Find values in 3D matrix with given indices

Hello everyone,

I'm trying to modify my matrix but I'm a little stuck. Maybe someone has an idea how it could be solved in a less complicated way :)

Basically, I have a 3D data set whereis the first two dimension represent coordinates, longitude and latitude and the third one is the time. The data is given for a set of coordinates in a range where longitude is from -180:180 and the latitude from 20:90. Now I want to change the data so it is not given for all coordinates but just section where the latitude is from 77:81 and the longitude is 0 with +-1 entry in the matrix. In the end my data should have the dimensions 5x3x48.

I tried to do it in a way that I find the indexes where in the grid data these values take place, but i don't know how do i search and define my data vector with these given indices.

Maybe someone has an Idea, I would really appreciate it.

Thank you for your time :)

Bianka

%3D data
data=rand(71,361,48);
%Longitude and latitude data
lon_oras=(-180:180);
lat_oras=(20:90);
%define longitude section
lon=find(lon_oras==0);
%for more robustness
po=lon+1;
ne=lon-1;
%index fof longitude 
idx_lon=[ne lon po];

%define latitude section 
lat_range=77:81;
%find the index where the matrix take the numbers written in lat_range
idx_lat = find(lat_oras>=min(lat_range) & lat_oras<=max(lat_range));

Answer 1

[Scott MacKenzie]

Unless I'm missing something, this is just a simple matter of finding the indices corresponding to the latitude 77:81 elements and the longitude -1:1 elements. Try this:

data=rand(71,361,48); % lat (20:90), long (-180:180), time

% find indices to extract subset of data
latIdx = (77:81) - 19; % remap so 20 is 1, 21 is 2, and so on
lonIdx = (-1:1) + 181; % remap so -180 is 1, -179 is 2, and so on

datanew = data(latIdx,lonIdx,:);
whos

  Name          Size                  Bytes  Class     Attributes

  data         71x361x48            9842304  double              
  datanew       5x3x48                 5760  double              
  latIdx        1x5                      40  double              
  lonIdx        1x3                      24  double              

[Bianka Markovic]

thank you very much sir! It works now

[darova]