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     • This Python program allows the user to enter any number and check whether the user entered is a leap year or not using the If statement. yr = int(input("Please Enter the Number you wish: ")) if ((yr%400 == 0) or ((yr%4 == 0) and (yr%100 != 0))): print("%d is a Leap year" %yr) else: print("%d is Not" %yr)
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     Python Program to Check Leap Year or Not - Tutorial Gateway

2. Top results related to is 1492 a leap year python code

   • Python: To check whether a given year is a leap year or not?

     Top Answer

     Answered Nov 13, 2022 · 0 votes

     Let's step through your code for 2024:

     • year%4==0 is True, so we enter the if block
     • year%100!=0 is also True, so we enter the if block again.

     And now you check for year%400==0. Um, why? 2024 isn't a century, so it can never be a multiple of 400. So your logic is broken here. 2024 is divisible by 4 and not divisible by 100. At this point we can simply return True!

     Where we need to check for divisibility by 400 is in the next case (where year%100!=0 is False).

     If we take all of this, we get this code:

     def is_leap(year): if(year%4==0): if(year%100!=0): return True else: if(year%400==0): return True else: return False else: return False-

     (note how I got rid of the completely unnecessary variable.)

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     1/5

   • Python how to check if it is a leap year with ==int

     Top Answer

     Answered Nov 01, 2017 · 1 votes

     Your issue is that int is a type. If you try to compare a number to a type of object, which is what you are doing when you write if year/400 == int :, it will always return False, because these can never be the same.

     A better way to check if year/400 is an integer would be:

     if year%400 == 0: return False

     This is saying: If the remainder of year/400 is equal to 0, return False, which is what you wanted.

     Some other things are:

     You should use elif instead of if in most cases. In this one, it doesn't matter, since the return statement terminates the execution of the function early, but you should still use it anyways. Otherwise, even when you have your final result, it will go through the rest of the if statements.

     The other thing isn't related to your code, but your leap year criteria are incorrect. If the year is divisible by 4, then it's a leap year. Unless the year is divisible by 100, then it's not. Unless it's divisible by 400, then it is.

     Improved code:

     def leapyear(year): if year%400 == 0: return True elif year%100 == 0: return False elif year%4 == 0: return True return False

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     2/5

   • How can I check if a year is a leap year using Python?

     Top Answer

     Answered May 10, 2020 · 4 votes

     Divisibility by 400 is an exception to the rule that years divisible by 100 are not leap years, which itself is an exception to the rule that years divisible by 4 are leap years. If you wrote it out in sequence, you might write

     if year % 400 == 0: # Some centuries are leap years... print("leap year")elif year % 100 == 0: # ... but most are not ... print("not leap year")elif year % 4 == 0: # ... even though other divisibly-by-four years are print("leap year")else: print("not leap year")

     or work your way up

     if year % 4 != 0: print("not a leap year")elif year % 100 != 0: print("leap year")elif year % 400 != 0: print("not a leap year")else: print("leap year")

     Combining these into a single test would be something like

     if year % 4 != 0 or (year % 100 == 0 and year % 400 != 0): print("not a leap year")else: print("leap year")

     which I find a little harder to follow than a series of simple tests.

     (Given the nature of a solar year, making every year divisible by 400 a leap year is also a problem, though not as big a one as making every century year a leap year. People wanting to avoid drifts over the millenia will propose additional exceptions like "years divisible by 4000, or 40,000, or something, should not be leap years"; no official rule exists yet, though.)

     Other Answers

     Answered May 10, 2020 · 1 votes

     Because 1900 % 4 is actually 0, so the first if conditional is True ( True or False is True) and then it doesn't execute the second if block because is an elif (else if, and since the first was True, there is no need to execute the else part).

     Other Answers

     Answered Jul 27, 2020 · 1 votes

     Try like this!

     year = int(input("Input year: "))-if year % 4 == 0: print("Year is leap.") if year % 100 == 0 and year % 400 != 0: print("Year is common.")else: print("Year is common.")

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     3/5

   • Python | Pandas PeriodIndex.is_leap_year

     Python is a great language for doing data analysis, primarily because of the fantastic ecosystem of data-centric python packages. Pandas is one of those packages and makes importing and analyzing data much easier.

     Pandas PeriodIndex.is_leap_year attribute return an array of boolean values corresponding to each element in the PeriodIndex object. It return True if the given year is a leap year else it return False if it is not a leap year.

     Syntax : PeriodIndex.is_leap_year

     Parameters : None

     Return : array of boolean values

     Example #1: Use PeriodIndex.is_leap_year attribute to check for each element in the given PeriodIndex object, whether it is a leap year or not.

     # importing pandas as pd import pandas as pd   # Create the PeriodIndex object pidx = pd.PeriodIndex(start ='2003-12-21 08:45 ',               end ='2009-12-21 11:55', freq ='Y')   # Print the PeriodIndex object print(pidx)

     Output :

     Now we will use the PeriodIndex.is_leap_year attribute to check if the given year is a leap year or not.

     # check for leap year pidx.is_leap_year -

     Output :

     As we can see in the output, the PeriodIndex.is_leap_year attribute has returned an...

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     4/5

   • Python | Pandas DatetimeIndex.is_leap_year

     Python is a great language for doing data analysis, primarily because of the fantastic ecosystem of data-centric python packages. Pandas is one of those packages and makes importing and analyzing data much easier.

     Pandas DatetimeIndex.is_leap_year attribute return a boolean indicator if the date belongs to a leap year. A leap year is a year, which has 366 days (instead of 365) including 29th of February as an intercalary day. Leap years are years which are multiples of four with the exception of years divisible by 100 but not by 400.

