Averaging z values for xy gridded data

I have a dataset of lat lon areas and corresponding values. This is a small sample:

   maxlat    minlat   minlon    maxlon    temp 
   60.0000   50.0000  170.0000  180.0000  -16.1680
   60.0000   50.0000  170.0000  180.0000   -8.6020
   70.0000   60.0000  170.0000  180.0000  -52.5874
   70.0000   60.0000  170.0000  180.0000  -54.0840
   70.0000   60.0000  170.0000  180.0000  -53.5014
         0  -10.0000  170.0000  180.0000   19.6848
         0  -10.0000  170.0000  180.0000   22.4549
   10.0000         0  170.0000  180.0000  -56.9327
   10.0000         0  170.0000  180.0000  -65.2633

I don't know how to average the temp values for each grid area where values of lat and lon are equal.

Answer 1

[Shounak Shastri]

Hello Ara,

So the brute force implementation would be to run a search through your data and check the lat and lon values. If they are equal, then copy the temp values in a separate array. Once you have gone through the completre data, you can use mean() to find the average. This is a lenghty process though and might take some time to complete.

An easier way to do this would be to use find() to get the indices of all the values of maxlat which are same, then check if those rows are equal.

for example,

k = find(maxlat == maxlat(1)); %Generate a vector of rows with same values of maxlat
%say k = [1, 2]
%the name of your table is "table"
isequal (table(k(1), :), table(k(2), :))% Check if all the coloums are equal

if the answer is true, then average, otherwise skip.

Good Luck!!

Answer 2

[KSSV]

A = [60.0000   50.0000  170.0000  180.0000  -16.1680
   60.0000   50.0000  170.0000  180.0000   -8.6020
   70.0000   60.0000  170.0000  180.0000  -52.5874
   70.0000   60.0000  170.0000  180.0000  -54.0840
   70.0000   60.0000  170.0000  180.0000  -53.5014
         0  -10.0000  170.0000  180.0000   19.6848
         0  -10.0000  170.0000  180.0000   22.4549
   10.0000         0  170.0000  180.0000  -56.9327
   10.0000         0  170.0000  180.0000  -65.2633];
   
   
   temp = A(:,end) ; 
   lat = A(:,1) ; 
   
   [c,ia,ib] = unique(lat) ; 
   temp_avg = zeros(length(c),1) ;
  for i = 1:length(c)
    temp_avg(i) = mean(temp(ib==i)) ; 
  end