问题
Input: # of seconds since January 1st, of Year 0001
Output: # of Full years during this time period
I have developed an algorithm that I do not think is the optimal solution. I think there should be a solution that does not involve a loop. See Code Block 1 for the algorithm which A) Determines the quantity of days and B) Iteratively subtracts 366 or 365 depending on Leap Years from the Day Total while incrementing the Year Total
It's not as simple as Dividing DayCount by 365.2425 and truncating, because we hit a failure point at on January 1, 0002 (31536000 Seconds / (365.2425 * 24 * 60 * 60)) = 0.99934.
Any idea on a non-looping method for extracting years from a quantity of seconds since January 1, 0001 12:00 AM?
I need to figure this out because I need a date embedded in a long (which stores seconds) so that I can track years out to 12+ million with 1-second precision.
Code block 1 - Inefficient Algorithm to get Years from Seconds (Including Leap Years)
Dim Days, Years As Integer
'get Days
Days = Ticks * (1 / 24) * (1 / 60) * (1 / 60) 'Ticks = Seconds from Year 1, January 1
'get years by counting up from the beginning
Years = 0
While True
'if leap year
If (Year Mod 4 = 0) AndAlso (Year Mod 100 <> 0) OrElse (Year Mod 400 = 0) Then
If Days >= 366 Then 'if we have enough days left to increment the year
Years += 1
Days -= 366
Else
Exit While
End If
'if not leap year
Else
If Days >= 365 Then 'if we have enough days left to increment the year
Years += 1
Days -= 365
Else
Exit While
End If
End If
End While
Return Years
Edit: My solution was to skip the memory savings of embedding a date within 8 bits and to store each value (seconds through years) in separate integers. This causes instant retrievals at the expense of memory.
Edit2: Typo in first edit (8bits)
回答1:
If you need accuracy to the very second, you'll probably want a commercial-grade datetime package; it's just too complex to do accurately with a simple algorithm. For instance:
- Many people have noted that we have leap years every 4 years, but did you know that every year divisible by 100 but not by 400 is not a leap-year? This has caused issues even in large commercial products.
- Some countries do not observe daylight-savings, and the ones who do observe it do so at different times of the year. This changes arbitrarily from year-to-year.
- Years before 1582 use a slightly different calendar
- There were only 355 days in the year 1582 (or 354 in the year 1752, depending on the country).
- There are major issues when countries switch timezones.
- Then there's leap-seconds. Some governing body decides arbitrarily whether or not we should add one (or sometimes, two) seconds to the clock each year. There's no way to know ahead of time when we'll have the next leap-second, and no pattern to past leap-seconds.
Because of these and more complications, you are better off not writing the code yourself, unless you can relax the constraint that you need accuracy to the very second over 12-million years.
"October 4, 1582 – Saint Teresa of Ávila dies. She is buried the next day, October 15."
回答2:
Wikipeda has an article on Julian Date with an algorithm which you could adapt to your needs.
回答3:
Const TICKS_PER_YEAR As Long = 315360000000000
Function YearsSinceBeginningOfTimeUntil(ByVal d As DateTime) As Integer
Return Math.Floor(d.Ticks / TICKS_PER_YEAR)
End Function
回答4:
You don't need a loop, calculate seconds from 1 Jan 0001 to unix epoch start (1 Jan 1970 00:00:00), and save somewhere. Then subtract it from your input, then use any tool available to convert unix timestamp (seconds from 1 Jan 1970) to years, then add 1970. I don't know much VB programming to post a detailed guide.
回答5:
The following assumes that the Gregorian calendar will remain in effect for the upcoming five hundred and eighty four and a half billion years. Be prepared for disappointment, though; the calendar may end up being scrapped as our sun begins to expand, changing the orbit of the Earth and the duration of the year, and it is very likely something else will be adopted when the Earth falls into the sun seven and a half billion years from now.
As an aside, I don't even try to handle dates prior to the adoption of the Gregorian calendar. I simply return the number of days the date occurred prior to the 15th of October, 1582, and the need to be able to express that sort of return value is the only reason the GetDateFromSerial function has an asString parameter.
Sub GetDateFromSerial(ByVal dateSerial As ULong, ByRef year As Long, ByRef month As Integer, ByRef dayOfMonth As Integer, ByRef secondsIntoDay As Integer, ByRef asString As String)
Const SecondsInOneDay As ULong = 86400 ' 24 hours * 60 minutes per hour * 60 seconds per minute
'Dim startOfGregorianCalendar As DateTime = New DateTime(1582, 10, 15)
'Dim startOfGregorianCalendarInSeconds As ULong = (startOfGregorianCalendar - New DateTime(1, 1, 1)).TotalSeconds
Const StartOfGregorianCalendarInSeconds As ULong = 49916304000
secondsIntoDay = dateSerial Mod SecondsInOneDay
If dateSerial < StartOfGregorianCalendarInSeconds Then
year = -1
month = -1
dayOfMonth = -1
Dim days As Integer = (StartOfGregorianCalendarInSeconds - dateSerial) \ SecondsInOneDay
asString = days & IIf(days = 1, " day", " days") & " before the adoption of the Gregorian calendar on October 15, 1582"
Else
'Dim maximumDateValueInSeconds As ULong = (DateTime.MaxValue - New DateTime(1, 1, 1)).TotalSeconds
Const MaximumDateValueInSeconds As ULong = 315537897600
If dateSerial <= MaximumDateValueInSeconds Then
Dim parsedDate As DateTime = DateTime.MinValue.AddSeconds(dateSerial)
year = parsedDate.Year
month = parsedDate.Month
dayOfMonth = parsedDate.Day
Else
' Move the date back into the range that DateTime can parse, by stripping away blocks of
' 400 years. Aim to put the date within the range of years 2001 to 2400.
