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In [275]: from pandas.tseries.holiday import get_calendar , HolidayCalendarFactory , USLaborDay In [276]: cal = get_calendar ( "ExampleCalendar" ) In [277]: cal . rules Out[277]: [Holiday: Memorial Day (month=5, day=31, offset=), Holiday: July 4th (month=7, day=4, observance=), Holiday: Columbus Day (month=10, day=1, offset=)] In [278]: new_cal = HolidayCalendarFactory ( "NewExampleCalendar" , cal , USLaborDay ) In [279]: new_cal . rules Out[279]: [Holiday: Labor Day (month=9, day=1, offset=), Holiday: Memorial Day (month=5, day=31, offset=), Holiday: July 4th (month=7, day=4, observance=), Holiday: Columbus Day (month=10, day=1, offset=)] Time Series-related instance methods # Shifting / lagging # BUSINESS_DATE = "[Previous (" + DAY_NAME[YESTERDAY.weekday()] + "):'" + YESTERDAY.strftime('%y%m%d') In [264]: pd . date_range ( .....: start = "7/1/2012" , end = "7/10/2012" , freq = pd . offsets . CDay ( calendar = cal ) .....: ) . to_pydatetime () .....: Out[264]: array([datetime.datetime(2012, 7, 2, 0, 0), datetime.datetime(2012, 7, 3, 0, 0), datetime.datetime(2012, 7, 5, 0, 0), datetime.datetime(2012, 7, 6, 0, 0), datetime.datetime(2012, 7, 9, 0, 0), datetime.datetime(2012, 7, 10, 0, 0)], dtype=object) In [265]: offset = pd . offsets . CustomBusinessDay ( calendar = cal ) In [266]: datetime . datetime ( 2012 , 5 , 25 ) + offset Out[266]: Timestamp('2012-05-29 00:00:00') In [267]: datetime . datetime ( 2012 , 7 , 3 ) + offset Out[267]: Timestamp('2012-07-05 00:00:00') In [268]: datetime . datetime ( 2012 , 7 , 3 ) + 2 * offset Out[268]: Timestamp('2012-07-06 00:00:00') In [269]: datetime . datetime ( 2012 , 7 , 6 ) + offset Out[269]: Timestamp('2012-07-09 00:00:00') Is there an easy fix so that I can replace the "D" with something else or do I have to rewrite everything? In [113]: dft2 = pd . DataFrame ( .....: np . random . randn ( 20 , 1 ), .....: columns = [ "A" ], .....: index = pd . MultiIndex . from_product ( .....: [ pd . date_range ( "20130101" , periods = 10 , freq = "12H" ), [ "a" , "b" ]] .....: ), .....: ) .....: In [114]: dft2 Out[114]: A 2013-01-01 00:00:00 a -0.298694 b 0.823553 2013-01-01 12:00:00 a 0.943285 b -1.479399 2013-01-02 00:00:00 a -1.643342 ... ... 2013-01-04 12:00:00 b 0.069036 2013-01-05 00:00:00 a 0.122297 b 1.422060 2013-01-05 12:00:00 a 0.370079 b 1.016331 [20 rows x 1 columns] In [115]: dft2 . loc [ "2013-01-05" ] Out[115]: A 2013-01-05 00:00:00 a 0.122297 b 1.422060 2013-01-05 12:00:00 a 0.370079 b 1.016331 In [116]: idx = pd . IndexSlice In [117]: dft2 = dft2 . swaplevel ( 0 , 1 ) . sort_index () In [118]: dft2 . loc [ idx [:, "2013-01-05" ], :] Out[118]: A a 2013-01-05 00:00:00 0.122297 2013-01-05 12:00:00 0.370079 b 2013-01-05 00:00:00 1.422060 2013-01-05 12:00:00 1.016331

