How do I separate the two connected circles in the image below, using MATLAB? I have tried using imerode, but this does not give good results. Eroding does not work, because in order to erode enough to separate the circles, the lines disappear or become mangled. In other starting pictures, a circle and a line overlap, so isolating the overlapping objects won't work either.
The image shows objects identified by bwboundaries, each object painted a different color. As you can see, the two light blue circles are joined, and I want to disjoin them, producing two separate circles. Thanks
I would recommend you use the Circular Hough Transform through imfindcircles. However, you need version 8 of the Image Processing Toolbox, which was available from version R2012a and onwards. If you don't have this, then unfortunately this won't work :(... but let's go with the assumption that you do have it. However, if you are using something older than R2012a, Dev-iL in his/her comment above linked to some code on MATLAB's File Exchange on an implementation of this, most likely created before the Circular Hough Transform was available: http://www.mathworks.com/matlabcentral/fileexchange/9168-detect-circles-with-various-radii-in-grayscale-image-via-hough-transform/
This is a special case of the Hough Transform where you are trying to find circles in your image rather than lines. The beauty with this is that you are able to find circles even when the circle is partially completed or overlapping.
I'm going to take the image that you provided above and do some post-processing on it. I'm going to convert the image to binary, and remove the border, which is white and contains the title. I'm also going to fill in any holes that result so that all of the objects are filled in with solid white. There is also some residual quantization noise after I do this step, so I'm going to a small opening with a 3 x 3 square element. After, I'm going to close the shapes with a 3 x 3 square element, as I see that there are noticeable gaps in the shapes. Therefore:
Therefore, directly reading in your image from where you've posted it:
im = imread('http://s29.postimg.org/spkab8oef/image.jpg'); %// Read in the image
im_gray = im2double(rgb2gray(im)); %// Convert to grayscale, then [0,1]
out = imclearborder(im_gray > 0.6); %// Threshold using 0.6, then clear the border
out = imfill(out, 'holes'); %// Fill in the holes
out = imopen(out, strel('square', 3));
out = imclose(out, strel('square', 3));
This is the image I get:
Now, apply the Circular Hough Transform. The general syntax for this is:
[centres, radii, metric] = imfindcircles(img, [start_radius, end_radius]);
img would be the binary image that contains your shapes, start_radius and end_radius would be the smallest and largest radius of the circles you want to find. The Circular Hough Transform is performed such that it will find any circles that are within this range (in pixels). The outputs are:
centres: Which returns the (x,y) positions of the centres of each circle detected
radii: The radius of each circle
metric: A measure of purity of the circle. Higher values mean that the shape is more probable to be a circle and vice-versa.
I searched for circles having a radius between 30 and 60 pixels. Therefore:
[centres, radii, metric] = imfindcircles(out, [30, 60]);
We can then demonstrate the detected circles, as well as the radii by a combination of plot and viscircles. Therefore:
imshow(out);
hold on;
plot(centres(:,1), centres(:,2), 'r*'); %// Plot centres
viscircles(centres, radii, 'EdgeColor', 'b'); %// Plot circles - Make edge blue
Here's the result:
As you can see, even with the overlapping circles towards the top, the Circular Hough Transform was able to detect two distinct circles in that shape.
Edit - November 16th, 2014
You wish to ensure that the objects are separated before you do bwboundaries. This is a bit tricky to do. The only way I can see you do this is if you don't even use bwboundaries at all and do this yourself. I'm assuming you'll want to analyze each shape's properties by themselves after all of this, so what I suggest you do is iterate through every circle you have, then place each circle on a new blank image, do a regionprops call on that shape, then append it to a separate array. You can also keep track of all of the circles by having a separate array that adds the circles one at a time to this array.
Once you've finished with all of the circles, you'll have a structure array that contains all of the measured properties for all of the measured circles you have found. You would use the array that contains only the circles from above, then use these and remove them from the original image so you get just the lines. You'd then call one more regionprops on this image to get the information for the lines and append this to your final structure array.
Here's the first part of the procedure I outlined above:
num_circles = numel(radii); %// Get number of circles
struct_reg = []; %// Save the shape analysis per circle / line here
%// For creating our circle in the temporary image
[X,Y] = meshgrid(1:size(out,2), 1:size(out,1));
%// Storing all of our circles in this image
circles_img = false(size(out));
for idx = 1 : num_circles %// For each circle we have...
%// Place our circle inside a temporary image
r = radii(idx);
cx = centres(idx,1); cy = centres(idx,2);
tmp = (X - cx).^2 + (Y - cy).^2 <= r^2;
% // Save in master circle image
circles_img(tmp) = true;
%// Do regionprops on this image and save
struct_reg = [struct_reg; regionprops(tmp)];
end
The above code may be a bit hard to swallow, but let's go through it slowly. I first figure out how many circles we have, which is simply looking at how many radii we have detected. I keep a separate array called struct_reg that will append a regionprops struct for each circle and line we have in our image. I use meshgrid to determine the (X,Y) co-ordinates with respect to the image containing our shapes so that I can draw one circle onto a blank image at each iteration. To do this, you simply need to find the Euclidean distance with respect to the centre of each circle, and set the pixels to true only if that location has its distance less than r. After doing this operation, you will have created only one circle and filtered all of them out. You would then use regionprops on this circle, add it to our circles_img array, which will only contain the circles, then continue with the rest of the circles.
