Dealing with leap years, creating arrays of yearly data

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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.

Accepted Answer

the cyclist
the cyclist on 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 Comments
the cyclist
the cyclist on 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 on 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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