Convert date to specific format julian date

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Charlie Wood on 3 Oct 2016
Edited: Charlie Wood on 4 Oct 2016
Need to convert calendar date to Julian date in a specific format (YYDDD). Ex: Day = (2016-05-04,2016-08-24,2016-12-31) J-Day = (16125,16237,16366)
Any help would be appreciated. Thanks.

Gareth Thomas on 3 Oct 2016
Datetime works nicely. I believe it came out in R2014b.
a = datetime('2016-05-04');
a.Format='yyddd'
Charlie Wood on 4 Oct 2016
Edited: Charlie Wood on 4 Oct 2016
Peter/Gareth, this is perfect. Switching the ddd to DDD was the fix needed in the a.Format.
>> a = datetime('2016-05-04');
>> a.Format='yyDDD'
a =
16125
This works for any year as well which is great. This will be new accepted answer vice the numerical version I put up.
Thanks a gain for the help!

Steven Lord on 3 Oct 2016
Create a datetime array from the char vectors containing the yyyy-MM-dd form of the dates. Use the juliandate function to compute the Julian date from the datetime array.
Charlie Wood on 3 Oct 2016
Edited: Charlie Wood on 3 Oct 2016
d = '2016-12-31';
dt = datetime(d)
dt = 31-Dec-2016
j = juliandate(dt)
j = 2.4578e+06
How do I get j into the yyddd format (to read 16366)?
Thanks.

Michael C. on 3 Oct 2016
My understanding was that Julian usually referred to the number of days since Jan 1 4713 BC, but judging by your example, you are using March 11th, 1972.
Either way, you can get number of days by doing a subtraction on two "datenum"s
datenum([2016 08 24 0 0 0]) - datenum([1972 3 11 0 0 0])
which gives the 16237. Passing in a string for the date will also work.
Charlie Wood on 3 Oct 2016
This worked pretty effectively. Unfortunately if I have a date from 2015 or 2014 (which I do), it doesn't work as well. I guess I can put an if statement in there to switch the second datenum to the right value. Thanks so much Michael.

Charlie Wood on 3 Oct 2016
Here's what I got combining some answers:
d = '2016-05-04';
dt = datime(d);
jday = (year(dt)-2000)*1000 + day(dt,'dayofyear')
The answer comes out with 16125. It's not pretty and not useful for dates before the year 2010 (maybe with some minor formatting it would be ok for 2000-2009). It does however work for my 2014-2016 dates so far.
Thanks to everyone for the help and input!!!

Ankit Kanthe on 3 Oct 2016
Here's what I got combining some answers:
d = '2016-05-04'; dt = datime(d); jday = (year(dt)-2000)*1000 + day(dt,'dayofyear') The answer comes out with 16125. It's not pretty and not useful for dates before the year 2010 (maybe with some minor formatting it would be ok for 2000-2009). It does however work for my 2014-2016 dates so far.