I want to convert the Julian date 273.791667 to calendar day and time

I want to convert the Julian date 273.791667 to calandar day and time

[James Tursa]

As a nit, that number should be referred to as "day of year" and not "Julian date". I know there are segments of the scientific community that call it a "Julian date", but that usage is confusing at best. E.g., the definition of the term from the USNO:

http://aa.usno.navy.mil/data/docs/JulianDate.php

You will find variations of this term in the literature depending on the underlying time scale (e.g., UT or TT) and 0 point, but even these are not "day of the year" types of measures.

Answer 1

[Chad Greene]

You'll have to specify a year, but for 2017 you can get a date vector like this:

datevec(datenum(2017,0, 273.791667))

or a date string like this:

datestr(datenum(2017,0, 273.791667))

[Thishan Dharshana Karandana Gamalathge]

Just wanted to make sure... The answer from the code provides the local time? Does it take in to account daylight saving time?

[Chad Greene]

Daylight savings time varies by region, so you'd have to perform that conversion yourself. Something like "if the julian day is between this range, add or subtract 1/24"

[Steven Lord]

FYI datetime lets you specify the TimeZone for your dates. You can also specify 'local' if you want to use your system's time zone.

>> x = datetime('today'); % No time zone specified
>> x.TimeZone % so x is unzoned
ans =
  0×0 empty char array
>> isdst(x)
ans =
  logical
   0

>> x = datetime('today', 'TimeZone', 'local');
>> x.TimeZone % Our Natick MA HQ is in New York's time zone
ans =
    'America/New_York'
>> isdst(x) % we are in Daylight Savings Time right now
ans =
  logical
   1

[Chad Greene]

Very cool!

[Thishan Dharshana Karandana Gamalathge]

Thanks.

Now when I import the local time data (converted from the Julian date), it is restricted to only two decimals. How can I have it for 6 decimals?

[Walter Roberson]

If you created a datetime object, then you can set its Format property

[Thishan Dharshana Karandana Gamalathge]

From a .mat file "X" with size(10x8), I am trying to calculate Z. But the following code gives an error. i cannot figure out the reason. Please help me with that. Thanks.

load X

for i=1:10
        Y(i)=X(i,1);
    UTC_start_time(i)=datevec(datenum(2010,0,Y(i,1)));
end

[James Tursa]

datevec returns a vector. You cannot stuff a vector into a single element.

Answer 2

[Steven Lord]

Use datetime. You can 'ConvertFrom' 'juliandate' (Number of days since noon UTC 24-Nov-4714 BCE (proleptic Gregorian calendar)) or 'modifiedjuliandate' (Number of days since midnight UTC 17-Nov-1858).

If you want to determine what the 273.791667th day of the year is, try something like this. Check the day of the year using the day function with the 'dayofyear' option:

x = datetime(2017, 1, 0) + days(273.791667)
day(x, 'dayofyear')

[Walter Roberson]

If you want to specify that this was UTC, you can use

   x = datetime(2017, 1, 0, 'TimeZone', 'UTC') + days(273.791667)

and if you then want to see it in local time,

x.TimeZone = 'local'

Note that this is not the same as

   x = datetime(2017, 1, 0, 'TimeZone', 'local') + days(273.791667)

In order for the timezone conversion to be done properly, you need to set to UTC first and then change the timezone; if you just set the timezone directly like in this second example then MATLAB will assume that you mean that the given date/time information is being stated relative to your timezone.

[Thishan Dharshana Karandana Gamalathge]

Now when I import the local time data (converted from the Julian date), it is restricted to only two decimals. How can I have it for 6 decimals?