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)
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)
def isow(jd):## iso week number
year = gregorian.jd_to(jd - 3)[0]
if jd>=iso_to_jd(year+1, 1, 1):
year += 1
return (jd - iso_to_jd(year, 1, 1)) // 7 + 1
def isow_year(jd):## iso week year
year = gregorian.jd_to(jd - 3)[0]
if jd>=iso_to_jd(year+1, 1, 1):
year += 1
return year
def isow(jd): ## iso week number
year = gregorian.jd_to(jd - 3)[0]
if jd >= iso_to_jd(year + 1, 1, 1):
year += 1
return (jd - iso_to_jd(year, 1, 1)) // 7 + 1
def isow_year(jd): ## iso week year
year = gregorian.jd_to(jd - 3)[0]
if jd >= iso_to_jd(year + 1, 1, 1):
year += 1
return year