In [256]: pd . Timestamp ( "2014-01-02" ) + pd . offsets . MonthBegin ( n = 0 ) Out[256]: Timestamp('2014-02-01 00:00:00') In [257]: pd . Timestamp ( "2014-01-02" ) + pd . offsets . MonthEnd ( n = 0 ) Out[257]: Timestamp('2014-01-31 00:00:00') In [258]: pd . Timestamp ( "2014-01-01" ) + pd . offsets . MonthBegin ( n = 0 ) Out[258]: Timestamp('2014-01-01 00:00:00') In [259]: pd . Timestamp ( "2014-01-31" ) + pd . offsets . MonthEnd ( n = 0 ) Out[259]: Timestamp('2014-01-31 00:00:00') Holidays / holiday calendars # When I try to list the holidays in a date range, I get the following error: In[11]: tradingCal.holidays(datetime(2014, 12, 31), datetime(2016, 12, 31))There are several time/date properties that one can access from Timestamp or a collection of timestamps like a DatetimeIndex. This is because one day’s business hour end is equal to next day’s business hour start. For example, def __add__ ( date ): date = rollforward ( date ) # does nothing if date is valid return date + < n number of periods >

Date times: A specific date and time with timezone support. Similar to datetime.datetime from the standard library.Cell In [ 484 ], line 1 ----> 1 rng_hourly . tz_localize ( 'US/Eastern' ) File ~/work/pandas/pandas/pandas/core/indexes/datetimes.py:291, in DatetimeIndex.tz_localize (self, tz, ambiguous, nonexistent) 284 @doc ( DatetimeArray . tz_localize ) 285 def tz_localize ( 286 self , ( ... ) 289 nonexistent : TimeNonexistent = "raise" , 290 ) -> Self : --> 291 arr = self . _data . tz_localize ( tz , ambiguous , nonexistent ) 292 return type ( self ) . _simple_new ( arr , name = self . name ) File ~/work/pandas/pandas/pandas/core/arrays/_mixins.py:80, in ravel_compat..method (self, *args, **kwargs) 77 @wraps ( meth ) 78 def method ( self , * args , ** kwargs ): 79 if self . ndim == 1 : ---> 80 return meth ( self , * args , ** kwargs ) 82 flags = self . _ndarray . flags 83 flat = self . ravel ( "K" ) File ~/work/pandas/pandas/pandas/core/arrays/datetimes.py:1066, in DatetimeArray.tz_localize (self, tz, ambiguous, nonexistent) 1063 tz = timezones . maybe_get_tz ( tz ) 1064 # Convert to UTC -> 1066 new_dates = tzconversion . tz_localize_to_utc ( 1067 self . asi8 , 1068 tz , 1069 ambiguous = ambiguous , 1070 nonexistent = nonexistent , 1071 creso = self . _creso , 1072 ) 1073 new_dates_dt64 = new_dates . view ( f "M8[ { self . unit } ]" ) 1074 dtype = tz_to_dtype ( tz , unit = self . unit ) File tzconversion.pyx:371, in pandas._libs.tslibs.tzconversion.tz_localize_to_utc () AmbiguousTimeError: Cannot infer dst time from 2011-11-06 01:00:00, try using the 'ambiguous' argument In [481]: pd . Timestamp ( .....: datetime . datetime ( 2019 , 10 , 27 , 1 , 30 , 0 , 0 ), .....: tz = "dateutil/Europe/London" , .....: fold = 0 , .....: ) .....: Out[481]: Timestamp('2019-10-27 01:30:00+0100', tz='dateutil//usr/share/zoneinfo/Europe/London') In [482]: pd . Timestamp ( .....: year = 2019 , .....: month = 10 , .....: day = 27 , .....: hour = 1 , .....: minute = 30 , .....: tz = "dateutil/Europe/London" , .....: fold = 1 , .....: ) .....: Out[482]: Timestamp('2019-10-27 01:30:00+0000', tz='dateutil//usr/share/zoneinfo/Europe/London') Ambiguous times when localizing #

