Python

Search

def is_leap(year): leap = False

def is_leap(year):
    leap = False
    if(year%4==0):
        #century year check
        if(year%100==0):
            #century leap year check
            if(year%400==0):
                leap=True
        else:
            leap=True
    return leap

Comment

PREVIOUS	NEXT

Code Example
Python :: python button click code
Python :: python len
Python :: python group by
Python :: add new column to pandas dataframe
Python :: math floor python
Python :: np.vstack python
Python :: list comprehension odd numbers python
Python :: python buffer
Python :: python inherit from objects
Python :: filter dictionary python
Python :: positive and negative number in python
Python :: python virtual env
Python :: socket.accept python
Python :: def rectangles
Python :: pygame draw square
Python :: fastest way to check odd or even in python
Python :: python default dictionary
Python :: django filter on related field
Python :: min max code in python
Python :: a int and float python
Python :: add to list python
Python :: open chrome console in selenium
Python :: csv reader url
Python :: part of a flower
Python :: Merge multiple dataframs
Python :: python suppress print output from function
Python :: How to sum a column in Python csv
Python :: python numpy euler
Python :: python3 vowels and consonants filter
Python :: slug

Search

PYTHON

def is_leap(year): leap = False

ADD CONTENT