Dealing with leap years, creating arrays of yearly data

조회 수: 3 (최근 30일)
SMA
SMA 2016년 4월 12일
댓글: dpb 2017년 8월 13일
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.

채택된 답변

the cyclist
the cyclist 2016년 4월 12일
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
  댓글 수: 3
the cyclist
the cyclist 2016년 4월 12일
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 2017년 8월 13일
"% 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;

댓글을 달려면 로그인하십시오.

추가 답변 (0개)

카테고리

Help CenterFile Exchange에서 Dates and Time에 대해 자세히 알아보기

태그

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by