Generating days based on leap years

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.

1960  1  366  366
1961  367  731  365
1962  732  1096  365
1963  1097  1461  365
1964  1462  1827  366
1965  1828  2192  365
1966  2193  2557  365
 .     .       .       .
2014

Does anybody have an idea?

Thanks in advance.

Answer 1

[Star Strider]

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];

[Damith]

Thanks a lot.

[Star Strider]

My pleasure!

[Honglei Chen]

Just to add one little remark, the algorithm definitely works but the leap year computation is a little more than an integer multiple of 4. It also cannot be an integer multiple of 100 yet an integer multiple of 400 is ok.

[Star Strider]

@Honglei — Correct. I considered that. We were lucky in this series in that 2000 was a leap year, so I didn’t include that logic here.

[Damith]

Thanks guys for valuable comments.

The new chanlenge I facing at the moment is to dynamically create the array for varying number of years (start year and end year are different)

Any ideas?

[Star Strider]

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]

Thanks again for the revised code. But now the challenge is my start year and end year changes in "yr" array for 123 files.

[Star Strider]

My pleasure!

The start and end years shouldn’t be a problem providing they are either (1) single years, or (2) an array of consecutive years. I tested it on a variety of year vectors to be certain it worked for various century years. Obviously, it only applies to the Gregorian Calendar, but that applies for the last few centuries, so you should be OK.

If you are having a problem, tell me what it is and I’ll do my best to adapt my code to it. Please be a specific as possible.

[Damith]

Thanks. I have prioratize this from the prevous challenge.

I have 50 years slots of 10,000 years of data (suppose starting from 1998) meaning 200 rows.

1      1  18262  (1998-2047)
2  18263  36524  (2048-2097)
.     .       .       . 
200

I have 'output" array has 20,000 years of daily data meaning output(7305000 x 8) array has 7305000 rows and 8 columns. I need to read 5th column of first 10,000 years of "output" meaning row 1 to 3652500 and store in "output2" matrix which has rows equivalent to 50 years (considering the leap years). So, the "output2" matrix has 18262 rows and 200 columns.

Hope you understand my question.

[Star Strider]

I am not certain I do understand what you want to do, but this revision of my previous code should work:

yr1 = 1998;
epok = 200;
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)];
end

The ‘epok’ variable is the number of 50-year segments (epochs) you want data for, and the ‘Mepok’ matrix is the output in the format you outlined in your previous Comment. I just incorporated my previous code and added the ‘Mepok’ matrix. I tested it on epok=5, but it should work on all 200.

[Damith]

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]

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.