How to determine whether a year is a leap year

Question

I am trying to make a simple calculator to determine whether or not a certain year is a leap year    By definition  a leap year is divisible by four  but not by one hundred  unless it is divisible by four hundred   Here is my code   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

User · Answer

The whole formula can be contained in a single expression   def is leap year year       return  year   4    0 and year   100    0  or year   400    0  print n    is a leap year  if is leap year n  else   is not a leap year

User · Answer

Your function doesn t return anything  so that s why when you use it with the print statement you get None  So either just call your function like this   leapyr 1900    or modify your function to return a value  by using the return statement   which then would be printed by your print statement   Note  This does not address any possible problems you have with your leap year computation  but ANSWERS YOUR SPECIFIC QUESTION as to why you are getting None as a result of your function call in conjunction with your print   Explanation   Some short examples regarding the above   def add2 n1  n2       print  the result is    n1   n2    prints but uses no  return  statement  def add2 New n1  n2       return n1   n2      returns the result to caller   Now when I call them   print add2 10  5    this gives   the result is  15 None   The first line comes form the print statement inside of add2    The None from the print statement when I call the function add2   which does not have a return statement  causing the None to be printed  Incidentally  if I had just called the  add2   function simply with  note  no print statement    add2     I would have just gotten the output of the print statement the result is  15 without the None  which looks like what you are trying to do    Compare this with   print add2 New 10  5    which gives   15   In this case the result is computed in the function add2 New   and no print statement  and returned to the caller who then prints it in turn

User · Answer

If you don t want to import calendar and apply  isleap method you can try this   def isleapyear year       if year   4    0 and  year   100    0 or year   400    0           return True     return False

User · Answer

A leap year is exactly divisible by 4 except for century years  years ending with 00   The century year is a leap year only if it is perfectly divisible by 400  For example   if   year   4     0       if    year   100      0            if    year   400     0                print   0  is a leap year  format year           else               print   0  is not a leap year  format year       else           print   0  is a leap year  format year    else       print   0  is not a leap year  format year

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In the Gregorian calendar  three conditions are used to identify leap years   The year can be evenly divided by 4  is a leap year  unless   The year can be evenly divided by 100  it is NOT a leap year  unless   The year is also evenly divisible by 400  Then it is a leap year       This means that in the Gregorian calendar  the years 2000 and 2400 are leap years  while 1800  1900  2100  2200  2300 and 2500 are NOT leap years  source def is leap year       leap   False     if year   4    0          leap   True         if year   4    0 and year   100    0              leap   False             if year   400    0                  leap   True      return leap  year   int input    leap   is leap year   if leap    print f quot  year  is a leap year quot   else    print f quot  year  is not a leap year quot

User · Answer

You test three different things on n  n   4 n   100 n   400  For 1900  1900   4    0 1900   100    0 1900   400    300  So 1900 doesn t enter the if clause because 1900   100    0 is False But 1900 also doesn t enter the else clause because 1900   4    0 is also False This means that execution reaches the end of your function and doesn t see a return statement  so it returns None  This rewriting of your function should work  and should return False or True as appropriate for the year number you pass into it    Note that  as in the other answer  you have to return something rather than print it   def leapyr n       if n   400    0          return True     if n   100    0          return False     if n   4    0          return True     return False print leapyr 1900    Algorithm from Wikipedia

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From 1700 to 1917  official calendar was the Julian calendar  Since then they we use the Gregorian calendar system  The transition from the Julian to Gregorian calendar system occurred in 1918  when the next day after January 31st was February 14th  This means that 32nd day in 1918  was the February 14th   In both calendar systems  February is the only month with a variable amount of days  it has 29 days during a leap year  and 28 days during all other years  In the Julian calendar  leap years are divisible by 4 while in the Gregorian calendar  leap years are either of the following   Divisible by 400   Divisible by 4 and not divisible by 100   So the program for leap year will be   def leap notleap year        yr          if year  lt   1917          if year   4    0              yr    leap          else              yr    not leap      elif year  gt   1919          if  year   400    0  or  year   4    0 and year   100    0               yr    leap          else              yr    not leap      else          yr    none actually  since feb had only 14 days       return yr

User · Answer

Use calendar isleap   import calendar print calendar isleap 1900

User · Answer

An alternative one liner      y   4     int  y -  y   100     y      y   400    100    - 1   lt  0  This was something that I put together for fun     that is also 1 1 compatible with C   y   4   gt  gt  gt It first checks if the year is a leap year via the typical mod-4 check   int  y -  y   100     y   gt  gt  gt It then accounts for those years divisible by 100  If the year is evenly divisible by 100  this will result in a value of 1  otherwise it will result in a value of 0    y   400    100     gt  gt  gt Next  the year is divided by 400  and subsequently 100  to return 1  2  or 3 if it is not  These two values  int y -  y   100     y   amp    y   400    100    are then multiplied together  If the year is not divisible by 100  this will always equal 0  otherwise if it is divisible by 100  but not by 400  it will result in 1  2  or 3  If it is divisible by both 100 and 400  it will result in 0  This value is added to  y   4   which will only be equal to 0 if the year is a leap year after accounting for the edge-cases  Finally  1 is subtracted from this remaining value  resulting in -1 if the year is a leap year  and either 0  1  or 2 if it is not  This value is compared against 0 with the less-than operator  If the year is a leap year this will result in True  or 1  if used in C   otherwise it will return False  or 0  if used in C   Please note  this code is horribly inefficient  incredibly unreadable  and a detriment to any code attempting to follow proper practices  This was an exercise of mine to see if I could do so  and nothing more  Further  be aware that ZeroDivisionErrors are a consequence of the input year equaling 0  and must be accounted for  For example  a VERY basic timeit comparison of 1000 executions shows that  when compared against an equivalent codeblock utilizing simple if-statements and the modulus operator  this one-liner is roughly 5 times slower than its if-block equivalent  That being said  I do find it highly entertaining

User · Answer

The logic in the  quot one-liner quot  works fine   From personal experience  what has helped me is to assign the statements to variables  in their  quot True quot  form  and then use logical operators for the result  A   year   4    0 B   year   100    0 C   year   400    0  I used      in the B statement instead of  quot    quot  and applied  not  logical operator in the calculation  leap   A and  not B or C   This comes in handy with a larger set of conditions  and to simplify the boolean operation where applicable before writing a whole bunch of if statements

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As a one-liner function  def is leap year year        quot  quot  quot Determine whether a year is a leap year  quot  quot  quot       return year   4    0 and  year   100    0 or year   400    0   It s similar to the Mark s answer  but short circuits at the first test  note the parenthesis   Alternatively  you can use the standard library s calendar isleap  which has exactly the same implementation  from calendar import isleap print isleap 1900

[python] How to determine whether a year is a leap year?

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