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
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
def iso_to_jd(year, week, day): """ Return Julian day of given ISO year, week, and 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
def test_to_jd(self):
for date, jd in self.dateToJdDict.items():
jdActual = gregorian.to_jd(*date)
self.assertEqual(
jdActual,
jd,
f"date={date}, jd={jd}, jdActual={jdActual}",
)
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 test_convert(self):
startYear = 1950
endYear = 2050
for year in range(startYear, endYear):
for month in range(1, 13):
monthLen = gregorian.getMonthLen(year, month)
for day in range(1, monthLen + 1):
date = (year, month, day)
jd = gregorian.to_jd(*date)
ndate = gregorian.jd_to(jd)
self.assertEqual(
ndate,
date,
f"jd={jd}, date={date}, ndate={ndate}",
)
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 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
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
