Mask making help - Divide a meshgrid in to 'boxes' and change the values of the cells in each 'box'

Hi,

I am trying to divide up a meshgrid of zeros in to equal sized boxes, then turn the values of a particular box to equal 1. The idea behind this is to make a mask which can be place on top of another meshgrid so that boxes with zero values are 'not visible' and boxes with ones are 'visible'. The goal it to keep some central box 'on' and cycle through all the other boxes, turning them on and then off again individually. I have written some code which makes the mesh of zeros, 680 by 600 and then divides the mesh in to 68 by 60 boxes and can set the value of each box equal to 1, see below.

The issue I am having is that I cannot figure out a way of:

• Firstly: adding the 1 values of a particular box back to the meshgrid i.e. box 2x3 corresponds to the mesh coordinates (21,31) to (30,40). So the Mesh would be all zeros apart from the cells the square outline by those coordinates.
• Secondly: A way to set up the code with the features described in above but so that I could select the box I wanted to turn on at will, so for example - turn on the values in box 2x3- do something interesting- turn 2x3 back off- turn 3x3 on- ect...

Thank you for any help you can provide. I feel like I am close but just can pull the last bit together.

 clear all; clc
  x = 680;
  y = 600;
  Mesh = zeros(x , y);
  [X,Y] = size(Mesh);
  L = 10;
  ki = 0;
  kj = 0;
  for i = 1 : L : X
      ki = ki + 1;
      for j = 1 : L : Y;
          kj = kj + 1;
          SubMesh = Mesh(i : (i + L - 1), j : (j + L - 1));
          SomeOnes = ones( size(SubMesh) );
          SomeOnesMesh = SubMesh * SomeOnes;
      end
  end
  kii = ki
  kjj = kj / kii

Answer 1

[Image Analyst]

Some observations:

Your Mesh and SubMesh are always zero. Even if you multiply by SOmeONes, it's still zero.

Your convention of having rows (the vertical directions, dimension) be called x and the columns (the horizontal direction) be called y is contrary to the way the rest of the world does it and you might want to change that to be consistent with the rest of us and have your code be more maintainable.

You're not even using meshgrid() function so that confused me for a bit.

Beyond that I'm not really sure what you want to do. The code appears to create small arrays of zeros. It does what you tell it to do. Do you want a full size matrix where you have a patch of 1's in some certain box location?

[Patrick Bevington]

Hi there,

Thanks for your response.

1. Yes you are correct, this multiplication should be an addition to make a small LxL matrix of 1's. This step is probably avoidable and was just trying to show that I want to change the zero values in the 600x680 matrix.

2. Apologies, for convention these should be the other way round.

3. Again apologies, this was simply poor word choice as I forgot there is a meshgrid function. I simply mean some 2D matrix.

4. Yes you have it in your last sentence there. The point is I want to divide my 600x680 matrix in to NxM LxL matrices, in this example 60x68 10x10 matrices, so that I can change the value of a 'box'. So yes a patch of 1's at a certain box location.

It may be easier to for get the loop and just change the values of certain cells in the matrix corresponding to position (i*L+1,j*L+1) to (i*L+10,j*L+10) such that is defines the 'box' ixj for example. Same as the example given for box 2x3 in my original question.

Please see attached the lines of code which need to be corrected.

4 - Mesh = zeros(y , x);

5 - [Y , X] = size(Mesh);

9 - for i = 1 : L : Y

11 - for j = 1 : L : X

15 - SomeOnesMesh = SubMesh + SomeOnes;

Thank you for any additional help you can provide

[Image Analyst]

Is this what you want:

clc;    % Clear the command window.
close all;  % Close all figures (except those of imtool.)
clear;  % Erase all existing variables. Or clearvars if you want.
workspace;  % Make sure the workspace panel is showing.
format long g;
format compact;
fontSize = 14;

columns = 680;
rows = 600;
ZeroImage = zeros(columns , rows);
[rows, columns] = size(ZeroImage);
L = 10;
counter = 1;
for row = 1 : L : rows
  for column = 1 : L : columns;
    SomeOnesMesh = ZeroImage; % Initialize
    % Assign a patch of 1's.
    SomeOnesMesh(row : (row + L - 1), column : (column + L - 1)) = 1;
    imshow(SomeOnesMesh);
    axis on;
    promptMessage = sprintf('This is patch #%d.\nDo you want to Continue processing,\nor Quit processing?', counter);
    titleBarCaption = 'Continue?';
    buttonText = questdlg(promptMessage, titleBarCaption, 'Continue', 'Quit', 'Continue');
    counter = counter + 1;
    if strcmpi(buttonText, 'Quit')
      return;
    end
  end
end

[Patrick Bevington]

Yes this is exactly what I was aiming to do, thank you. I will post the final code use.

Answer 2

[Guillaume]

I would use mat2cell and cell2mat to divide your mesh and reconstruct it.

height = 680;  %much better name than x
width = 600;   %and y
Mesh = false(height , width);
ngrid = 10;
boxheight = height / ngrid;
boxwidth = width / ngrid;
assert(mod(boxheight, 1) == 0 && mod(boxwidth, 1) == 0, 'Cannot divide Mesh evenly');

griddedmesh = mat2cell(Mesh, ones(1, ngrid) * boxheight, ones(1, ngrid) * boxwidth);

%to set a cell of griddedmesh to 0:
griddedmesh{5, 4} = false(size(griddedmesh{5, 4}));

%to set a cell of griddedmesh to 1:
griddedmesh{6, 2} = true(size(griddedmesh{5, 4}));

%to get back the full mesh:
fullmesh = cell2mat(griddedmesh);
imshow(fullmesh)

[Patrick Bevington]

This is really useful also, thank you. Smart way to deal with it.

In the line:

 griddedmesh{6, 2} = true(size(griddedmesh{5, 4}))

I see that the first part 'griddedmesh{6, 2}' sets the location of the 'on' square but what does the 'griddedmesh{5, 4}' part do? I thought this might set the size of the square but if I change 5 and 4 then nothing seems to change

Thanks again,

Paddy

[Patrick Bevington]

Sorry I have just figured it out, it set the box size to the size of cell 5, 4 for example. In this case all the cells are the same size (10x10) so it doesn't make a difference. If that correct?

[Guillaume]

No, the griddedmesh{5, 4} was just an example on how to turn off a box after you've turned it on. Since all the boxes start off, it does nothing in the example.

Here is a more complete demo:

height = 680;  %much better name than x
width = 600;   %and y
Mesh = false(height , width);
ngrid = 10;
boxheight = height / ngrid;
boxwidth = width / ngrid;
assert(mod(boxheight, 1) == 0 && mod(boxwidth, 1) == 0, 'Cannot divide Mesh evenly');
griddedmesh = mat2cell(Mesh, ones(1, ngrid) * boxheight, ones(1, ngrid) * boxwidth);

sequence = spiral(ngrid);
figure;
for box = 1:numel(sequence)
  %turn new box on
  [r, c] = find(sequence == box); %not the most efficient, good enough for demo
  griddedmesh{r, c} = true(size(griddedmesh{r, c}));
  imshow(cell2mat(griddedmesh));
  drawnow;
  pause(0.05);
  %then turn it back off
  griddedmesh{r, c} = false(size(griddedmesh{r, c}));   
end
imshow(cell2mat(griddedmesh));

I find it's much easier to let matlab do the gridding itself rather than figuring which coordinate of the whole mesh correspond to a particular grid cell.