decimal day

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

Answer 1

[Walter Roberson]

Not taking leap-years and leap seconds in to account:

(year * 365 + day) + (hour / 24 + minute / (24*60))

Answer 2

[Fangjun Jiang]

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)

[Walter Roberson]

This is probably more reliable than the version I gave.

Answer 3

[Andrei Bobrov]

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);

Answer 4

[ricco]

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

[Fangjun Jiang]

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)

Answer 5

[Moyo Ajayi]

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