Writing a script for multiple arrays with a set criteria for each array and identifying certain value locations

I am working with arrays of data, where each value is a latitude and longitude grid point - columns are longitude and rows are latitude.

There are three arrays of equal size, representing the same latitude-longitude grid points over the Earth. Each array however contains a different type of data: the first, temperature; the second, wind; the third, height of the atmosphere.

I want to develop a script which identifies certain grid points according to some criteria: The process:

1. scanning the first array for if certain values are 20 units less than the surrounding grid points
2. For these locations of where minimum values are, scanning the second array to check whether the value two places above it is negative.
3. For these locations where the both the minimum values are (for the first array) and two places above is negative (for the second array), scanning the third array for where the value to the right of the location must be greater.

I presume this will require if statements and for loops.

For Example:

1st array

30 30 30 30 30 30 30 30
30 30 30 30 30 10 30 30
30 20 30 30 30 30 30 30
30 30 30 10 30 30 30 30
30 10 30 30 30 30 30 30 
30 30 30 30 30 30 30 30

2nd array

22 22 22 22 22 22 22 22
22 22 22 -5 22 22 22 22
22 -5 22 22 22 22 22 22
22 22 22 22 22 22 22 22
22 22 22 22 22 22 22 22 
22 22 22 22 22 22 22 22

3rd array

100 100 100 100 100 100 100 100
100 100 100 100 100 100 100 100
100 100 100 100 100 100 100 100
100 100 100 100 200 100 100 100
100 100 200 100 100 100 100 100 
100 100 100 100 100 100 100 100

The script would scan array 1, find the values, scan array 2 at those values lcoations, if it matches critera, it scans array 3 at these lcoations, if it matches the criteria, creates a lsit of the row and column where the lcoations match all 3 criteria. for the above example it would be row and column ([r c]):

4 4
5 2

because in row 5, column 2:

1)10 is less than surrounding eight grid points, 2) the -5 two above it is negative and 3) the 200 is greater than 100.

Thanks for any help

[Oliver]

Thank you for your answers, all methods work fine :)

One extra thing

For multiple time steps, how do I create a table which calculates the occurrences

e.g.

for n = 1:10
endii = imerode(A(:,:,1),[1 1 1;1 0 1;1 1 1]) > A(:,:,1) &...
   imerode(A(:,:,2),[1;0;1;0;0]) < 0 &...
   imdilate(A(:,:,3),[1,1,0]) > A(:,:,3);
[r,c] = find(ii);
end

Instead of it just displaying a the r and c value for every n value in the command window, could another matrix be created which fills in r and n values as it goes along e.g.

1 4 34
2 3 90
2 7 75
3 14 89

Where in the above the first column is n, the second r and the third c.

Answer 1

[Star Strider]

This doesn’t search diagonally — that requires 2D filtering or convolution that could take some time to implement, and given the constraints may not be possible to do with the requisite accuracy — but it does search adjacent rows and columns.

A1 = [30 30 30 30 30 30 30 30
    30 30 30 30 30 10 30 30
    30 20 30 30 30 30 30 30
    30 30 30 10 30 30 30 30
    30 10 30 30 30 30 30 30
    30 30 30 30 30 30 30 30];

A2 = [22 22 22 22 22 22 22 22
    22 22 22 -5 22 22 22 22
    22 -5 22 22 22 22 22 22
    22 22 22 22 22 22 22 22
    22 22 22 22 22 22 22 22
    22 22 22 22 22 22 22 22];

A3 = [100 100 100 100 100 100 100 100
    100 100 100 100 100 100 100 100
    100 100 100 100 100 100 100 100
    100 100 100 100 200 100 100 100
    100 100 200 100 100 100 100 100
    100 100 100 100 100 100 100 100];

drA1 = diff([A1(:,1) A1],1,2);
dcA1 = diff([A1(1,:); A1],1,1);
[A1r,A1c] = find((drA1 == -20) & (dcA1 == -20));        % Step #1
for k1 = 1:length(A1r)                                  % Step #2
    A2q = circshift(A2, [3-A1r(k1) 0]);
    A2rc(k1,:) = (A2q(1,A1c(k1))<0)*[A1r(k1) A1c(k1)];
end
A2rc = reshape(A2rc(A2rc>0),[],2);     % Remove Zero Elements
for k1 = 1:size(A2rc,1)                                 % Step #3
    A3q = circshift(A3, [0 1-A2rc(k1,2)])
    A3rc(k1,:) = (A3q(A2rc(k1,1),2) > A3q(A2rc(k1,1),1))*A2rc(k1,:);    
end

The ‘A3rc’ array contains the rows and columns that meet the criteria (except for the diagonal search in ‘A1’).