YESTERDAY = (datetime.today() - timedelta(max(1,(TODAY.weekday() + 6) % 7 - 3))) - 1 * US_BUSINESS_DAY If you just want to get the pandas Holiday Calendar that can be used in other pandas functions that take that as an argument: holidays = nyse.holidays() Here is what it can do by creating a pandas DatetimeIndex of all of the valid open hours for the NYSE: import pandas_market_calendars as mcal In certain countries, such as the United States, there are laws ( Uniform Monday Holiday Act of 1968), whose rules are included in Pandas, print(USFederalHolidayCalendar.rules) as an example for developing other calendars. In [82]: pd . date_range ( start , periods = 1000 , freq = "M" ) Out[82]: DatetimeIndex(['2011-01-31', '2011-02-28', '2011-03-31', '2011-04-30', '2011-05-31', '2011-06-30', '2011-07-31', '2011-08-31', '2011-09-30', '2011-10-31', ... '2093-07-31', '2093-08-31', '2093-09-30', '2093-10-31', '2093-11-30', '2093-12-31', '2094-01-31', '2094-02-28', '2094-03-31', '2094-04-30'], dtype='datetime64[ns]', length=1000, freq='M') In [83]: pd . bdate_range ( start , periods = 250 , freq = "BQS" ) Out[83]: DatetimeIndex(['2011-01-03', '2011-04-01', '2011-07-01', '2011-10-03', '2012-01-02', '2012-04-02', '2012-07-02', '2012-10-01', '2013-01-01', '2013-04-01', ... '2071-01-01', '2071-04-01', '2071-07-01', '2071-10-01', '2072-01-01', '2072-04-01', '2072-07-01', '2072-10-03', '2073-01-02', '2073-04-03'], dtype='datetime64[ns]', length=250, freq='BQS-JAN')In [379]: idx = pd . period_range ( "2014-07-01 09:00" , periods = 5 , freq = "H" ) In [380]: idx Out[380]: PeriodIndex(['2014-07-01 09:00', '2014-07-01 10:00', '2014-07-01 11:00', '2014-07-01 12:00', '2014-07-01 13:00'], dtype='period[H]') In [381]: idx + pd . offsets . Hour ( 2 ) Out[381]: PeriodIndex(['2014-07-01 11:00', '2014-07-01 12:00', '2014-07-01 13:00', '2014-07-01 14:00', '2014-07-01 15:00'], dtype='period[H]') In [382]: idx = pd . period_range ( "2014-07" , periods = 5 , freq = "M" ) In [383]: idx Out[383]: PeriodIndex(['2014-07', '2014-08', '2014-09', '2014-10', '2014-11'], dtype='period[M]') In [384]: idx + pd . offsets . MonthEnd ( 3 ) Out[384]: PeriodIndex(['2014-10', '2014-11', '2014-12', '2015-01', '2015-02'], dtype='period[M]') whenever the dob is greater than now. You may want to subtract a few years to now in the condition df['dob'] < now since it may be slightly more likely to have a 101 year old worker than a 1 year old worker... If it helps, I had a similar need for exchange trading calendars. There was some excellent code buried in the Zipline project by Quantopian. I extracted out the relevant part and created a new project for creating market exchange trading calendars in pandas. The links are here, with some of the functionality described below.

In [13]: friday = pd . Timestamp ( "2018-01-05" ) In [14]: friday . day_name () Out[14]: 'Friday' # Add 1 day In [15]: saturday = friday + pd . Timedelta ( "1 day" ) In [16]: saturday . day_name () Out[16]: 'Saturday' # Add 1 business day (Friday --> Monday) In [17]: monday = friday + pd . offsets . BDay () In [18]: monday . day_name () Out[18]: 'Monday'pandas.bdate_range # pandas. bdate_range ( start = None, end = None, periods = None, freq = 'B', tz = None, normalize = True, name = None, weekmask = None, holidays = None, inclusive = 'both', ** kwargs ) [source] #