Panda Bear 2nd Birthday Girl 2 Year Old Birthday Pandas Bday Tank Top

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The bins of the grouping are adjusted based on the beginning of the day of the time series starting point. This works well with frequencies that are multiples of a day (like 30D) or that divide a day evenly (like 90s or 1min). This can create inconsistencies with some frequencies that do not meet this criteria. To change this behavior you can specify a fixed Timestamp with the argument origin. will increment datetimes to the same time the next day whether a day represents 23, 24 or 25 hours due to daylight

Finally, there is the option of writing rules for calculating public holidays (as these are usually carried over to the next working day).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. Moreover, I also want to mark the days around the third Fridays with different values, e.g. day +1 after third_friday is 1 and day+2 is 2. To do that, I wrote a second for loop. Here the full example: for beg in pd.bdate_range("2000-01-01", "2017-05-01"): 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

The CustomBusinessDay class has now been merged into the upcoming 0.12 release of Pandas where you will be able to do something like the following: >>> from pandas.tseries.offsets import CustomBusinessDay In [396]: ps [ "2011" ] Out[396]: 2011-01 -2.916901 2011-02 0.514474 2011-03 1.346470 2011-04 0.816397 2011-05 2.258648 2011-06 0.494789 2011-07 0.301239 2011-08 0.464776 2011-09 -1.393581 2011-10 0.056780 2011-11 0.197035 2011-12 2.261385 Freq: M, dtype: float64 In [397]: dfp = pd . DataFrame ( .....: np . random . randn ( 600 , 1 ), .....: columns = [ "A" ], .....: index = pd . period_range ( "2013-01-01 9:00" , periods = 600 , freq = "T" ), .....: ) .....: In [398]: dfp Out[398]: A 2013-01-01 09:00 -0.538468 2013-01-01 09:01 -1.365819 2013-01-01 09:02 -0.969051 2013-01-01 09:03 -0.331152 2013-01-01 09:04 -0.245334 ... ... 2013-01-01 18:55 0.522460 2013-01-01 18:56 0.118710 2013-01-01 18:57 0.167517 2013-01-01 18:58 0.922883 2013-01-01 18:59 1.721104 [600 rows x 1 columns] In [399]: dfp . loc [ "2013-01-01 10H" ] Out[399]: A 2013-01-01 10:00 -0.308975 2013-01-01 10:01 0.542520 2013-01-01 10:02 1.061068 2013-01-01 10:03 0.754005 2013-01-01 10:04 0.352933 ... ... 2013-01-01 10:55 -0.865621 2013-01-01 10:56 -1.167818 2013-01-01 10:57 -2.081748 2013-01-01 10:58 -0.527146 2013-01-01 10:59 0.802298 [60 rows x 1 columns] 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))By having trial and error, I could only find the correct output by combining the method as below: from datetime import datetime, timedelta In [260]: from pandas.tseries.holiday import ( .....: Holiday , .....: USMemorialDay , .....: AbstractHolidayCalendar , .....: nearest_workday , .....: MO , .....: ) .....: In [261]: class ExampleCalendar ( AbstractHolidayCalendar ): .....: rules = [ .....: USMemorialDay , .....: Holiday ( "July 4th" , month = 7 , day = 4 , observance = nearest_workday ), .....: Holiday ( .....: "Columbus Day" , .....: month = 10 , .....: day = 1 , .....: offset = pd . DateOffset ( weekday = MO ( 2 )), .....: ), .....: ] .....: In [262]: cal = ExampleCalendar () In [263]: cal . holidays ( datetime . datetime ( 2012 , 1 , 1 ), datetime . datetime ( 2012 , 12 , 31 )) Out[263]: DatetimeIndex(['2012-05-28', '2012-07-04', '2012-10-08'], dtype='datetime64[ns]', freq=None) hint :

In [142]: ts2 . iloc [[ 0 , 2 , 6 ]] . index Out[142]: DatetimeIndex(['2011-01-02', '2011-01-16', '2011-02-13'], dtype='datetime64[ns]', freq=None) Time/date components # 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... under the default business hours (9:00 - 17:00), there is no gap (0 minutes) between 2014-08-01 17:00 and Holiday: New Years Day (month=1, day=1, observance=), the next business hour start or previous day’s end. Different from other offsets, BusinessHour.rollforward

