Dealing with leap years, creating arrays of yearly data

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SMA
SMA am 12 Apr. 2016
Kommentiert: dpb am 13 Aug. 2017
I have a time series in a Nx1 array where N are the number of days (different for each dataset, but lets take days from 1951 till 2007). Leap years are included (1952, 1956, ...). I am trying to convert the array into 366x57 (57 being the number of years between 1951-2007), by adding Nan at the Feb 29th (60th) position for non-leap years. I am sure there is a simple solution which I am having a hard time devising. Is there a way to accomplish this without loops.

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the cyclist
the cyclist am 12 Apr. 2016
Here is an example using two years, one leap and one not. Should be easy for you to see how to generalize.
% Time series with one leap year and one not
T = rand(365+366,1);
% The years
Y = 1951:1952;
numberYears = numel(Y);
% Slick way to identify the leap years
isLeapYear = datenum(Y,2,29)~=datenum(Y,3,1);
T_new = nan(366,numberYears);
for ny = 1:numberYears
numberDaysThisYear = 365+isLeapYear(ny);
T_new(1:numberDaysThisYear,ny) = T(1:numberDaysThisYear);
T(1:numberDaysThisYear) = []; % Deletes T as you go. Could do this differently
end
  3 Kommentare
the cyclist
the cyclist am 12 Apr. 2016
I think I got the alignment right here, but you should double-check:
% Time series with one leap year and one not
T = rand(365+366,1);
% The years
Y = 1951:1952;
% Slick way to identify the leap years
isLeapYear = datenum(Y,2,29)~=datenum(Y,3,1);
numberYears = numel(Y);
T_new = nan(366,numberYears);
for ny = 1:numberYears
numberDaysThisYear = 365+isLeapYear(ny);
if isLeapYear(ny)
T_new(1:366,ny) = T(1:366);
else
T_new(1:59, ny) = T(1:59);
T_new(61:366,ny) = T(60:365);
end
T(1:numberDaysThisYear) = []; % Deletes T as you go. Could do this differently.
end
dpb
dpb am 13 Aug. 2017
"% Slick way to identify the leap years"
My favorite is one of my standard utilities...
function is=isleapyr(yr)
% returns T for input year being a leapyear
is=eomday(yr,2)==29;

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