Python 2.5 Help

woody2371 · 05-10-2007, 06:45 AM

Hey, i got this code to work out whether or not it is a leap year, i defined it in a module.

def leapyear(year):
ret = 0
if(year % 1000):
ret = 1
elif(year % 400):
ret = 1
elif(year % 4):
ret = 1
if (ret == 1):
isleapyear = "Yes it is a leap year"
elif (ret == 0):
isleapyear = "No it is not a leap year"
return isleapyear

Now when i run it, inputting 4 as the year var makes it say correctly it is a leap year, as does 8,12,16 etc etc.

However, when i input 1000 or something around there, it says it is NOT a leap year. Anyone know why?

Cheers

**Kalim** · 05-12-2007, 04:11 AM

I don't know much of Python, but a friend I visited was using this script for what you're wanting to do:

Code:

def isLeapYear (year):
    result = year % 4
    if result > 0:
        print "It is NOT a leap year"
    else:
        print "It IS a leap year"

        result = (year * 100) % 4
        if result >0:
            print "It is NOT a leap year"

isLeapYear(400)
isLeapYear(800)
isLeapYear(1600)
isLeapYear(1601)
isLeapYear(2004)
isLeapYear(2012)
isLeapYear(3010)
isLeapYear(3330)

I know it had initial problems with calculating figures over a few centuries so she later had to tweak it more. It worked well, but IDK if it was for Python 2.4, earlier or 2.5.1, sorry.

Maybe it will be of some help.

**Deckard** · 05-12-2007, 09:53 AM

I believe the leap year formula for the Gregorian calendar (post-1582) is:

Quote:

Originally Posted by http://en.wikipedia.org/wiki/Leap_year#Leap_year_algorithms

if year modulo 400 is 0 then leap
else if year modulo 100 is 0 then no_leap
else if year modulo 4 is 0 then leap
else no_leap

Before 1582, you'd use the Julian calendar.

05-10-2007, 06:45 AM	#1 (permalink)
woody2371 Registered User Join Date: May 2007 Posts: 2 OS: Windows XP Pro	Python 2.5 Help Hey, i got this code to work out whether or not it is a leap year, i defined it in a module. def leapyear(year): ret = 0 if(year % 1000): ret = 1 elif(year % 400): ret = 1 elif(year % 4): ret = 1 if (ret == 1): isleapyear = "Yes it is a leap year" elif (ret == 0): isleapyear = "No it is not a leap year" return isleapyear Now when i run it, inputting 4 as the year var makes it say correctly it is a leap year, as does 8,12,16 etc etc. However, when i input 1000 or something around there, it says it is NOT a leap year. Anyone know why? Cheers

05-12-2007, 04:11 AM	#2 (permalink)
Kalim Roaming To Help Join Date: Nov 2006 Posts: 5,642 OS: Many	I don't know much of Python, but a friend I visited was using this script for what you're wanting to do: Code: def isLeapYear (year): result = year % 4 if result > 0: print "It is NOT a leap year" else: print "It IS a leap year" result = (year * 100) % 4 if result >0: print "It is NOT a leap year" isLeapYear(400) isLeapYear(800) isLeapYear(1600) isLeapYear(1601) isLeapYear(2004) isLeapYear(2012) isLeapYear(3010) isLeapYear(3330) I know it had initial problems with calculating figures over a few centuries so she later had to tweak it more. It worked well, but IDK if it was for Python 2.4, earlier or 2.5.1, sorry. Maybe it will be of some help.