Esempio n. 1
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def to_jd(year, month, day):
    ## The calendar is closely synchronized to the Gregorian Calendar, always starting on the same day
    ## We can use this and the regular sequence of days in months to do a simple conversion by finding
    ## the Julian Day number of the first day of the year and adding on the required number of months
    ## and days to get the final Julian Day number

    ## Calculate the jd of 1 Chaitra for this year and how many days are in Chaitra this year
    ## If a Leap Year, then 1 Chaitra == 21 March of the Gregorian year and Chaitra has 31 days
    ## If not a Leap Year, then 1 Chaitra == 22 March of the Gregorian year and Chaitra has 30 days
    ## Need to use dateToJulianDay() to calculate instead of setDate() to avoid the year 9999 validation
    if isLeap(year):
        jdFirstDayOfYear = gregorian.to_jd(year+78, 3, 21)
        daysInMonth1 = 31
    else:
        jdFirstDayOfYear = gregorian.to_jd(year+78, 3, 22)
        daysInMonth1 = 30

    ## Add onto the jd of the first day of the year the number of days required
    ## Calculate the number of days in the months before the required month
    ## Then add on the required days
    ## The first month has 30 or 31 days depending on if it is a Leap Year (determined above)
    ## The second to sixth months have 31 days each
    ## The seventh to twelth months have 30 days each
    ## Note: could be expressed more efficiently, but I think this is clearer
    if month==1:
        jd = jdFirstDayOfYear + day - 1
    elif month<=6:
        jd = jdFirstDayOfYear + daysInMonth1 + (month-2)*31 + day - 1
    else: ## month > 6
        jd = jdFirstDayOfYear + daysInMonth1 + 5*31 + (month-7)*30 + day - 1
    return jd
Esempio n. 2
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def jd_to(jd):
    ## The calendar is closely synchronized to the Gregorian Calendar, always starting on the same day
    ## We can use this and the regular sequence of days in months to do a simple conversion by finding
    ## what day in the Gregorian year the Julian Day number is, converting this to the day in the
    ## Indian year and subtracting off the required number of months and days to get the final date

    gregorianYear, gregorianMonth, gregorianDay = gregorian.jd_to(jd)
    jdGregorianFirstDayOfYear = gregorian.to_jd(gregorianYear, 1, 1)
    gregorianDayOfYear = jd - jdGregorianFirstDayOfYear + 1

    ## There is a fixed 78 year difference between year numbers, but the years do not exactly match up,
    ## there is a fixed 80 day difference between the first day of the year, if the Gregorian day of
    ## the year is 80 or less then the equivalent Indian day actually falls in the preceding    year
    if gregorianDayOfYear > 80:
        year = gregorianYear - 78
    else:
        year = gregorianYear - 79

    ## If it is a leap year then the first month has 31 days, otherwise 30.
    if isLeap(year):
        daysInMonth1 = 31
    else:
        daysInMonth1 = 30

    ## The Indian year always starts 80 days after the Gregorian year, calculate the Indian day of
    ## the year, taking into account if it falls into the previous Gregorian year
    if gregorianDayOfYear>80:
        indianDayOfYear = gregorianDayOfYear - 80
    else:
        indianDayOfYear = gregorianDayOfYear + daysInMonth1 + 5*31    + 6*30 - 80

    ## Then simply remove the whole months from the day of the year and you are left with the day of month
    if indianDayOfYear <= daysInMonth1:
        month = 1
        day = indianDayOfYear
    elif indianDayOfYear <= daysInMonth1 + 5*31:
        month = (indianDayOfYear-daysInMonth1-1) // 31 + 2
        day = indianDayOfYear - daysInMonth1 - (month-2)*31
    else:
        month = (indianDayOfYear - daysInMonth1 - 5*31 - 1) // 30 + 7
        day = indianDayOfYear - daysInMonth1 - 5*31 - (month-7)*30
    return (year, month, day)
Esempio n. 3
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def iso_to_jd(year, week, day):
    #assert week>0 and day>0 and day<=7
    jd0 = gregorian.to_jd(year-1, 12, 28)
    return day + 7*week + jd0 - jd0%7 - 1