How to divide an area having multiple random points and work on each part separately?

I have asked this question before but couldnt get a satisfactory answer, so let me explain it a bit more, maybe i can make myself more clear.

I have scattered 100 points randomly in a 100x100 plot and divided that area into 9 grids.

a=rand(2,100)*100;
scatter(a(1,:),a(2,:))
grid on
set(gca,'xtick',[0:33.33333:100])
set(gca,'ytick',[0:33.33333:100])

Now I want to get some idea that how can I make groups for each grid (having points within) so that I can work on each grid separately.

Any help will be appreciated, thanks a lot!

[KL]

You should rather continue with the old thread with your explantions. Duplicating a question is not a good idea.

[Asim Ismail]

There was no response on my comment, and this thread is a kind of explanation of that one.

[Jan]

What exactly does "make groups" mean?

[Asim Ismail]

@ Jan, if we have random points scattered in a plot, then i want to find which grid have which points. The group means the group of points belong to a grid. For example

Grid1=[p1, p3, p4]

Grid2=[p6, p7, p8, p9] .... and so on

Answer 1

[Jan]

Perhaps something like this simple loops:

a    = rand(2,100)*100;
grid = 0:33.33333:100;
Result = cell(numel(grid) - 1, numel(grid) - 1);
for i1 = 1:numel(grid) - 1
  for i2 = 1:numel(grid) - 1
    index = (grid(i1) <= a(1, :) & a(1, :) < grid(i1 + 1) & ...
             grid(i2) <= a(2, :) & a(2, :) < grid(i2 + 1));
    Result{i1, i2} = a(index);
  end
end

[EDITED] Now the points in the grid element [i,j] are store in the cell Result{i,j}.

[Asim Ismail]

@Jan thank you for your kind responses, the main thing i am asking for is to separate each grid along with its point, so that i can work on each of them by calling there names. Like grid1 which has [.....] points, grid2 which has[.....] points, as i have mentioned in answer of your previous comment.

[Jan]

See [EDITED]

[Asim Ismail]

@Jan thanks a lot, can we see the result on a plot?

Answer 2

[Image Analyst]

Use max, min, and linspace.

[Asim Ismail]

Can you please explain it little more

[Jan]

@asim: Accepting an answer means, that it has solved your problem.

[Asim Ismail]

Oh sorry i didnt knew

[Salma Hassan]

mr.image analyst i sent you a comment in this link : https://www.mathworks.com/matlabcentral/answers/177316-how-to-show-my-extracted-features#comment_451853

plz can you see it