Dim dateSerialInDays As ULong = dateSerial \ SecondsInOneDay
Const DaysInFourHundredYears As Integer = 365 * 400 + 97 ' Three multiple-of-4 years in each 400 are not leap years.
Dim fourHundredYearBlocks As Integer = dateSerialInDays \ DaysInFourHundredYears
Dim blocksToFactorInLater As Integer = fourHundredYearBlocks - 5
Dim translatedDateSerialInDays As ULong = dateSerialInDays - blocksToFactorInLater * CLng(DaysInFourHundredYears)
' Parse the date as normal now.
Dim parsedDate As DateTime = DateTime.MinValue.AddDays(translatedDateSerialInDays)
year = parsedDate.Year
month = parsedDate.Month
dayOfMonth = parsedDate.Day
' Factor back in the years we took out earlier.
year += blocksToFactorInLater * 400L
End If
asString = New DateTime(2000, month, dayOfMonth).ToString("dd MMM") & ", " & year
End If
End Sub
Function GetSerialFromDate(ByVal year As Long, ByVal month As Integer, ByVal dayOfMonth As Integer, ByVal secondsIntoDay As Integer) As ULong
Const SecondsInOneDay As Integer = 86400 ' 24 hours * 60 minutes per hour * 60 seconds per minute
If (year < 1582) Or _
((year = 1582) And (month < 10)) Or _
((year = 1582) And (month = 10) And (dayOfMonth < 15)) Then
Throw New Exception("The specified date value has no meaning because it falls before the point at which the Gregorian calendar was adopted.")
End If
' Use DateTime for what we can -- which is years prior to 9999 -- and then factor the remaining years
' in. We do this by translating the date back by blocks of 400 years (which are always the same length,
' even factoring in leap years), and then factoring them back in after the fact.
Dim fourHundredYearBlocks As Integer = year \ 400
Dim blocksToFactorInLater As Integer = fourHundredYearBlocks - 5
If blocksToFactorInLater < 0 Then blocksToFactorInLater = 0
year = year - blocksToFactorInLater * 400L
Dim dateValue As DateTime = New DateTime(year, month, dayOfMonth)
Dim translatedDateSerialInDays As ULong = (dateValue - New DateTime(1, 1, 1)).TotalDays
Const DaysInFourHundredYears As ULong = 365 * 400 + 97 ' Three multiple-of-4 years in each 400 are not leap years.
Dim dateSerialInDays As ULong = translatedDateSerialInDays + blocksToFactorInLater * DaysInFourHundredYears
Dim dateSerial As ULong = dateSerialInDays * SecondsInOneDay + secondsIntoDay
Return dateSerial
End Function
回答6:
I know this question is old now, but I see ones like it often and there weren't any easy answers in here.
My solution uses an old trick of writing two dates as if they were numbers (e.g. 'Dec 12th 2013' as 20131212) and then subtracting one from the other and discarding the last four digits. I dug up my implementation in F#, you can paste it into LinqPad to check answers. Takes leap years etc into account too:
let dateFrom = new DateTime(1,1,1)
let dateTo = dateFrom.AddSeconds(100000000.0)
let yearsSince dateFrom dateTo =
let intRepresentation (date: DateTime) =
date.ToString "yyyy.MMdd" |> Convert.ToDouble
let dateToNum = intRepresentation dateTo
let dateFromNum = intRepresentation dateFrom
int (dateToNum - dateFromNum)
yearsSince dateFrom dateTo |> Dump
let dob = DateTime(1985, 4, 16)
let calculateAge = yearsSince dob
calculateAge DateTime.Today |> Dump
Please note that this is quite naive: it doesn't take timezones or historical timezone changes into account beyond those that are already handled by .NET's DateTime class. The actual grunt work is being done by the DateTime.AddSeconds method. Hope this helps.
回答7:
I think that this will work for you:
function foo(days):
count = days
year = 0
while (count > 0):
if leap_year(year)
count = count - 366
else
count = count - 365
year ++
return year
来源：https://stackoverflow.com/questions/2144845/need-a-formula-extracting-years-from-seconds-since-january-1-0001-1200-am