Generating days based on leap years

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Damith 2014-9-15

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评论： Star Strider 2014-9-19

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Hi,

I wanted to generate a matrix like below. Basically, years in the first column, start days, end days for the year by conditional check of first column. Fourth row is basically MOD function to check if a leap year or not.

1  366  366
367  731  365
732  1096  365
1097  1461  365
1462  1827  366
1828  2192  365
2193  2557  365
 .     .       .       .
2014

Does anybody have an idea?

Thanks in advance.

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Star Strider 2014-9-16

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There may be more efficient approaches, but this works:

yr = [1960:2014]';
lpyr = (mod(yr,4)==0);
dpy = ones(size(yr))*365+lpyr;
csd = cumsum(dpy);
M = [yr [1; csd(1:end-1)+1]  csd dpy];

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Star Strider 2014-9-17

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If the same format holds, and if I understand your new objective correctly, my code should work regardless. If you are including other centuries, Honglei’s comment applies: years divisible by 100 are not leap years unless they are also divisible by 400. So 1900 was not a leap year, but 2000 was.

This should work for those exceptions as well:

yr = [1960:2014]';
cyr = mod(yr,100)==0;                   % Century Years
clpyr = mod(yr,400)==0;                 % Century Leap Year
lpyr = (mod(yr,4)==0) + (clpyr - cyr);  % Leap Years  Vector
dpy = ones(size(yr))*365+lpyr;          % Create ‘Days/Year’ + Leap Years Vector
csd = cumsum(dpy);                      % Sum ‘Days/Year’
M = [yr [1; csd(1:end-1)+1]  csd dpy]

The construction ‘+(clpyr - cyr)’ removes century years but adds years divisible by 400 to the leap year vector. I tested it with various years and it works for them.

I don’t know what you mean by ‘dynamically create the array’. (My code works only for single years and for vectors of consecutive years.) You could likely wrap it in a function m-file if you want to. The input argument is the year vector ‘yr’, and the output is the matrix ‘M’.

Damith 2014-9-19

data_arrange.xlsx

Thanks again but this is not the result I want. The output matrix needs to have 18263 rows and 200 columns. Please see the attached excel file. Number of rows changes from 18262 to 18263 alternatively for 50 year segments from 1998 to 10000 years.

Star Strider 2014-9-19

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Your Excel sheet doesn’t exactly match your description, but this seems to do what you want:

yr1 = 1998;
epok = 50;
Mprev = [0  0];
for k1 = 1:epok
    yr = yr1:yr1+49;
    cyr = mod(yr,100)==0;                   % Century Years
    clpyr = mod(yr,400)==0;                 % Century Leap Year
    lpyr = (mod(yr,4)==0) + (clpyr - cyr);  % Leap Years  Vector
    dpy = ones(size(yr))*365+lpyr;          % Create ‘Days/Year’ + Leap Years Vector
    csd = cumsum(dpy);                      % Sum ‘Days/Year’
    M = [[0 csd(1:end-1)]'+1  csd'];
    Mepok(k1,:) = sum([Mprev; [M(1,1) M(end,2)]],1);
    Mprev = [Mepok(max([1 k1]),2) Mepok(end,2)];
    Mout(k1,:) = [k1 Mepok(k1,:) (yr+k1-1):50:(yr+k1-1)+200*(epok-1)];
end
Mout                                        % Display Output Matrix

I tweaked my previous code, and created a new output matrix ‘Mout’.

I would never have understood what you want without the spreadsheet.

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