp or equivalent) of the last day of this [week/month/quarter/year] with datetime, pandas or other date & time utils?

Upvotes: 8

Views: 9722

Answers (3)

unutbu
unutbu

Reputation: 879651

You could add a pandas.DateOffset to a DateTimeIndex, Timestamp or datetime.date or datetime.datetime:

dates = pd.date_range('2015-8-13', periods=4, freq='3D')
# DatetimeIndex(['2015-08-13', '2015-08-16', '2015-08-19', '2015-08-22'],
# dtype='datetime64[ns]', freq='3D', tz=None)

Snap to the last day of the week (for example, Sunday):

In [232]: dates+offsets.Week(weekday=6)
Out[232]: DatetimeIndex(['2015-08-16', '2015-08-23', '2015-08-23', '2015-08-23'], dtype='datetime64[ns]', freq=None, tz=None)

Snap to the last day of the month:

In [207]: dates+offsets.MonthEnd()
Out[207]: DatetimeIndex(['2015-08-31', '2015-08-31', '2015-08-31', '2015-08-31'], dtype='datetime64[ns]', freq=None, tz=None)

Snap to the last day in the quarter:

In [212]: dates+offsets.QuarterEnd()
Out[215]: DatetimeIndex(['2015-09-30', '2015-09-30', '2015-09-30', '2015-09-30'], dtype='datetime64[ns]', freq=None, tz=None)

Snap to the last day of the year:

In [219]: dates+offsets.YearEnd()
Out[222]: DatetimeIndex(['2015-12-31', '2015-12-31', '2015-12-31', '2015-12-31'], dtype='datetime64[ns]', freq=None, tz=None)

Notice that adding an offset always advances the date. For example, 2015-08-16 is a Sunday, and adding an offsets.Week(weekday=6) advances it it 2015-08-23:

In [233]: pd.Timestamp('2015-8-16')+offsets.Week(weekday=6)
Out[233]: Timestamp('2015-08-23 00:00:00')

To prevent that from happening, you could subtract one day from dates:

In [234]: dates - offsets.Day() + offsets.Week(weekday=6)
Out[237]: DatetimeIndex(['2015-08-16', '2015-08-16', '2015-08-23', '2015-08-23'], dtype='datetime64[ns]', freq=None, tz=None)

Upvotes: 5

zopieux
zopieux

Reputation: 2901

import datetime
from dateutil.relativedelta import relativedelta, SU

def last_day_of_week(year, week_num):
    # assuming sunday is the last day of week
    # weeks= is an offset, hence the minus one
    return datetime.date(year, 1, 1) + relativedelta(weekday=SU(1), weeks=week_num - 1)

def last_day_of_month(year, month):
    return datetime.date(year, month, 1) + relativedelta(months=1, days=-1)

def last_day_of_year(year):
    return datetime.date(year, 1, 1) + relativedelta(years=1, days=-1)

Upvotes: 4

Ben
Ben

Reputation: 2472

Using datetime only.

>>> d = datetime.date.today()

Last day of week:

# This was wrong earlier.
>>> d + datetime.timedelta(days=5 - d.weekday())
datetime.date(2015, 8, 15)

Last day of month:

>>> datetime.date(year=(d.year + int(d.month % 12 == 0)), month=(d.month + 1) % 12, day=1) - datetime.timedelta(days=1)
datetime.date(2015, 8, 31)

Last day of quarter:

>>> datetime.date(year=d.year, month=((d.month % 3) + 1) * 3 + 1, day=1) - datetime.timedelta(days=1)
datetime.date(2015, 9, 30)

Last day of year:

>>> datetime.date(year=d.year, month=12, day=31)
datetime.date(2015, 12, 31)

EDIT: This is all pretty ugly and using a higher level third party library is probably best, unless there is a compelling reason not to (and there does not seem to be here).

Upvotes: 7

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