How to label data index with count using 3D histogram in Matlab - matlab

Question

I have a set of data points (around 20000) with their x,y values and I want to remove the points that not very close to other points. I try to approach by 'digitizing' and I think the closest way to implement it in Matlab is a 3D histogram so I can remove the points in the low-count bins. I used hist3() but the problems is I couldn't get the index of the points labeled with counts (like the output 'ind' from histc()). The only way I can think of is a nested for loop which is the last thing I want to try. Is there any way I can label the points index or any other approach to do this?
Thanks
I feel like I need some clarification
I have the histogram graph from the data generated by #rayryeng
There are some bins have N=0 or N=1 so I want to remove the data in these bins.
For histc() there is a form of output [bincounts,ind]= histc( ) where ind returns the bin numbers the data falls into. So I can find the index of bins which less/equal or larger than 1, then find the data in the particular bins. Is there any similar thing I can do for a 2D inputs?
Thanks Again

Answer 1

hist3 should be able to accomplish this for you. I'm not quite sure where the problem is. You can call hist3 like so:
[N,C] = hist3(X);
This will automatically partition your dataset into a 10 x 10 grid of equally spaced containers. You can override this behaviour by doing:
[N,C] = hist3(X, NBINS);
NBINS is a 2 element array where the first element tells you how many bins you want vertically and the second element is how many bins you would like horizontally.
N will tell you how many elements fall within each location of the grid and C will give you a 1 x 2 cell array where the first element of the cell array gives you the X co-ordinates of each centre of the bin while the second element of the cell array gives you the Y co-ordinates of each centre of the bin.
To be explicit, if we have a 10 x 10 grid, C will contain a two element cell array where each element is 10 elements long. For each X co-ordinate of the centre found in C{1}, we will have 10 corresponding Y co-ordinates that relate to the a bin's centre in C{2}. This means that the first 10 bin centres are located at C{1}(1), C{2}(1), C{1}(1), C{2}(2), C{1}(1), C{2}(3), ..., C{1}(1), C{2}(10), then the next 10 bin centres are located at: C{1}(2), C{2}(1), C{1}(2), C{2}(2), C{1}(2), C{2}(3), ..., C{1}(1), C{2}(10).
As a quick example, let's do this on a grid between [0,1] on the x-axis and [0,1] on the y-axis. I'm going to generate 100 2D points. Let's also decompose the image into 10 bins horizontally and 10 bins vertically (as per the default of hist3).
rng(100); %// Set seed for reproducibility
A = rand(100,2);
[N,C] = hist3(A);
disp(N);
celldisp(C);
We thus get:
N =
1 2 0 1 2 0 1 0 1 1
0 1 1 1 1 1 0 0 2 5
0 4 1 1 1 1 1 4 0 1
2 0 3 2 2 1 1 0 2 1
0 0 0 0 1 1 1 0 0 1
1 1 1 2 1 1 0 2 0 1
1 0 2 1 2 0 3 1 1 1
0 1 0 0 0 1 1 0 0 1
1 0 1 2 3 3 0 0 0 2
0 2 1 1 0 1 0 3 0 1
C{1} =
Columns 1 through 7
0.0541 0.1528 0.2516 0.3503 0.4491 0.5478 0.6466
Columns 8 through 10
0.7453 0.8440 0.9428
C{2} =
Columns 1 through 7
0.0513 0.1510 0.2508 0.3505 0.4503 0.5500 0.6498
Columns 8 through 10
0.7495 0.8493 0.9491
This tells us that the first grid located at the top left corner of our point distribution only has 1 value logged into it. The next grid after that has 2 bins logged in it and so on and so forth. We also have our bin centres for each of the bins shown in C. Remember, We have 10 x 10 possible bin centres. If we want to display our data with the bin locations, this is what we can do:
[X,Y] = meshgrid(C{1},C{2});
plot(A(:,1), A(:,2), 'b*', X(:), Y(:), 'r*');
grid;
We thus get:
The red stars denote the bin centres while the blue stars denote our data points within the grid. Because our origin is on the bottom left corner of our plot, but the origin of the N matrix is at the top left corner (i.e. the first bin that is decomposed is at the top left while in our data it's at the bottom left corner), we need to rotate N by 90 degrees counter-clockwise so that the origins of each of the matrices agree with each other, and also agree with the plot. As such:
Nrot = rot90(N);
disp(Nrot);
Nrot =
1 5 1 1 1 1 1 1 2 1
1 2 0 2 0 0 1 0 0 0
0 0 4 0 0 2 1 0 0 3
1 0 1 1 1 0 3 1 0 0
0 1 1 1 1 1 0 1 3 1
2 1 1 2 1 1 2 0 3 0
1 1 1 2 0 2 1 0 2 1
0 1 1 3 0 1 2 0 1 1
2 1 4 0 0 1 0 1 0 2
1 0 0 2 0 1 1 0 1 0
As you can see from the picture, this agrees with what we see within the (rotated) N matrix as well as the bin centres C. Using N (or Nrot if you get the convention correct), you can now figure out which points to eliminate from your array of points. Any points that have low membership within N, you would find those points that are the closest to that bin centre that is associated to the grid location in N and remove them.
As an example, supposing that the bin in the first row, second column (of the rotated result) is the one you want to filter out. This corresponds to (C{1}(2), C{2}(10)). We also know that we need to filter out 5 points as they belong to this bin centre. Therefore:
numPointsToRemove = N(2,10); %//or Nrot(1,2);
%// Computes Euclidean distance between this bin centre with every point
dists = sqrt(sum(bsxfun(#minus, A, [C{1}(2) C{2}(10)]).^2, 2));
%// Find the numPointsToRemove closest points to the bin centre and remove
[~,ind] = sort(dists);
A(ind(1:numPointsToRemove,:)) = [];
We sort our distances in ascending order, then determine the numPointsToRemove closest points to this bin centre. We thus remove them from our data matrix.
If you want to remove those bins that have either a 0 or a 1 for the count, we can find those locations, then run a for loop and filter accordingly. However, any bins that have 0 means that we don't even need to run through and filter anything, because no points were mapped to there! You really need to filter out those values that have just 1 in the bins. In other words:
[rows, cols] = find(N == 1);
for index = 1 : numel(rows)
row = rows(index);
col = cols(index);
%// Computes Euclidean distance between this bin centre with every point
dists = sqrt(sum(bsxfun(#minus, A, [C{1}(row) C{2}(col)]).^2, 2));
%// Finds the closest point to the bin centre and remove
[~,ind] = min(dists);
A(ind,:) = [];
end
As you can see, this is similar the same procedure as above. As we wish to filter out those bins that only have 1 assigned to a bin, we just need to find the minimum distance. Remember, we don't need to process any bins that have a count of 0 so we can skip those.