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Converting Day of Year to month and day

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Hamza
Hamza 2012-9-25
评论： michele paoletti 2019-6-14
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!!
  1 个评论
Star Strider
Star Strider 2012-9-25
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.

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采纳的回答

Andrei Bobrov
Andrei Bobrov 2012-9-25
编辑：Andrei Bobrov 2012-9-25
% 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
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Hamza
Hamza 2012-9-25
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
Hamza 2012-9-25
I tried
str= datestr(datenum([yy mm dd HH MM SS]),0)
and works good

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更多回答（2 个）

Jan
Jan 2012-9-25
编辑：Jan 2012-9-25
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]
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Hamza
Hamza 2012-9-25
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
Jan 2012-9-25
编辑：Jan 2012-9-25
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]

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michele paoletti
michele paoletti 2019-6-14
I need convert from year and day of year to day and month. How can i do?

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