decimal day
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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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Respuestas (5)
Walter Roberson
el 4 de Nov. de 2011
Not taking leap-years and leap seconds in to account:
(year * 365 + day) + (hour / 24 + minute / (24*60))
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Fangjun Jiang
el 4 de Nov. de 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)
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Andrei Bobrov
el 4 de Nov. de 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);
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ricco
el 4 de Nov. de 2011
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Fangjun Jiang
el 4 de Nov. de 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)
Moyo Ajayi
el 24 de Oct. de 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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