Help in averaging climate data

I am a novice in matlab and looking for help with calculating average monthly data from hourly data.
I have the data for full year in one file and have multiple txt files for different years. Column 1 is year, column 2 is month of year (1 to 12), column 3 is day of month and column 4 is hour of day (0 to 23). Looking forward to some guidance on how to do this.
Thanks in advance

5 Kommentare

Kye Taylor
Kye Taylor am 6 Jun. 2013
Have you imported the data into MATLAB? I.e. do you have the data stored in a N-by-4 dimensional matrix yet?
So, does a row of the data looks like
[1990, 2, 28, 23] % for 11pm Feb 28 1990
Where is the climate data? Is that stored in another column? or a separate vector?
Sam
Sam am 6 Jun. 2013
This is how I imported the data to matlab:
D = importdata('metdata1.txt')
D =
data: [8760x13 double]
textdata: {8760x1 cell}
rowheaders: {8760x1 cell}
Kye Taylor
Kye Taylor am 6 Jun. 2013
So, which columns of the 13 are the four you describe above, and which one is the climate data?
Sam
Sam am 6 Jun. 2013
And this is what the first row looks like if I am doing it right:
D.data(1,:)
ans =
1.0e+03 *
Columns 1 through 12
1.9730 0.0010 0.0010 0 -0.9990 -0.0999 -0.0090 -0.9999 -0.9999 -0.9999 -0.9999 -0.0990
Column 13
-0.0990

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Antworten (2)

Sam
Sam am 6 Jun. 2013

0 Stimmen

Yes, I have imported them into matlab
Roger Stafford
Roger Stafford am 6 Jun. 2013

0 Stimmen

I'll suppose you have the climate data in column five of an array, A, with the year, month, day, and hour in its columns 1, 2, 3, and 4, respectively.
[u,~,ix] = unique(A(:,1:2),'rows');
B = [u,accumarray(ix,A(:,5))./accumarray(ix,1)];

5 Kommentare

Sam
Sam am 6 Jun. 2013
Bearbeitet: Sam am 6 Jun. 2013
Hi Roger, Thanks, this seems to produce a matrix for the 12 months. But how not to take into account the "999.9" data i.e. the missing value data in the column?
Roger Stafford
Roger Stafford am 6 Jun. 2013
It should produce a row for every year-month combination, not just for 12 months.
If you can fully(!) describe the "missing value" situation, perhaps a solution could be found. As things stand, I can only guess at what you mean.
Sam
Sam am 6 Jun. 2013
Actually data for different years are in different files in the data column, a data value of "999.9" means a missing value so not to be considered in the a calculation. Now it is averaging over the entire column for unique values of month.
You could initially remove all rows with the "999.9" data values:
A = A(A(:,5)~=999.9,:);
However, it might be better to allow some tolerance for rounding differences:
A = A(abs(A(:,5)-999.9)<100*eps(999.9),:);
Noor Mohd Safar
Noor Mohd Safar am 14 Jul. 2015
Thank you guys.

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Sam
am 6 Jun. 2013

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