In [485]: rng_hourly . tz_localize ( "US/Eastern" , ambiguous = "infer" ) Out[485]: DatetimeIndex(['2011-11-06 00:00:00-04:00', '2011-11-06 01:00:00-04:00', '2011-11-06 01:00:00-05:00', '2011-11-06 02:00:00-05:00'], dtype='datetime64[ns, US/Eastern]', freq=None) In [486]: rng_hourly . tz_localize ( "US/Eastern" , ambiguous = "NaT" ) Out[486]: DatetimeIndex(['2011-11-06 00:00:00-04:00', 'NaT', 'NaT', '2011-11-06 02:00:00-05:00'], dtype='datetime64[ns, US/Eastern]', freq=None) In [487]: rng_hourly . tz_localize ( "US/Eastern" , ambiguous = [ True , True , False , False ]) Out[487]: DatetimeIndex(['2011-11-06 00:00:00-04:00', '2011-11-06 01:00:00-04:00', '2011-11-06 01:00:00-05:00', '2011-11-06 02:00:00-05:00'], dtype='datetime64[ns, US/Eastern]', freq=None) Nonexistent times when localizing # 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') This is because one day’s business hour end is equal to next day’s business hour start. For example, 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

command refers to 20 days INCLUDING weekends, but I want it to refer to 20 WEEKDAYS; e.g. something like this: df["window"].loc[beg: beg + pd.to_timedelta(20, "Weekdays_only")] = 2 In [34]: dates = [ ....: pd . Timestamp ( "2012-05-01" ), ....: pd . Timestamp ( "2012-05-02" ), ....: pd . Timestamp ( "2012-05-03" ), ....: ] ....: In [35]: ts = pd . Series ( np . random . randn ( 3 ), dates ) In [36]: type ( ts . index ) Out[36]: pandas.core.indexes.datetimes.DatetimeIndex In [37]: ts . index Out[37]: DatetimeIndex(['2012-05-01', '2012-05-02', '2012-05-03'], dtype='datetime64[ns]', freq=None) In [38]: ts Out[38]: 2012-05-01 0.469112 2012-05-02 -0.282863 2012-05-03 -1.509059 dtype: float64 In [39]: periods = [ pd . Period ( "2012-01" ), pd . Period ( "2012-02" ), pd . Period ( "2012-03" )] In [40]: ts = pd . Series ( np . random . randn ( 3 ), periods ) In [41]: type ( ts . index ) Out[41]: pandas.core.indexes.period.PeriodIndex In [42]: ts . index Out[42]: PeriodIndex(['2012-01', '2012-02', '2012-03'], dtype='period[M]') In [43]: ts Out[43]: 2012-01 -1.135632 2012-02 1.212112 2012-03 -0.173215 Freq: M, dtype: float64 This particular day contains a day light savings time transition In [146]: ts = pd . Timestamp ( "2016-10-30 00:00:00" , tz = "Europe/Helsinki" ) # Respects absolute time In [147]: ts + pd . Timedelta ( days = 1 ) Out[147]: Timestamp('2016-10-30 23:00:00+0200', tz='Europe/Helsinki') # Respects calendar time In [148]: ts + pd . DateOffset ( days = 1 ) Out[148]: Timestamp('2016-10-31 00:00:00+0200', tz='Europe/Helsinki') In [149]: friday = pd . Timestamp ( "2018-01-05" ) In [150]: friday . day_name () Out[150]: 'Friday' # Add 2 business days (Friday --> Tuesday) In [151]: two_business_days = 2 * pd . offsets . BDay () In [152]: friday + two_business_days Out[152]: Timestamp('2018-01-09 00:00:00') In [153]: ( friday + two_business_days ) . day_name () Out[153]: 'Tuesday' In [234]: bhour_mon = pd . offsets . CustomBusinessHour ( start = "10:00" , weekmask = "Tue Wed Thu Fri" ) # Monday is skipped because it's a holiday, business hour starts from 10:00 In [235]: dt + bhour_mon * 2 Out[235]: Timestamp('2014-01-21 10:00:00') Offset aliases #In [218]: bh = pd . offsets . BusinessHour ( start = "17:00" , end = "09:00" ) In [219]: bh Out[219]: In [220]: pd . Timestamp ( "2014-08-01 17:00" ) + bh Out[220]: Timestamp('2014-08-01 18:00:00') In [221]: pd . Timestamp ( "2014-08-01 23:00" ) + bh Out[221]: Timestamp('2014-08-02 00:00:00') # Although 2014-08-02 is Saturday, # it is valid because it starts from 08-01 (Friday). In [222]: pd . Timestamp ( "2014-08-02 04:00" ) + bh Out[222]: Timestamp('2014-08-02 05:00:00') # Although 2014-08-04 is Monday, # it is out of business hours because it starts from 08-03 (Sunday). In [223]: pd . Timestamp ( "2014-08-04 04:00" ) + bh Out[223]: Timestamp('2014-08-04 18:00:00') 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] #