using datenum over several years

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Richard
Richard 2012 年 3 月 22 日
Say if I want to find the decimal day of year I would use:
clear all
StartDate = '2011-01-01 00:00';
EndDate = '2011-12-31 23:57';
Resolution = 60;
DateTime=datestr(datenum(StartDate,'yyyy-mm-dd HH:MM'):Resolution/(60*24):...
datenum(EndDate,'yyyy-mm-dd HH:MM'),...
'yyyy-mm-dd HH:MM');
DateTime=cellstr(DateTime);
JDay = cellfun(@(x)datenum(x),DateTime,'un',0);
DecDay = cell2mat(cellfun(@(x)x - datenum(2011,0,0),JDay,'un',0));
How is this achieved if the data extends over one year? So, say that my start data remains the same but my end data changes to '2013-12-31 23:57' how would I convert my time series into decimal day of year.
  1 件のコメント
Oleg Komarov
Oleg Komarov 2012 年 3 月 22 日
Note that datenum(2011,0,0) is not '2011-01-01 00:00' but:
datestr(datenum(2011,0,0),'dd-mmm-yyyy HH:MM:SS PM')
ans =
31-Dec-2010 12:00:00 AM

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採用された回答

Oleg Komarov
Oleg Komarov 2012 年 3 月 22 日
InDate = '2011-01-01 00:00';
FiDate = '2021-01-31 23:57';
% Create hourly spaced datetimes
v = datenum(InDate):1/24:datenum(FiDate);
% Shift with respect to InDate
v - datenum(InDate)
If you need to start from 1 just add '+ 1' to the last statement

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