leap-year

The following sequence is an extract of the pandas DataFrame that I've got: >>> df_t value 2011-01-31 -5.575000 2011-03-31 7.700000 2011-05-31 15.966667 2011-07-31 10.683333 2011-08-31 10.454167 2011-10-31 9.320833 2011-12-31 -0.358333 2012-01-31 -11.550000 2012-03-31 1.700000 2012-05-31 12.333333 2012-07-31 12.816667 2012-08-31 11.837500 2012-10-31 2.733333 2012-12-31 4.075000 2013-01-31 2.450000 2013-03-31 -4.262500 2013-05-31 11.491667 2013-07-31 14.812500 2013-08-31 13.920833 2013-10-31 4.125000 2013-12-31 0.075000 How can I delete March 31st in every leap year? I tried something like: def

In the following query the leap year is not taken into account. SELECT e.id, e.title, e.birthdate FROM employers e WHERE DAYOFYEAR(curdate()) <= DAYOFYEAR(e.birthdate) AND DAYOFYEAR(curdate()) +14 >= DAYOFYEAR(e.birthdate) So in this query the birthdate of someone who is born in a leap year has got a different dayofyear in a non leap year. How can i adjust the query to make sure it also works in a leap year? The mysql version i have is: 5.0.67 Where NOW() is a non-leap year 2011 , the problem arises from anybody born on a leap year after February 29 will have an extra day because you are using

Is there an easy way to determine if a year is a leap year? Use Date#leap? . now = DateTime.now flag = Date.leap?( now.year ) e.g. Date.leap?( 2018 ) # => false Date.leap?( 2016 ) # => true Aldrine Einsteen For your understanding: def leap_year?(year) if year % 4 == 0 if year % 100 == 0 if yearVar % 400 == 0 return true end return false end return true end false end This could be written as: def leap_year?(year) (year % 4 == 0) && !(year % 100 == 0) || (year % 400 == 0) end Pesto Try this: is_leap_year = year % 4 == 0 && year % 100 != 0 || year % 400 == 0 Here is my answer for the exercism.io

This question already has an answer here: Java Code for calculating Leap Year 20 answers Here is what I have: Scanner input = new Scanner(System.in); System.out.print("Enter a year: "); int Year = input.nextInt(); System.out.print("Enter a month (first three letters with the first" + " letter uppercase): "); String Month = input.next(); String ThirtyOne = "Jan" + "Mar" + "May" + "Jul" + "Aug" + "Oct" + "Dec"; String DaysThirtyOne = ThirtyOne.substring(21) + "31"; String Thirty = "Apr" + "Jun" + "Sep" + "Nov"; String DaysThirty = Thirty.substring(12) + "30"; String TwentyEight = "Feb"; String

I am assuming Java has some built-in way to do this. Given a date, how can I determine the date one day prior to that date? For example, suppose I am given 3/1/2009. The previous date is 2/28/2009. If I had been given 3/1/2008, the previous date would have been 2/29/2008. Use the Calendar interface. Calendar cal = Calendar.getInstance(); cal.setTime(myDate); cal.add(Calendar.DAY_OF_YEAR,-1); Date oneDayBefore= cal.getTime(); Doing "addition" in this way guarantees you get a valid date. This is valid for 1st of the year as well, e.g. if myDate is January 1st, 2012, oneDayBefore will be December

In PHP given a UTC timestamp I would like to add exactly N number of years. This should take into consideration leap years. Thank you. $newTimestamp = strtotime('+2 years', $timestamp); Replace "+2 years" as required. ref: http://php.net/manual/en/function.strtotime.php $date = new DateTime(); $date->add(new DateInterval('P10Y')); adds 10 years ( 10Y ) to "today". DateTime's only in PHP 5.3, though. One thing you should consider when you do this. $newTimestamp = strtotime('+2 years', $timestamp); This adds up 2 years ( 720 or 721 days). In case you just want to keep the same day and month and