Detecting a US Holiday

2019-03-14 22:26发布

What's the simplest way to determine if a date is a U.S. bank holiday in Python? There seem to be various calendars and webservices listing holidays for various countries, but I haven't found anything specific to banks in the U.S.

标签: python date bank
5条回答
虎瘦雄心在
2楼-- · 2019-03-14 22:47

The Pandas package provides a convenient solution for this:

from pandas.tseries.holiday import USFederalHolidayCalendar
cal = USFederalHolidayCalendar()
holidays = cal.holidays(start='2014-01-01', end='2014-12-31').to_pydatetime()
if datetime.datetime(2014,01,01) in holidays:
    print True 
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Ridiculous、
3楼-- · 2019-03-14 22:49

I've actually worked recently on a problem much like this one. The static holidays are rather trivial to generate (e.g. New Years Eve - December 31st, (cycle through years)). There are well defined algorithms out there to generate the floating holidays. Essentially you have a starting date (e.g. January 1st, 1900) and work from there. What I ended up implementing was a nth day of nth week of nth month algorithm (e.g. MLK day = 3rd Monday of January). Easter is a bit different to do, but again there are well defined algorithms for that already out there (Good Friday is trivial after you have January).

There's a fairly decent book on this out there you may want to check out: Calendrical Calculations

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何必那么认真
4楼-- · 2019-03-14 22:50

Use the holiday library in python.

pip install holidays

For USA holiday:

1. To check a date holiday or not.

from datetime import date
import holidays

# Select country
us_holidays = holidays.US()

# If it is a holidays then it returns True else False
print('01-01-2018' in us_holidays)
print('02-01-2018' in us_holidays)

# What holidays is it?
print(us_holidays.get('01-01-2018'))
print(us_holidays.get('02-01-2018'))

2. To list out all holidays in US:

from datetime import date
import holidays

# Select country
us_holidays = holidays.US()

# Print all the holidays in US in year 2018
for ptr in holidays.US(years = 2018).items():
    print(ptr)

You can find holidays for any country you like the list of countries listed on my Blog. My Blog on Holidays

Github Link of Holidays Python

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何必那么认真
5楼-- · 2019-03-14 22:51

Some general comments:

I don't think that @ast4 really means "nth day of nth week of nth month algorithm". The notion of "nth week in nth month" is mind-snapping (like the "ISO calendar"). I've never seen a holiday defined in terms of "nth week". Martin Luther King Day is an example of the"Nth weekday in month" type of holiday:

MONDAY, ...., SATURDAY = range(7)
JAN, ....., DEC = range(1, 12)

Holiday("Martin L King's Birthday", type='floating',
    ordinal=3, weekday=MON, month=JAN)
Holiday("Memorial Day", type='floating',
    ordinal=-1, weekday=MON, month=MAY)

The USA doesn't have Easter-related holidays. Definition is not difficult:

Holiday("Good Friday", type='moveable',
    base='gregorian_easter', delta_days=-2)
Holiday("Easter Monday", etc, delta_days=1)
# Some states in Australia used to have Easter Tuesday (no kidding)
Holiday("Easter Tuesday", etc, delta_days=2)

The 'base' idea can be used to cater for lunar new year, in fact any holiday that is an offset from a base date that needs a special procedure to calculate it.

The so-called "static" holidays are not fixed when the "fixed" date is a Saturday or Sunday and may even vanish (no alternative day off work):

# Americans will get a day off on Friday 31 Dec 2010
# because 1 Jan 2011 is a Saturday.
Holiday("New Year's Day", type='fixed',
    day=1, month=JAN, sat_adj=-1, sun_adj=????)

# Australia observes ANZAC Day on the day, with no day off
# if the fixed date falls on a weekend.
Holiday("ANZAC Day", type='fixed', day=25, month=APR, sat_adj=0, sun_adj=0)

# Two consecutive "fixed" holidays is OK; three would need re-thinking.
# Australia again:
Holiday("Christmas Day", type='fixed', day=25, month=DEC, sat_adj=2, sun_adj=1)
Holiday("Boxing Day",    type='fixed', day=26, month=DEC, sat_adj=2, sun_adj=2)

I'm sure there are ways of specifying holidays that aren't catered for by the above rules ... any such news is welcome.

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对你真心纯属浪费
6楼-- · 2019-03-14 23:11
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