max and min value of matrix

I have AB matrix that contains 3000x2000 grids data.

AB =
  (1,1) (1,2) (1,3) (1,4) ... (1,3000)
  (2,1) (2,2) (2,3) (2,4) ... (2,3000)
  (3,1) (3,2) (3,3) (3,4) ... (3,3000)
  (4,1) (4,2) (4,3) (4,4) ... (4,3000)
  (5,1) (5,2) (5,3) (5,4) ... (5,3000)
   ...
   ...
   ...
  (2000,1)

I want to calculate (MAX-MIN) value for each 3x5 grids in matrix AB

For Example :

If target grid is (3,2), I use 15 grids data (1~5,1~3) for calculation of MAX and MIN.

And if MAX is (1,1) and MIN is (2,3), then (MAX-MIN) at (3,2) is ((1,1)-(2,3)).

In the same way, if target grid is (3,3), I use (1~5,2~4).

Can you help me to make automatic script to calculate this??

[Stephen23]

@Ashvin Hamzah Driwantara: your question is not clear. Please edit your question and provide us with complete input and output examples so that we can write and test code.

[Ashvin Hamzah Driwantara]

thankyou,, i have edited it

Answer 1

[Stephen23]

blockproc might do what you want:

>> X = rand(3000,2000);
>> Xmax = blockproc(X,[3,5],@(s)max(s.data(:)));
>> Xmin = blockproc(X,[3,5],@(s)min(s.data(:)));
>> Xdif = Xmax - Xmin;

[Ashvin Hamzah Driwantara]

thanks my friend,, I will try later,, please help me if I face some problem

How to calculate average and standard deviation at same case?

[Stephen23]

Use blockproc with mean and std.

[Ashvin Hamzah Driwantara]

so i just replace @min with @mean or @std?

[Stephen23]

@Ashvin Hamzah Driwantara: have a look at my revised answer. You will need to change the function inside the anonymous function:

 @(s)std(s.data(:))

[Ashvin Hamzah Driwantara]

I'm using 5x5 grids, and calculation start from grid (3,3)...

how to calculate grid on (1,1)?

Answer 2

[Guillaume]

I believe Ashvin is looking for nlfilter rather than blockproc. I.e., he wants sliding overlapping blocs instead of adjacent blocks. So:

 nlfilter(X, [3 5], @(block) std(block(:)));

[Ashvin Hamzah Driwantara]

it works,, but it takes a long time...

how about using colfilt?

can you help me, sir?

Answer 3

[Ashvin Hamzah Driwantara]

I get an error

BLOCKPROC encountered an error while evaluating the user-supplied function handle, FUN.

The cause of the error was:

Undefined function 'min' for input arguments of type 'struct'.

Error in blockprocFunDispatcher (line 13)
        output_block = fun(block_struct);

Error in blockprocInMemory (line 80)
[ul_output fun_nargout] = blockprocFunDispatcher(fun,block_struct,...

Error in blockproc (line 242)
    result_image = blockprocInMemory(source,fun,options);

please help

[Stephen23]

@Ashvin Hamzah Driwantara: read my answer and comments again.

[Ashvin Hamzah Driwantara]

okay thankyou, the script works...

but it decreases the matrix...

how can I calculate the MAX-MIN without decrease the matrix?

for example : matrix AB = 3000x2000, I want to get the output 3000x2000 too... Is it possible?

[Stephen23]

@Ashvin Hamzah Driwantara: do you wish for all pixels within one block to have the same value?

[Ashvin Hamzah Driwantara]

no, I think each pixel has different block, but maybe some pixel can't calculate, like pixel (1,1)

[Stephen23]

@Ashvin Hamzah Driwantara: your explanation is not clear. If each block does not generate one value, then you will need to explain in more detail, and provide input and output sample matrices so that we can test code. Until you provide a clear explanation of what you want there is nothing more that I can do for you.

[Ashvin Hamzah Driwantara]

yes, I'm sorry... the matrix is from nc file...

do you know how to solve the problem using colfilt?

because nlfilter takes a long time

Answer 4

[Ashvin Hamzah Driwantara]

I have solved my problem using this script

c13 = matrix 3000x2000

matrix_WV=c13;
[tinggi,lebar]=size(matrix_WV);
maxmin=zeros(tinggi,lebar);

for i=1:tinggi
    for j=1:lebar
        %fprintf('hitung grid ke (%d,%d)\n',i,j);

          n=0;
          for tempi=-2:2
              for tempj=-2:2
                  if i+tempi>=1 && i+tempi<=tinggi && j+tempj>=1 && j+tempj<=lebar
                      n=n+1;
                      area5x5(n)=matrix_WV(i+tempi,j+tempj);
                  end
              end
          end

          max=area5x5(1);
          min=area5x5(1);
          for k=1:n
              if max<area5x5(k)
                  max=area5x5(k);
              end
              if min>area5x5(k)
                  min=area5x5(k);
              end
          end

          maxmin(i,j)=max-min;
      end
  end
  %fprintf('done\n');

Answer 5

[Zeeshan Salam]

i=3:3:smax what is meaning of this?