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,
4 个评论
David Hill
2020-3-3
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
2020-3-3
编辑：per isakson
2020-3-3
David Hill
2020-3-3
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
2020-3-4
采纳的回答
per isakson
2020-3-4
编辑：per isakson
2020-3-5
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 个评论
TESFALEM ALDADA
2020-3-4
per isakson
2020-3-5
I fixed a copy&paste error in the elseif-statement, see the answer.
TESFALEM ALDADA
2020-3-5
编辑：per isakson
2020-3-5
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