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column to multiple row conversion

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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,

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TESFALEM ALDADA
TESFALEM ALDADA on 3 Mar 2020
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
David Hill on 3 Mar 2020
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
TESFALEM ALDADA on 4 Mar 2020
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!

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Accepted Answer

per isakson
per isakson on 4 Mar 2020
Edited: per isakson on 5 Mar 2020
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

  3 Comments

TESFALEM ALDADA
TESFALEM ALDADA on 4 Mar 2020
Ja, this pefectly works for one year data.
Then what could be changed if i am applying thus for long years?
per isakson
per isakson on 5 Mar 2020
I fixed a copy&paste error in the elseif-statement, see the answer.

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