column to multiple row conversion

Hi,

I wanted to convert a column vector in to multiple rows. The target is i have column of 1 year precipitation data (365,1) and i wonted to convert this in to 12*31 (month by days) matrix including leap years.

Who can can help me with this?

With kind regards,

Tesfalem,

[David Hill]

You will either need to add some zeros to the months that don't have 31 days to keep it in a matrix, or use a cell array. For the case using a cell array:

y=cumsum([0 31 28 31 30 31 30 31 31 30 31 30 31]);
ly=cumsum([0 31 29 31 30 31 30 31 31 30 31 30 31]);
if length(data)==365
    for k=1:12
        m{k}=data(y(k)+1:y(k+1))';
    end
else
    for k=1:12
        m{k}=data(ly(k)+1:ly(k+1))';
    end
end

[TESFALEM ALDADA]

Thank you, for the information

But, the above script doesn't convert it into 12*31 matrix of a year. I tried it using 365 days of column matrix...whats next i have to do?

Best

[David Hill]

The above is for a cell array. If you want a matrix you are going to have to pad the months that have less than 31 days with some value (zero maybe or nan). What are you trying to do? Why would a cell array not work?

[TESFALEM ALDADA]

Okey lets consider I have 31 days for all months (teplacing it by nan), then how would look like the new method to convert it in to row for one year,and also how can i apply it for long years.

Thank you!

Answer 1

[per isakson]

Try

col = rand(366,1); % sample data
mat = col2mat( col, 2020 );

and

col = rand(365,1); % sample data
mat = col2mat( col );

where

function    mat = col2mat( col, year )
    
    narginchk(1,2)
    if nargin==1
        if numel(col)==365
            year = 2001;    % not a leap year  
        elseif numel(col)==366                          % Fixed error; replaced 365 by 366
            year = 2000;    % a leap year
        else
            error( 'Wrong length of col, %d.', numel(col) )
        end
    end
    
    mat = nan( 31, 12 );        % pad with NaNs
    len = eomday( year, 1:12 );
    
    ix1 = [ 1, cumsum(len(1:11))+1 ];   % first day of month
    ix2 = [ ix1(2:end)-1, sum(len) ];   % last day of month
    
    for jj = 1 : 12
        mat( 1:(ix2(jj)-ix1(jj)+1), jj ) = col(ix1(jj):ix2(jj)); 
    end
end

[TESFALEM ALDADA]

Ja, this pefectly works for one year data.

Then what could be changed if i am applying thus for long years?

[per isakson]

I fixed a copy&paste error in the elseif-statement, see the answer.

[TESFALEM ALDADA]

ok