How do you convert Month/Day to Julian Day?

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Tom W
Tom W 2014-7-29
回答： Steven Lord 2016-11-10
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)

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采纳的回答

Tom W
Tom W 2016-11-10
I figured it out. Had to use:
julianday = hms2h(hr,min,sec) + day;
Can check by creating a matrix
greg2(julianday-1, year);

更多回答（2 个）

dpb
dpb 2014-7-29
>> 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
Steven Lord
Steven Lord 2016-11-10
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.

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