Grouping boundary points of several ROIs

Hi,

I have 6 region of interests (ROI) and their boundary points and centroids. I got this boundary points randomly through MATLAB programming. Do you have an idea how to group them according to corresponding ROIs in MATLAB? I need individual ROI's boundary points for further calculations.

I have applied Kmeans and it didn't work.

[DGM]

What exactly do you mean by boundary points? I ask because if you're trying to find the perimeters of the objects, you shouldn't need to resort to random processes to do so. While bwperim() can find perimeters of objects in a binary image, regionprops() could succinctly find both centroids and perimeters, all sorted neatly in a struct so that it's obvious to which object each entry relates.

[Sunetra Banerjee]

I used

[y,x] = find(inpict);

idx = boundary(x,y) ;

to find the boundary points (perimeter co-ordinate points) of the ROIs. But it is mixed up for all six ROIs. I need separatly for six different ROIs. I don't need to calculate perimeter. Rather I need the perimeter co-ordinate points, so that I can calculate the red marked points' co-ordinates shown in the image.

[DGM]

Ah okay.

You could segment the image first. That way you're only finding the boundary of one object at a time.

inpict = imread('lumber.png')>128;

[L N] = bwlabel(inpict);
idx = cell(N,1);
for n = 1:N
	[y,x] = find(L==n); 
	idx{n} = boundary(x,y);  
end
idx

That would give you a cell array idx containing one boundary index vector for each object in the image.

The way bwlabel() sorts results should be the same as how regionprops() does, but not generally in what the apparent order might seem to be from looking at the image. If you used regionprops() to get the centroids, they should be in the same order as the idx array. If you want both in a different order (e.g. top to bottom), you can use the centroid properties to sort both.

[Sunetra Banerjee]

Thank you so much. I got

idx =

6×1 cell array

{170×1 double}

{210×1 double}

{190×1 double}

{243×1 double}

{182×1 double}

{162×1 double}

So, That means the first ROI has 170 boundry points, 2nd has 210 boundary points and so on. Right?

[DGM]

Yes, but like I said, bear in mind which one is "first", "second", etc. The label array L contains a map of each object it found in the image, each labeled with their ordinal value. In this case, the objects are numbered [5 6 4 3 2 1] from top to bottom.

Answer 1

[DGM]

This may be extraneous, but if you don't need the boundary coordinates for anything other than finding the extrema, consider this instead:

inpict = imread('lumber.png')>128;

% you'll need the centroid anyway
S = regionprops(inpict,'centroid','extrema');

imshow(inpict); hold on
for n = 1:numel(S)
    % calculate the right and left extrema of this object
    rextrema = mean(S(n).Extrema(3:4,:),1);
    lextrema = mean(S(n).Extrema(7:8,:),1);
    % for sake of demonstration, show where the extrema are
    plot(rextrema(1),rextrema(2),'*')
    plot(lextrema(1),lextrema(2),'*')
end

[Sunetra Banerjee]

Awesome. I just needed the extremas. Great!

Thanks a lot.

[Image Analyst]

For next time, look up bwboundaries() - better than boundary(). Also look up bwferet() if you want extrema locations and distances.

[DGM]

That's a good point. I was assuming the goal was to find the horizontal extrema instead of the maximal diameter.