EDIT — The 2D filter can replace these lines:

drA1 = diff([A1(:,1) A1],1,2);
dcA1 = diff([A1(1,:); A1],1,1);
[A1r,A1c] = find((drA1 == -20) & (dcA1 == -20));        % Step #1

with these lines, now searching the diagonals as well:

h =- [-1 -1 -1; -1 20 -1; -1 -1 -1];
Y = filter2(h, A1);
[A1r,A1c] = find(Y > 0);                                % Step #1

This can be done with core MATLAB functions, not requiring the Image Processing Toolbox.

[Oliver]

Thanks :)

Would this work for the Step #1 also:

S1= imregionalmin(imhmin(A1,20)).*A1;
[Ar1,A1c] = find(S1 > 0);

and then so on through the remaining steps...

[Star Strider]

My pleasure!

The Image Processing Toolbox functions would certainly work, but since I don’t know what toolboxes people have, unless they mention specific ones or specific functions from them, I always go with core MATLAB functions in my answers. It tends to avoid frustration.

[Oliver]

I am having a problem with the script, am running it over a large for loop to repeat the process for different datasets

In some of these, there are no differences and it displays

Undefined function or variable 'A2rc'.

Is there an if statement that can be used that if nothing is found, it just moves onto the the next dataset. Thanks :)

[Star Strider]

Probably the easiest way to do this is to insert a test for ‘A2rc’ being empty just before the reshape call:

if ~isempty(A2rc)
    A2rc = reshape(A2rc(A2rc>0),[],2); % Remove Zero Elements
else
    continue
end

That will cause it to stop executing further statements in the loop and continue to the next iteration.

This is obviously untested, so if it causes further problems, let me know.

Answer 2

[Andrei Bobrov]

A = cat(3,[30 30 30 30 30 30 30 30
  30 30 30 30 30 10 30 30
  30 20 30 30 30 30 30 30
  30 30 30 10 30 30 30 30
  30 10 30 30 30 30 30 30
  30 30 30 30 30 30 30 30],...
  [22 22 22 22 22 22 22 22
  22 22 22 -5 22 22 22 22
  22 -5 22 22 22 22 22 22
  22 22 22 22 22 22 22 22
  22 22 22 22 22 22 22 22
  22 22 22 22 22 22 22 22],...
  [100 100 100 100 100 100 100 100
  100 100 100 100 100 100 100 100
  100 100 100 100 100 100 100 100
  100 100 100 100 200 100 100 100
  100 100 200 100 100 100 100 100
  100 100 100 100 100 100 100 100]);

With Image Processing Toolbox

ii = imerode(A(:,:,1),[1 1 1;1 0 1;1 1 1]) > A(:,:,1) &...
     imerode(A(:,:,2),[1;0;1;0;0]) < 0 &...
     imdilate(A(:,:,3),[1,1,0]) > A(:,:,3);

[r,c] = find(ii);

Without Image Processing Toolbox

n = size(A);
f1 = false(1,n(2));
f2 = false(n(1),1);
l1 = [f1;diff(A(:,:,1))<0] & [flipud(diff(A(end:-1:1,:,1)))<0;f1] &...
     [f2,diff(A(:,:,1),1,2)<0] & [fliplr(diff(A(:,end:-1:1,1),1,2))<0,f2];
l2 = sign(A(rem(n(1)-k +(0:n(1)-1),n(1))+1,:,2)) == -1;
l3 = [diff(A(:,:,3),1,2) > 0, f2];

[r,c] = find(l1 & l2 & l3);

[Oliver]

for the first step

imerode(A(:,:,1),[1 1 1;1 0 1;1 1 1]) > A(:,:,1) &...

Is this 20 less or just less than?

Also I presume using separate array names such as A1, A2 and A3 in place of A(:,:,1), A(:,:,2) and A(:,:,3) works equally well?

[Oliver]

Thank you for your answer, I have posted an additional bit to the question above, do you mind having a look at it for me, thanks :)