![]() If from_in_this_year =0 and days=0, years More robust function - calculates difference in years (age) and days: def get_diff_in_years_and_days(from_date, to_date):įrom_in_this_year = date(to_date.year, from_date.month, from_date.day)įrom_in_this_year = date(to_date.year, from_date.month, from_date.day-1) # today is feb in leap year Adding a difference to another date will result in a value that can be converted back into a useful date.The difference between any dates with the same month and day and time will be an exact number of years.I'm not suggesting that this is the solution, because I don't think a perfect solution is possible. Return dt.year + days_from_jan1.days * size_of_day + days_from_conds * size_of_secondĭifference_in_years = date_as_float(end_time) - date_as_float(start_time) If not isleap(dt.year) and days_from_jan1.days >= 31+28: Second is that if you try to coerce such a number back to a date, you'll have to account for non-leap years and check for the special case of Feb. ![]() 29 - "Sorry Johnny, you don't get a birthday this year, there's no Feb. The anomalies will firstly be that there can't be a date which is exactly a year (or 2 or 3) from Feb. Thus I am proposing yet another encoding, based on each year containing 366 days whether it is a leap year or not. The problem with this scheme is that you still can't make sense of (date2-date1)+date3, because the fraction can't be resolved back to a day with any consistency. If you treat each of the previous conditions as an integral year, all you have to do is decide on which fraction to assign to any number of days left over. My assertion that you needed to break this down into integral years and fractional years was an attempt to get around this problem. This is the case for your original code which assumed a day was 1/365.2425 of a year, or indeed for any code which assumes a constant fraction of a year per day, even if the size of a day accounts for the years which are leap years. If you choose to represent a date as a year plus a fraction of a year based on the day, it is impossible to make both of these statements true. How about the difference between -01? Again, most people would agree that it should be exactly 1.0 years. What's the difference between -28? Most people would agree that it should be exactly 1.0 years. Any solution you come up with is going to have anomalies because of this, and you're going to have to choose which anomalies you can live with. The basic problem is that calendar years are not a constant size, but you're coercing them to be constant by setting them to 1.0. You want the fractional part to increment in equal amounts until it equals 1.0 at the next anniversary date, so it should be based on the number of days in the year after the end date.ĭo you want your date range to include 1900 or 2100? Things get a little easier if you don't.Įdit: It has taken me a long time to reason this through. Both need to deal with leap years, but in different ways - the integral needs to deal with a starting date of February 29, and the fractional must deal with the differing number of days in a year. To make sense of leap years, you are almost forced to break this into two parts: an integral number of years, and a fractional part. ![]() In this case that's a difference of 0.0012322917425568528 years, or 0.662 days, considering that this is not a leap year. To be correct you have to do this: from datetime import datetimeĭiffyears = end_date.year - start_date.yearĭifference = end_date - start_date.replace(end_date.year)ĭays_in_year = isleap(end_date.year) and 366 or 365ĭifference_in_years = diffyears + (difference.days + conds/86400.0)/days_in_year So the question doesn't really make much sense. Well, yeah, averaged over a thousand years, yeah. And how long is a day in years? You say 1/365.2425. Do you want the difference in leap years or normal years? :-) As you calculate you are always going to get a slightly incorrect answer. But it depends on how correct you want to be. More efficient? No, but more correct, probably.
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