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decimal day

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ricco
ricco on 4 Nov 2011
Answered: Moyo Ajayi on 24 Oct 2019
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

Answers (5)

Walter Roberson
Walter Roberson on 4 Nov 2011
Not taking leap-years and leap seconds in to account:
(year * 365 + day) + (hour / 24 + minute / (24*60))

Fangjun Jiang
Fangjun Jiang on 4 Nov 2011
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 Comment
Walter Roberson
Walter Roberson on 4 Nov 2011
This is probably more reliable than the version I gave.

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Andrei Bobrov
Andrei Bobrov on 4 Nov 2011
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 on 4 Nov 2011
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 Comment
Fangjun Jiang
Fangjun Jiang on 4 Nov 2011
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 on 24 Oct 2019
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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