Creating a cell for each plot grid

Hi guys! I'm struggling with this problem.

At first I would like to display an empty graph spaced by 2. For this purpose both axes have value from 0 to 10 stepped by 2.

Secondly, I would like to associate to each subgrid to a square polygon let's assume P1=(0,0) P2=(2,0) P3=(0,2) P4=(2,2).

At this point I would like to check for whichever point in the space if they belong or not to this polygon.

If yes, I need to save in a cell/table all information related to the point namely X and Y coordinates and timestamp.

Thanks to all in advance.

Answer 1

[Adam Danz]

At first I would like to display an empty graph spaced by 2. For this purpose both axes have value from 0 to 10 stepped by 2.

After generating the axes, set the XTick and YTick to 0:2:10.

Secondly, I would like to associate to each subgrid to a square polygon let's assume P1=(0,0) P2=(2,0) P3=(0,2) P4=(2,2).

I interpret this as I would like to add a polygon to the axes. If the polygon is rectangular and parallel to the axes use

rectangle('position',[L,B,W,H])

where L and B are the (x,y) coordinates for the left, bottom corner and W and H are width and height.

Otherwise you can use polyshape or patch.

At this point I would like to check for whichever point in the space if they belong or not to this polygon.

see inpolygon().

If yes, I need to save in a cell/table all information related to the point namely X and Y coordinates and timestamp.

Use the outputs of inpolygon in a conditional statement. Set up a table that has a column for the variables you listed above and add values to the table as needed.

If you get stuck on any of the steps above, share your updated code and point out the problem so we can help get you unstuck.

[Riccardo Tronconi]

@Adam Danz thank you very much I did not faced any issues but at this point I have more "advanced" questions.

Since my area is 10x10 and I have to divide in equal square of 2x2 dimension, instead of repeating this command rectangle('position',[L,B,W,H]) is there a much speedy way to do this?

Furthermore since I have 100 000+ point is X and Y, I was wondering whether is efficient doing a foor loop over the rows and then, before assign to a cell/table, check to which polygon the point belong.

Many thanks!

[Adam Danz]

After re-reading your questions, my new interpretation is that you don't really need to plot polygons. Instead, you just want to know which quadrant a coordinate is in. Is that correct? I don't understand the new question you asked in the comment above. Perhaps you could provide an illustration.

[Riccardo Tronconi]

Yes, in this early stage I do not need any visualization so just to know where the point lies.

the second question is how to assign a point in a quadrant. To do so I have created two arrays as it follows:

for i = 1:length(p) 
    
    pos(:,i)= get(p(i), 'Position'); 
    xv(:,i)=[pos(1,i)+0, pos(1,i)+pos(3,i), pos(1,i)+pos(3,i), pos(1,i)+0]'
    yv(:,i)=[pos(2,i)+0, pos(2,i)+0, pos(2,i)+pos(4,i), pos(2,i)+pos(4,i)]';
    
end 

xv and yv cointais the vertices related to each polygon.

At this point, I have a cell (3x1) in which each row represent represent a position in X and Y of a given subject (1,2,3) and the related timestamp. The following code is useful to assign each point to its related subpolygon.

for j=1:n % loop over each element of the cell T{1}, ..., T{3}
    for k= 1:height(T{j}) % loop over each row of T{j} 
        for z = 1:length(p) % z=1:25 which is the total number of my possible vertices
            
            if inpolygon (T{j}{k,2}, T{j}{k,3}, xv(:,z)', yv(:,z)')==1 % check where the point is contained
                Spg{z}{k,1}= T{j}{k,1}; 
                 Spg{z}{k,2}= T{j}{k,2};
                  Spg{z}{k,3}= T{j}{k,3};
            end
        % Spg{z} is a cell array for all subpolygons and for this purpose
        % would contain every point which belong to a particular polygon. 
        % es. P1=(0,54; 0.99) is contained in Spg{1}
        
        end
        
        
    end 
    
    
end 

This code with several nested loops works but is very very slow to perform. Is there a much faster way to do that?

[Adam Danz]

Since you don't need to display the rectangles, you just need to know which rectangle a dot belongs to.

That's what histcounts2() does. Here's a demo.

% Create random variable within the x & y extent
rng(210731) % for reproducibility
xy = rand(1,2)*10 

% Determine where the random coordinate is within 
% a grid of 2x2 rectangles extending from 0 to 10.
N = histcounts2(xy(1), xy(2), 0:2:10, 0:2:10)

This shows that coorinate (4.15, 7.06) falls into grid (3,4).