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decimal day

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ricco
ricco 2011-11-4
回答： Moyo Ajayi 2019-10-24
Any ideas of how to convert year, day of year(1:365), and time (i.e. 17:59 is given as 1759) to decimal day.
I've already changed the time into hours and minutes so now have: year; day(1:365); hour(1:24); min(1:60) But how would i go about converting this to decimal day?
thanks

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回答（5 个）

Walter Roberson
Walter Roberson 2011-11-4
Not taking leap-years and leap seconds in to account:
(year * 365 + day) + (hour / 24 + minute / (24*60))
Fangjun Jiang
Fangjun Jiang 2011-11-4
Year=2011;
Day=250;
Hour=23;
Minute=35;
D=datenum(Year,1,1)+Day-1;
[Year,Month,Day]=datevec(D);
D=datenum(Year,Month,Day,Hour,Minute,0)
  1 个评论
Walter Roberson
Walter Roberson 2011-11-4
This is probably more reliable than the version I gave.

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Andrei Bobrov
Andrei Bobrov 2011-11-4
for array d
d =[ 2002 190 17 16
2003 345 7 14
2004 233 17 41
2003 350 17 51
2007 88 2 21];
out = datenum([d(:,1), zeros(size(d,1),2)])+d(:,2)+sum(bsxfun(@rdivide,d(:,3:end),[24 60]),2);
ricco
ricco 2011-11-4
All of these seem to work i.e. they give the same answer, but i though that decimal day would be given in the form of day 1.23344 etc, these values are in the range of 7.325572215277777e+05? what does this refer to?
thanks
  1 个评论
Fangjun Jiang
Fangjun Jiang 2011-11-4
That is the number of days since Jan 0, year 0000. See the help document of datenum, datestr, datevec. Also, try the following
datestr(0)
datestr(1)
now
datestr(now)

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Moyo Ajayi
Moyo Ajayi 2019-10-24
A simple way to do this is by using dec_doy. You will not have to worry about leap year and instead. I am still working on formatting the repository, which will have more useful functions. However, for now, it should be helpful for your problem. All you need to do before hand is convert your dates into datetime objects by using the datetime() function.
Example:
>> dt = datetime(2019, 10, 23, 21, 50, 05, 123);
>> decimal_doy = dec_doy(dt)
decimal_doy =
296.9098

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