Converting Day of Year to month and day

Dear All

I have the following data:

epoch=    '96318.74847837'
      '96319.62211352'
      '96319.62351606'
      '96319.62356237'
      '96320.05952563'
      '96320.49676119'

the firs 2 elements of the data is year '96' and rest is the day of year '318.74847837'. I got the following code to separate them.

year=cellfun(@(y) y(1:2),epoch,'uni',false);
day=cellfun(@(y) y(3:5),epoch,'uni',false);

And the following code to convert day of year to date and month.

doy=318;
dayspermonth = [0 31 28 31 30 31 30 31 31 30 31 30 31];
mm = ones(size(doy))*NaN;
dd = ones(size(doy))*NaN;

for im = 1:12
   I = find(doy > sum(dayspermonth(1:im)) & doy <= sum(dayspermonth(1:im+1)));
   mm(I) = ones(size(I)).*im;
   dd(I) = doy(I) - ones(size(I))*sum(dayspermonth(1:im));
end

I would like to take the input of day into doy so that I have a single array of doy with all the results from day.

I know this code is not taking care of leap year. but if someone can help me with that I would be grateful.

Thanks!!

[Star Strider]

LEAP YEARS — To find and work with leap years, see the documentation for eomday. You can use it to create your dayspermonth vector for any given year, including correct values for leap years.

Answer 1

[Andrei Bobrov]

% Let y - our year
y = 1996;
doy = 318.25;
[yy mm dd HH MM] = datevec(datenum(y,1,doy));

variant of full solution

x=    {'96318.74847837'
      '96319.62211352'
      '96319.62351606'
      '96319.62356237'
      '96320.05952563'
      '96320.49676119'} % initial array

a = str2double(x);
y = fix(a/1000);
[yy mm dd HH MM SS] = datevec(datenum(y,0,a - y*1000)); % corrected

[Hamza]

great man! working good..though I don't need the year. I already get it by

    year=cellfun(@(y) y(1:2),epoch,'uni',false);
So i just need to convert day of year to month and date.

Any suggestion on how to convert the terms after '.' to hours and seconds? e.g: 0.74847837?

[Andrei Bobrov]

no, you need the year, this "taking care of leap year".

About hours and seconds see row "% corrected" in my answer.

[Hamza]

Thanks Andrei! in the end, how to convert your output [yy mm dd HH MM SS] into dd-mm-yy HH:MM:SS format?

[Hamza]

I tried

str= datestr(datenum([yy mm dd HH MM SS]),0)

and works good

Answer 2

[Jan]

When I understand you correctly, this is the question:

Convert "doy" to "day" and "month":
day = {'318.74847837', ...
       '319.62211352', ...
       '319.62351606', ...
       '319.62356237', ...
       '320.05952563', ...
       '320.49676119'};

Solution:

daynum = sscanf(sprintf('%s*', day{:}), '%g*');
datev  = datevec(daynum);
day    = datev(:, 3);   % [EDITED]
month  = datev(:, 2);   % [EDITED]

Now this does not consider the leap year. To do this:

daynum  = sscanf(sprintf('%s*', day{:}), '%g*');
yearnum = sscanf(sprintf('%s*', year{:}), '%g*');
yearvec = zeros(numel(yearnum), 6);
yearvec(:, 1) = yearnum;   % [EDITED], no transpose
yearvec(:, 1) = 1;  
% Perhaps also: yearvec(:,3) = 1;
yearsn  = datenum(yearvec);
datesn  = yearsn + daynum;

datev = datevec(datesn);
day   = datev(:, 3);   % [EDITED]
month = datev(:, 2);   % [EDITED]

[Hamza]

Hey Jan Thanks again. when i run this

year=cellfun(@(y) y(1:2),epoch,'uni',false);
day=cellfun(@(y) y(3:5),epoch,'uni',false);
num   = sscanf(sprintf('%s*', day{:}), '%g*');
date  = datevec(num);
day   = date(3)
month = date(2)

it gives me day = 0 and month =0

[Hamza]

datesn = yearsn + daynum; daynum undefined

Perhaps num=daynum???

[Hamza]

by the way

epoch=    '96318.74847837'
      '96319.62211352'
      '96319.62351606'
      '96319.62356237'
      '96320.05952563'
      '96320.49676119'

and

year=cellfun(@(y) y(1:2),epoch,'uni',false);
day=cellfun(@(y) y(3:5),epoch,'uni',false);

is doing the work good. I only want to convert the day to month and date.

[Jan]

@Hamza: I do not agree: The CELLFUN(@anonymous) method is slow and does not reply the values in the required numerical format. Therefore I do not think, that it does its work good here.

You cannot simply mix the different approachs. Please try the code I've posted or feel free to ignore it.

Yes, "num" should be "daynum" here. This is fixed in the code now.

Btw. Please post the data such that it can be accessed by copy&paste. It is not clear if your "epoch" is a CHAR matrix or a cell string.

[Hamza]

Apologies for not clarifying. epoch is actually a cell string. I am trying to run your approach but it's not working.

It runs if i'm taking input from:

year=cellfun(@(y) y(1:2),epoch,'uni',false);
day=cellfun(@(y) y(3:5),epoch,'uni',false);

But still the output is just '1' for both. And as you said that it won't do good if i mix two different approaches. Any suggestions where I am going wrong?

[Jan]

Yes, use CELLFUN to get "day" and "year", when you like to. Afterwards I had a bug in my code. Replace:

day    = datev(3);
month  = datev(2);

by:

day    = datev(:, 3);
month  = datev(:, 2);

See [EDITED]

Answer 3

[michele paoletti]

I need convert from year and day of year to day and month. How can i do?

[Andrei Bobrov]

Let:

year1 = 2019;
dayofyear = 241;

then:

out = datetime(2019,1,dayofyear)

[michele paoletti]

Thank you so much!