How to convert dates into Julian dates?

I have read the dates in yyyymmdd format as follows;

dates = 20230409 20230414 20230504 20230514 20230519 20230608 20230613 20230723 20230817 20230827 20230901 20231006 20231011 20231026 20231115 20231125 20231205 20231210 20240129

I want to convert these dates into julian dates. Please suggest me how to do it?

I will appreciate your kind cooperation.

Deva

[Stephen23]

Make sure that you import the data as DATETIME, then use:

https://www.mathworks.com/help/matlab/ref/datetime.juliandate.html

Do not import dates as scalar numerics whose digits just happen to look like those of some unrelated date units.

Answer 1

[the cyclist]

Assuming your dates are currently stored as a numeric array, then

dates = [20230409    20230414    20230504    20230514    20230519    20230608    20230613    20230723    20230817      20230827    20230901    20231006    20231011    20231026    20231115    20231125    20231205    20231210     20240129];

gdt = datetime(dates,"ConvertFrom","yyyymmdd") % Gregorian

gdt = 1x19 datetime array
Columns 1 through 13

   09-Apr-2023   14-Apr-2023   04-May-2023   14-May-2023   19-May-2023   08-Jun-2023   13-Jun-2023   23-Jul-2023   17-Aug-2023   27-Aug-2023   01-Sep-2023   06-Oct-2023   11-Oct-2023

Columns 14 through 19

   26-Oct-2023   15-Nov-2023   25-Nov-2023   05-Dec-2023   10-Dec-2023   29-Jan-2024

jdt = juliandate(gdt) % Julian

jdt = 1x19

1.0e+06 * 2.4600 2.4600 2.4601 2.4601 2.4601 2.4601 2.4601 2.4601 2.4602 2.4602 2.4602 2.4602 2.4602 2.4602 2.4603 2.4603 2.4603 2.4603 2.4603

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See the documentation for juliandate for details on the meaning of that output.

[Devendra]

thank you very much for your valuable suggestions. It worked well.

First, How these julian dates can be converted into julian dates something like 273,287,298,etc?

Second, I want to store these julian dates into an array consisting already 10 real number variables than only julian dates are stored while the values of all other 10 variables become zero. Why it is so? Is it due to some format issue ? Please suggest me how to store all the variables including julian dates into array.

Deva

[Steven Lord]

all other 10 variables becomes zero.

Do those other elements become zero or are they simply displayed as zero because they're much, much smaller than the Julian dates?

J = juliandate(datetime('today'))

J = 2.4604e+06

x = 1e-4

x = 1.0000e-04

V = [J, x] % V(2) _displayed_ as 0

V = 1x2

1.0e+06 * 2.4604 0.0000

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y = V(2) % but it's not actually _stored_ as 0

y = 1.0000e-04

If so, changing the display format may be sufficient for your needs.

format shortg
V

V = 1x2

1.0e+00 * 2.4604e+06 0.0001

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format longg
V

V = 1x2

1.0e+00 * 2460408.5 0.0001

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[Devendra]

Thanks for your kind help. Yes they are displayed as zero because they're much, much smaller than the Julian dates? Yes it displayed all the variables. I appreciate your kind support.

Deva

Answer 2

[James Tursa]

Based on your comment that you are expecting numbers in the range of 273,287,298, etc., if what you are after is actually "day of year" and not "Julian Date", then

dates = [20230409    20230414    20230504    20230514    20230519    20230608    20230613    20230723    20230817      20230827    20230901    20231006    20231011    20231026    20231115    20231125    20231205    20231210     20240129];
gdt = datetime(dates,"ConvertFrom","yyyymmdd") % Gregorian

gdt = 1x19 datetime array
Columns 1 through 13

   09-Apr-2023   14-Apr-2023   04-May-2023   14-May-2023   19-May-2023   08-Jun-2023   13-Jun-2023   23-Jul-2023   17-Aug-2023   27-Aug-2023   01-Sep-2023   06-Oct-2023   11-Oct-2023

Columns 14 through 19

   26-Oct-2023   15-Nov-2023   25-Nov-2023   05-Dec-2023   10-Dec-2023   29-Jan-2024

days(gdt - datetime(year(gdt),1,1)) + 1 % day of year, with Jan 1 being day number 1

ans = 1x19

99 104 124 134 139 159 164 204 229 239 244 279 284 299 319 329 339 344 29

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I am aware that some fields of study refer to "day of year" as "Julian Date", but this is incorrect terminology.

[Devendra]

Thank you very much that is what I was looking for. I am very much grateful to you for your kind help.

Deva

[Steven Lord]

You don't need to write your last expression. Just ask the day function for the 'dayofyear'.

dates = [20230409    20230414    20230504    20230514    20230519    ...
    20230608    20230613    20230723    20230817      20230827    ...
    20230901    20231006    20231011    20231026    20231115    ...
    20231125    20231205    20231210     20240129];
gdt = datetime(dates,"ConvertFrom","yyyymmdd"); % Gregorian
day(gdt, 'dayofyear')

ans = 1x19

99 104 124 134 139 159 164 204 229 239 244 279 284 299 319 329 339 344 29

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[Devendra]

Thank you very much for your kind help.

Deva

[James Tursa]

@Steven Lord I thought I remembered there was a function for this but couldn't find it in datetime methods. Didn't think to look in day( ) doc. Thanks for the update.