Convert hourly data to daily data in Matlab - matlab

Question

We have two matrices. Name one of them "Date", And another name is "Data"
There are several columns in the Date matrix included:
year month day julusi hour
1951 1 1 1 0
1951 1 1 1 3
1951 1 1 1 6
1951 1 1 1 9
1951 1 1 1 12
1951 1 1 1 15
1951 1 1 1 18
1951 1 1 1 21
1951 1 2 2 0
1951 1 2 2 3
1951 1 2 2 6
1951 1 2 2 9
1951 1 2 2 12
1951 1 2 2 15
1951 1 2 2 18
1951 1 2 2 21
.... . . . .
.... . . . .
1951 12 30 364 0
1951 12 30 364 3
1951 12 30 364 6
1951 12 30 364 9
1951 12 30 364 12
1951 12 30 364 15
1951 12 30 364 18
1951 12 30 364 21
1951 12 31 365 0
1951 12 31 365 3
1951 12 31 365 6
1951 12 31 365 9
1951 12 31 365 12
1951 12 31 365 15
1951 12 31 365 18
1951 12 31 365 21
.... .. . .. .
2018 12 31 365 0
2018 12 31 365 3
2018 12 31 365 6
2018 12 31 365 9
2018 12 31 365 12
2018 12 31 365 15
2018 12 31 365 18
2018 12 31 365 21
In my Data matrix, there are 410 columns(198696*410).The size of my Date matrices is equal. "198696*1". I want to convert the "Data Matrix on basis the Date Matrix to daily data
I use the following code
N=0;
for year=1951:2018;
for Juliusi=1:365;
cxa=(Date(:,4)==Juliusi);
cxb=(Date(:,1)==year);
a=cxa & cxb;
N=N+1;
dayy(N,:)=nanmean(Data(a,:));
end;end;
The conversion result is correct, but the size of the matrix is not the same
198696/8=24837 is correct but my matrix 24820 is incorrect
Where is the problem?
What to do to consider leap days?

Answer 1

Since I recently learned from Luis Mendo, that convolution is the key to success, I came up with the following idea: If your data is complete, i.e. you can guarantee, that there are always 8 entries for each day, you can just simply use the following approach:
% Some test data.
Date = [
1951 1 1 1 0;
1951 1 1 1 3;
1951 1 1 1 6;
1951 1 1 1 9;
1951 1 1 1 12;
1951 1 1 1 15;
1951 1 1 1 18;
1951 1 1 1 21;
1952 1 2 2 0;
1952 1 2 2 3;
1952 1 2 2 6;
1952 1 2 2 9;
1952 1 2 2 12;
1952 1 2 2 15;
1952 1 2 2 18;
1952 1 2 2 21]
% Temporary result for convolution.
temp = conv2(Date, ones(8, 1)) / 8;
% Extract values of interest.
dayy = temp(8:8:end, :)
Output:
Date =
1951 1 1 1 0
1951 1 1 1 3
1951 1 1 1 6
1951 1 1 1 9
1951 1 1 1 12
1951 1 1 1 15
1951 1 1 1 18
1951 1 1 1 21
1952 1 2 2 0
1952 1 2 2 3
1952 1 2 2 6
1952 1 2 2 9
1952 1 2 2 12
1952 1 2 2 15
1952 1 2 2 18
1952 1 2 2 21
dayy =
1951.0000 1.0000 1.0000 1.0000 10.5000
1952.0000 1.0000 2.0000 2.0000 10.5000
If you need the year and day information, then these could be obtained separately. But in your original post, these information seemed to be unneeded.
Just to be sure: I DO know, I used the Date matrix in my example. But since, Date follows the same format as Data, and you can easily verify the results of the wanted mean operation, I used it as an example.