How do you convert Month/Day to Julian Day?

I need to convert [yyyy MMM dd (hh mm ss)] into 3-digit (decimal) julian day, but every function I try takes numbers 1-31 (for dd "day") and doesn't increment the month, so julian day output is always less than 32. I need the ability to use the "MMM" portion...

I do not have major toolboxes (i.e. one solution requires the Financial Toolbox, not viable for me)

There is a solution I have by adding monthly # of days, but I'm not sure how to account for LEAP years (when # of days in February changes).

Output should be 001-365 (with floating decimal)

Answer 1

[Tom W]

I figured it out. Had to use:

julianday = hms2h(hr,min,sec) + day;

Can check by creating a matrix

greg2(julianday-1, year);

Answer 2

[dpb]

>> dn=datenum(2008,1,[1:2:400].',0,0,0);  % make up some date nums
>> d=dn-dn(1);                            % get the day from beginning
>> [d(1:10) d(180:189)]                   % display some results
ans =
   0   358
   2   360
   4   362
   6   364
   8   366
  10   368
  12   370
  14   372
  16   374
  18   376
>>

NB: datenum handles leap years transparently--2008 was selected specifically because it was a leap year; note that day 366 did show up.

This is cumulative over years as can be observed.

A handy little utility routine is

function is=isleapyr(yr) % returns T for given year being a leapyear

is=(datenum(yr+1,1,1)-datenum(yr,1,1))==366;

Put in m-file isleapyr.m and place on matlabpath

Answer 3

[Steven Lord]

Use the day method for datetime objects.

D = datetime('today')
DOY = day(D, 'dayofyear')

Since 2016 is a leap year and today is November 10th, DOY should be (and is) 315 according to Wikipedia.

datetime objects also have a juliandate method but that's slightly different.