RRule set for days and hours in day

Question

I'm using python's rrule to work out trading hours. It's easy for days, i'm using a slightly modified example I found on this very site:

def get_rset(start_date):    
    # Create a rule to recur every weekday starting today
    r = rrule.rrule(rrule.DAILY,
                    byweekday=[rrule.MO, rrule.TU, rrule.WE, rrule.TH, rrule.FR],
                    dtstart=start_date)
    # Create a rruleset
    rs = rrule.rruleset()
    # Attach our rrule to it
    rs.rrule(r)
    # Add holidays as exclusion days
    for exdate in holidays:
        rs.exdate(exdate)
    return rs

The problem is while this works great for shares, I need to calculate forex dates differently. I need to work on an hourly basis, add in public holidays etc.

In UTC I believe the markets are open from 10pm Sunday to 10pm the following Friday.

To make this into a rrule I now need 6 different days with Sunday and Friday needing special hours and the remaining weekdays to be considered all hours. I'm pretty sure I need to mix rrule byday and byhour, but I can't any good examples on doing this.

Any help very welcome!

joeButler · Accepted Answer · 2013-03-18T14:20:47.877

I've figured out a simpler way, following some time alone with Google, the code and class docs. It uses a slight (but appropriate) cheat. Please see example solution below.

from dateutil import rrule
from datetime import  timedelta , datetime
holidays = [] # This is just a list of dates to exclude

def datetime_in_x_trading_hours(start_dt,future_hours):
    # First we add two hours. This is because its simpler to view the timeset
    # as 24hrs MON - FRI. (This also helps align the dates for the holidays)
    print start_dt
    start_dt += timedelta(hours=2)
    rs = get_fx_rset(start_dt)
    # Now using the set get the desired time and put the the missing hours
    future_time = rs[future_hours]
    future_time -= timedelta(hours=2)
    return future_time

def get_fx_rset(start_date_time):

    # Create a rule to recur every weekday starting today
    r = rrule.rrule(rrule.HOURLY,
                    byweekday=[rrule.MO, rrule.TU, rrule.WE, rrule.TH, rrule.FR],
                    byhour=[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23],
                    byminute=[0],
                    bysecond=[0],
                    dtstart=start_date_time)

    # Create a rruleset
    rs = rrule.rruleset()
    # Attach our rrule to it
    rs.rrule(r)
    # Add holidays as exclusion days
    for exdate in holidays:
        rs.exdate(exdate)

    return rs

today = datetime.now() - timedelta(days=2)
print datetime_in_x_trading_hours(today, 7)

RRule set for days and hours in day

1 Answers1