How to extract values from a 3-d .mat file?

Hi,

I loaded a netcdf file and derived a .mat file which is a 3-D matrix of lon x lat x value (360 x 40 x 111). I wanted to extract the third dimension as entire column.

lon=ncread('sstarc.nc','lon');
lat=ncread('sstarc.nc','lat');
summean=ncread('sstarc.nc','sst');

I used the above code and got the mat file. The third dimension has 111 values which I wanted to extract the "111" values for the entire 360 x 40 grid as a column (eg. time series). Attached .mat file!

Looing forward to your help. Thanks!

Answer 1

[Cris LaPierre]

Your mat file contains variables. Loading the mat file loads the variables back into the workspace.

load summean.mat

If there are multiple variables stored in the mat file, you can tell it to load specific variables instead of all of them.

load summean.mat filteredA

As for the values to extract, you can do that using indexing on the variable containing the values.

The third dimension does not have 111 values. Instead, you have 111 matrices of dimension 360 x 40 stacked on top of each other. To extract the last one, you could do the following

vals = filteredA(:,:,111);

[Keegan Carvalho]

Apologies as I may have not explained my problem correctly.

So 360 x 40 is the longitude x latitude (for context: this pretty much covers the Northern Hemisphere from latitudes 50.5 to 89.5). The 111 are different values of a variable for 111 months. So I wanted to extract the 111 values as a time series or column for the entire area of 360 x 40. I don't want to index any coordinate since I am studying the entire region.

I tried

vals=filteredA(lon,lat,:);

But got an error:"Index in position 1 is invalid. Array indices must be positive integers or logical values."

Hopeful that I could explain my problem well :)

[Cris LaPierre]

So if I understand, the row and column indices represent lat and lon, and all the values in the matrix are the values for 111 months in the corresponding location. Sound correct?

Your code for doing this is correct, so I suspect your values for lon or lat are not. You are indexing the matrix, so the indices must be postive integer values greater than 0. I suspect at least some of your lon and lat values contain decimal values. Hence, the error message.

So this works: vals=filteredA(60,10,:);

This does not: vals=filteredA(50.5,20.2,:);

If you have lon and lat values, you will want to find a way to map those to the rows and columns of filteredA.

[Cris LaPierre]

One comment - to get vals to be a vector, use the squeeze function.

vals = squeeze(filteredA(60,10,:));

[Cris LaPierre]

The more I read your reply, the more I wonder if you are picturing the data correctly. There are not 111 values in the third dimension. There are 1,598,400. That is, 111 pages of 360x40 values.

If what you want is all 1,598,400 values in a single column vector, that can be done like this:

allVals = filteredA(:);

The result is column-ordered. The first values are the first column on the first page, followed by the second column on the first page, etc until all values on the first page are added. It then adds the values in the first column of the second page, then the second column, etc. This pattern repeats until it gets to the last column on the last page.

[Keegan Carvalho]

https://drive.google.com/open?id=1eVce8POk2T_UWv34Q7Llo-_NNKr55I9z

I have attached the file. So the file is simlar 512 x 50 x 444 (just a change with longitude and latitude). These are monthly data for temperature from January 1982 to December, 2018. I decided to get summer monthly means for the time period and hence got 111 values using the following code:

lon=ncread('sst.nc','longitude');
lat=ncread('sst.nc','latitude')';
sst=ncread('sst.nc','variable');
summerfrom1982 = 1:size(sst, 3);
monthsnumber = mod(summerfrom1982 - 1, 12) + 1;
tokeep = monthsnumber >= 6 & monthsnumber <= 8;
filteredA = sst(:, :, tokeep);

The resulting file is 512 x 50 x 111. So I wanted to get the 111 values as a time series for the entire grid. Also the file has NaN values which I want to ignore as it may affect "mean", etc. calcualtions...

(I understood your reply above. I have attached the orignal file. Would be grateful if you could look into it)

[Cris LaPierre]

Ok, so there are 37 years of data. Each page corresponds to one month (37 x 12 = 444). Here's a quick visualiation of the January 1982 data. So far so good?

Here's the code I used

[LAT, LON] = meshgrid(lat,lon);
sstJan=sst(:,:,1);

geoscatter(LAT(:),LON(:),sstJan(:))

[Cris LaPierre]

The variable filteredA contains the data for Jun-Aug for all 37 years (512 x 50 x 111). Now you want this data as a time series. What does that mean to you? Knowing there are now 2,841,600 values, how do you envision this data being organized in a vector? Do you want to preserve location information? Month? Year?

I would be inclined to create a timetable, thus preseving the month and year info. I'd add variables (columns) for sst value, lon and lat.

Following your example above, this is what the code might look like.

% extract data from nc file
lon=ncread('sst.nc','longitude');
lat=ncread('sst.nc','latitude')';
sst=ncread('sst.nc','variable');

% identify summer months
startDate = datetime(1982,1,1,"Format","MMM-uuuu");
endDate = datetime(2018,12,31,"Format","MMM-uuuu");
dates = startDate:calmonths(1):endDate;
summerMonth = month(dates) >= 6 & month(dates) <= 8;

% Set up variables for creating timetable
datesTT = permute(repmat(dates,512,1,50),[1,3,2]); % create a date for each value
datesTT = datesTT(:,:,summerMonth);
[LAT, LON] = meshgrid(lat,lon);
latTT = repmat(LAT,1,1,sum(summerMonth)); % create matrix same size as datesTT
lonTT = repmat(LON,1,1,sum(summerMonth)); % create matrix same size as datesTT
sstSummer = sst(:,:,summerMonth);   % Extract data for summer months

% Create timeTable
summerTT = timetable(datesTT(:),sstSummer(:),lonTT(:),latTT(:));
summerTT.Properties.VariableNames = {'sst','lon','lat'};

Here's a preview of what the table looks like (first 8 rows).

      Time       sst        lon       lat  
    ________    ______    _______    ______

    Jun-1982    271.46          0    89.463
    Jun-1982    271.46    0.70313    89.463
    Jun-1982    271.46     1.4063    89.463
    Jun-1982    271.46     2.1094    89.463
    Jun-1982    271.46     2.8125    89.463
    Jun-1982    271.46     3.5156    89.463
    Jun-1982    271.46     4.2188    89.463
    Jun-1982    271.46     4.9219    89.463

[Cris LaPierre]

As for taking the mean when there are NaN entries, you can use the following syntax

mean(A,'omitnan')

You can see an example here.

[Keegan Carvalho]

Thank you Cris. Very grateful for the explanations. I actually have another query on the same. I am trying to create a map of seasonal means and thereby want to have 37 years of data (360 x 40 x37). I used the following code:

lon=ncread('sstarc.nc','lon');
lat=ncread('sstarc.nc','lat');
sst=ncread('sstarc.nc','sst');
DJF = sort([[1:12:444],[2:12:444],[12:12:444]]);
Wacc = cumsum(sst(:,:,DJF));
Wavg = Wacc(:,:,3:3:end)/3;

meanDJF=mean(Wavg,3); 
h=pcolor(lon,lat,meanDJF);

However I seem to get odd temperature values of 100 degrees and above at certain places. I asked this question: (https://se.mathworks.com/matlabcentral/answers/512283-how-to-extract-seasonal-means)

I know this may seem to be going too far, but I would gladly appreciate your help as I am working on a project.