     Syntax: DatetimeIndex.is_leap_year

     Returns: numpy array containing logical values.

     Example #1: Use DatetimeIndex.is_leap_year attribute to check if the dates present in the DatetimeIndex object belongs to a leap year.

     # importing pandas as pd import pandas as pd   # Create the DatetimeIndex didx = pd.DatetimeIndex(['2014-01-01', '2008-12-31', '2017-03-31', '2000-12-31'])   # Print the DatetimeIndex print(didx)

     Output :

     Now we want to find if the dates contained in the given DatetimeIndex object belongs to a leap year or not.

     # find if the dates belong to leap year didx.is_leap_year -

     Output : ...

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     5/5

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3. People also ask

   How to program a leap year in Python?

   • To program a leap year, you need to check if a given year meets specific conditions. In Python, you can use an algorithm that checks if the year is divisible by 4, except for years divisible by 100. However, years divisible by 400 are still considered leap years. By implementing this logic in your program, you can accurately determine leap years.

   Python Program For Leap Year (With Code) - Python Mania

   pythonmania.org/python-program-for-leap-year/

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   How often do leap years occur in Python?

   • Leap years occur every four years, except for years that are divisible by 100. However, we consider the years divisible by 400 as leap years as well. For example, the years 2000 and 2004 are leap years, while 1900 is not. Q: Can I use the Python program for any year? Yes, you can use this Python program for leap year for any year.

   Python Program For Leap Year (With Code) - Python Mania

   pythonmania.org/python-program-for-leap-year/

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   How to tell if a year is a leap year?

   • Also, if the year is divisible by 4 and not divisible by 100, then that is considered a leap year. Let’s create a function that accepts the year and tells if the given year is a leap year or not. if year % 4 == 0: if year % 100 == 0: if year % 400 == 0: print(year, "is a leap year") else: print(year, "is not a leap year") else:

   Leap Year Program In Python Using Function - Python Guides

   pythonguides.com/python-program-to-check-leap-year/

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   Is there a “def” function for leap year in Python?

   • There is no specific “DEF” function for leap year in Python. However, you can define a function using the def keyword, as shown in the example provided earlier. The def keyword is used to define user-defined functions in Python.

   Python Program For Leap Year (With Code) - Python Mania

   pythonmania.org/python-program-for-leap-year/

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4. www.geeksforgeeks.org › python-program-to-check-ifPython Program to Check if given year is Leap Year ...

   www.geeksforgeeks.org › python-program-to-check-if

   • Cached

   Feb 8, 2024 · Using Lambda Function. Using any () Method. Check If A Given Year Is Leap Year Using Modulo Operator. In this example, The function `is_leap_year_modulo` checks if a given year is a leap year using the modulo operator to verify divisibility by 4, excluding years divisible by 100 unless divisible by 400.

5. stackoverflow.com › questions › 11621740python - How to determine whether a year is a leap year ...

   stackoverflow.com › questions › 11621740

   • Cached

   Jul 24, 2012 · def leapyr(n): if n%4==0 and n%100!=0: if n%400==0: print(n, "is a leap year.") elif n%4!=0: print(n, "is not a leap year.") print(leapyr(1900)) When I try this inside the Python IDLE, the module returns None. I am pretty sure that I should get 1900 is a leap year.

   Code sample

   import calendar

   print(calendar.isleap(1900))

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6. www.programiz.com › examples › leap-yearPython Program to Check Leap Year

   www.programiz.com › examples › leap-year

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   # Python program to check if year is a leap year or not year = 2000 # To get year (integer input) from the user # year = int(input("Enter a year: ")) # divided by 100 means century year (ending with 00) # century year divided by 400 is leap year if (year % 400 == 0) and (year % 100 == 0): print("{0} is a leap year".format(year)) # not divided ...

7. pythonguides.com › python-program-to-check-leap-yearLeap Year Program in Python using Function

   pythonguides.com › python-program-to-check-leap-year

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   May 9, 2024 · def is_leap_year(): year = int(input("Enter the year: ")) if year % 4 == 0: if year % 100 == 0: if year % 400 == 0: print(year, "is a leap year") else: print(year, "is not a leap year") else: print(year, "is a leap year") else: print(year, "is not a leap year")

8. pythonmania.org › python-program-for-leap-yearPython Program For Leap Year (With Code) - Python Mania

   pythonmania.org › python-program-for-leap-year

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   user_input = int(input("Enter the year: ")) if(is_leap_year(user_input)): print(user_input, "is a leap year.") else: print(user_input, "is not a leap year.") You can run this code on our free Online Python Compiler. Output. Enter the year: 2020.

9. blog.finxter.com › 5-best-ways-to-check-for-a-leap5 Best Ways to Check for a Leap Year in Python Using datetime

   blog.finxter.com › 5-best-ways-to-check-for-a-leap

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   Feb 28, 2024 · from datetime import datetime from dateutil.relativedelta import relativedelta def is_leap_year(year): feb_28 = datetime(year=year, month=2, day=28) mar_01 = datetime(year=year, month=3, day=1) return (mar_01 - feb_28).days > 1 print(is_leap_year(2020)) print(is_leap_year(2019))

10. beginnersbook.com › 2018 › 01Python Program to Check whether Year is a Leap Year or not

    beginnersbook.com › 2018 › 01

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    Jan 3, 2018 · # User enters the year year = int (input ("Enter Year: ")) # Leap Year Check if year % 4 == 0 and year % 100 != 0: print (year, "is a Leap Year") elif year % 100 == 0: print (year, "is not a Leap Year") elif year % 400 ==0: print (year, "is a Leap Year") else: print (year, "is not a Leap Year") Output: About the Author.