Making a monthly mean
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Hi
I have data for the geopotential for every 6 hours over a whole month. I want to get a mean over the data for the month. For example for January, I have 124 timesteps, each with their data, but I only want data for the whole month (1 dataset). I have tried, but I'm not sure if I've done it right? Pleas can anyone help me?
lon = ncread(filename,'longitude') ; nx = length(lon) ;
lat = ncread(filename,'latitude') ; ny = length(lat) ;
time = ncread(filename,'time') ; nt = length(time);
zmean = zeros([nx ny]);
for n = 1:nt
z = ncread(filename,'z',[1 1 nt],[nx ny 1]);
zx(:,1:ny) = z(:,ny:-1:1);
zmean = zmean + zx;
end
zmean = zmean/nt;
Here is (lon = 360x1), (lat = 181x1), (time = 124x1 int32), (z = 360x181)
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6 个评论
Jonas Damsbo
2018-10-27
Jonas Damsbo
2018-10-27
I will take a look at your data.
What kind of format is this?
time =
124×1 int32 column vector
700512
700518
700524
You should read all your data at the same time, each month. Then you simply put it all in a timetable and use retime. No for loop required. The only problem is that I don't know how to parse your time format. I understand that it's all from the same months, but what does e.g. 700512 mean?
Jonas Damsbo
2018-10-27
Jonas Damsbo
2018-10-27
编辑：Jonas Damsbo
2018-10-27
After looking at your data I think perhaps a timetable is in fact not the way to go, as I suspect that you want to retain the gridded nature of your data. If the problem is to average the monthly data, then I think there is a very simple solution. Simply,
z = ncread(filename,'z');
z_monthly(:,:,1) = mean(z,3);
This takes the average over the third dimension (time). Then you can simply put february's data in z_monthly(:,:,2) etc... and you will end up with a lat x lon x n matrix where n is the number of months.
If the data is uniformly spaced, then I see no point in using retime. If some data is missing, however, then you'd probably want to interpolate before taking the average.
I would also suggest you take your time-vector and convert it to datetime, to make life easier for you.
t = datetime(1900,1,1,0,0,0)+hours(time);
t =
124×1 datetime array
01-Dec-1979 00:00:00
01-Dec-1979 06:00:00
01-Dec-1979 12:00:00
Did I understand the issue?
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Jonas Damsbo
2018-10-27
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2 个评论
jonas
2018-10-27
If you load the data one month at a time, then you can just average all the data along the time dimension. If you load more than one month, then you need to split the data in n segments when taking the mean (where n is the number of months).
Jonas Damsbo
2018-10-27
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