35. Finish daysBetweenDates
Finish daysBetweenDates
Question:
Start Quiz:
# Credit goes to Websten from forums
#
# Use Dave's suggestions to finish your daysBetweenDates
# procedure. It will need to take into account leap years
# in addition to the correct number of days in each month.
def nextDay(year, month, day):
"""Simple version: assume every month has 30 days"""
if day < 30:
return year, month, day + 1
else:
if month == 12:
return year + 1, 1, 1
else:
return year, month + 1, 1
def dateIsBefore(year1, month1, day1, year2, month2, day2):
"""Returns True if year1-month1-day1 is before year2-month2-day2. Otherwise, returns False."""
if year1 < year2:
return True
if year1 == year2:
if month1 < month2:
return True
if month1 == month2:
return day1 < day2
return False
def daysBetweenDates(year1, month1, day1, year2, month2, day2):
"""Returns the number of days between year1/month1/day1
and year2/month2/day2. Assumes inputs are valid dates
in Gregorian calendar."""
# program defensively! Add an assertion if the input is not valid!
assert not dateIsBefore(year2, month2, day2, year1, month1, day1)
days = 0
while dateIsBefore(year1, month1, day1, year2, month2, day2):
year1, month1, day1 = nextDay(year1, month1, day1)
days += 1
return days
def test():
test_cases = [((2012,1,1,2012,2,28), 58),
((2012,1,1,2012,3,1), 60),
((2011,6,30,2012,6,30), 366),
((2011,1,1,2012,8,8), 585 ),
((1900,1,1,1999,12,31), 36523)]
for (args, answer) in test_cases:
result = daysBetweenDates(*args)
if result != answer:
print "Test with data:", args, "failed"
else:
print "Test case passed!"
test()
User's Answer:
(Note: The answer done by the user is not guaranteed to be correct)
# Credit goes to Websten from forums
#
# Use Dave's suggestions to finish your daysBetweenDates
# procedure. It will need to take into account leap years
# in addition to the correct number of days in each month.
def is_leap_year(year):
if year % 4 == 0:
return True
return False
def days_in_month(year, month):
if month in [1, 3, 5, 7, 8, 10, 12]:
return 31
elif month == 2:
if is_leap_year(year):
return 29
else:
return 28
else:
return 30
def nextDay(year, month, day):
"""Simple version: assume every month has 30 days"""
if day < days_in_month(year, month):
return year, month, day + 1
else:
if month == 12:
return year + 1, 1, 1
else:
return year, month + 1, 1
def dateIsBefore(year1, month1, day1, year2, month2, day2):
"""Returns True if year1-month1-day1 is before year2-month2-day2. Otherwise, returns False."""
if year1 < year2:
return True
if year1 == year2:
if month1 < month2:
return True
if month1 == month2:
return day1 < day2
return False
def daysBetweenDates(year1, month1, day1, year2, month2, day2):
"""Returns the number of days between year1/month1/day1
and year2/month2/day2. Assumes inputs are valid dates
in Gregorian calendar."""
# program defensively! Add an assertion if the input is not valid!
assert not dateIsBefore(year2, month2, day2, year1, month1, day1)
days = 0
while dateIsBefore(year1, month1, day1, year2, month2, day2):
year1, month1, day1 = nextDay(year1, month1, day1)
days += 1
return days
def test():
test_cases = [((2012,1,1,2012,2,28), 58),
((2012,1,1,2012,3,1), 60),
((2011,6,30,2012,6,30), 366),
((2011,1,1,2012,8,8), 585 ),
((1900,1,1,1999,12,31), 36523)]
for (args, answer) in test_cases:
result = daysBetweenDates(*args)
if result != answer:
print "Test with data:", args, "failed. Result is ", result, " but the answer is ", answer
else:
print "Test case passed!"
test()
INSTRUCTOR NOTE:
The answer to this quiz starts in the next video and progresses through the rest of the lesson. Click the navigation bar to move on to it.