At this point, we will have saved all of our circles. This is what circles_img looks like so far:
You'll notice that the circles drawn are clean, but the actual circles in the original image are a bit jagged. If we tried to remove the circles with this clean image, you will get some residual pixels along the border and you won't completely remove the circles themselves. To illustrate what I mean, this is what your image looks like if I tried to remove the circles with circles_img by itself:
... not good, right?
If you want to completely remove the circles, then do a morphological reconstruction through imreconstruct where you can use this image as the seed image, and specify the original image to be what we're working on. The job of morphological reconstruction is essentially a flood fill. You specify seed pixels, and an image you want to work on, and the job of imreconstruct is from these seeds, flood fill with white until we reach the boundaries of the objects that the seed pixels resided in. Therefore:
out_circles = imreconstruct(circles_img, out);
Therefore, we get this for our final reconstructed circles image:
Great! Now, use this and remove the circles from the original image. Once you do this, run regionprops again on this final image and append to your struct_reg variable. Obviously, save a copy of the original image before doing this:
out_copy = out;
out_copy(out_circles) = false;
struct_reg = [struct_reg; regionprops(out_copy)];
Just for sake of argument, this is what the image looks like with the circles removed:
Now, we have analyzed all of our shapes. Bear in mind I did the full regionprops call because I don't know exactly what you want in your analysis... so I just decided to give you everything.
Hope this helps!
erosion is the way to go. You should probably use a larger structuring element.
How about
1 erode
2 detect your objects
3 dilate each object for itself using the same structuring element
Related
I am struggling to find a good contour detection function that would count the number of contour in bw images that I have processed using some previous tools. As you can see, my profile picture is an example of such images,
,
In this image, ideally, I wish to have a function which counts four closed contour.
I don't mind if it also detects the really tiny contours in between, or the entire shape itself as extra contours. As long as it counts the medium sized ones, I can fix the rest by applying area threshold. My problem is that any function I have tried detects only one contour - the entire shape, as it cannot separate it to the su-conours which are connected to one another.
Any suggestions?
Here is my shot at this, although your question might get closed because it's off-topic, too broad or a possible duplicate. Anyhow I propose another way to count the number of contours. You could also do it using bwboundaries as was demonstrated in the link provided by #knedlsepp in the possible duplicate. Just for the sake of it here is another way.
The idea is to apply a morphological closure of your image and actually count the number of enclosed surfaces instead instead of contours. That might end up being the same thing but I think it's easier to visualize surfaces.
Since the shapes in your image look like circle (kind of...) the structuring element used to close the image is a disk. The size (here 5) is up to you but for the image you provided its fine. After that, use regionprops to locate image regions (here the blobs) and count them, which comes back to counting contours I guess. You can provide the Area parameter to filter out shapes based on their area. Here I ask the function to provide centroids to plot them.
Whole code:
clear
clc
close all
%// Read, threshold and clean up the image
Im = im2bw(imread('ImContour.png'));
Im = imclearborder(Im);
%// Apply disk structuring element to morphologically close the image.
%// Play around with the size to alter the output.
se = strel('disk',5);
Im_closed = imclose(Im,se);
%// Find centroids of circle-ish shapes. Youcan also get the area to filter
%// out those you don't want.
S = regionprops(~Im_closed,'Centroid','Area');
%// remove the outer border of the image (1st output of regioprops).
S(1) = [];
%// Make array with centroids and show them.
Centro = vertcat(S.Centroid);
imshow(Im)
hold on
scatter(Centro(:,1),Centro(:,2),40,'filled')
And the output:
So as you see the algorithm detected 5 regions, but try playing a bit with the parameters and you will see which ones to change to get the desired output of 4.
Have fun!
I have this image (8 bit, pseudo-colored, gray-scale):
And I want to create an intensity band of a specific measure around it's border.
I tried erosion and other mathematical operations, including filtering to achieve the desired band but the actual image intensity changes as soon as I use erosion to cut part of the border.
My code so far looks like:
clear all
clc
x=imread('8-BIT COPY OF EGFP001.tif');
imshow(x);
y = imerode(x,strel('disk',2));
y1=imerode(y,strel('disk',7));
z=y-y1;
figure
z(z<30)=0
imshow(z)
The main problem I am encountering using this is that it somewhat changes the intensity of the original images as follows:
So my question is, how do I create such a band across image border without changing any other attribute of the original image?
Going with what beaker was talking about and what you would like done, I would personally convert your image into binary where false represents the background and true represents the foreground. When you're done, you then erode this image using a good structuring element that preserves the roundness of the contours of your objects (disk in your example).
The output of this would be the interior of the large object that is in the image. What you can do is use this mask and set these locations in the image to black so that you can preserve the outer band. As such, try doing something like this:
%// Read in image (directly from StackOverflow) and pseudo-colour the image
[im,map] = imread('http://i.stack.imgur.com/OxFwB.png');
out = ind2rgb(im, map);
%// Threshold the grayscale version
im_b = im > 10;
%// Create structuring element that removes border
se = strel('disk',7);
%// Erode thresholded image to get final mask
erode_b = imerode(im_b, se);
%// Duplicate mask in 3D
mask_3D = cat(3, erode_b, erode_b, erode_b);
%// Find indices that are true and black out result
final = out;
final(mask_3D) = 0;
figure;
imshow(final);
Let's go through the code slowly. The first two lines take your PNG image, which contains a grayscale image and a colour map and we read both of these into MATLAB. Next, we use ind2rgb to convert the image into its pseudo-coloured version. Once we do this, we use the grayscale image and threshold the image so that we capture all of the object pixels. I threshold the image with a value of 10 to escape some quantization noise that is seen in the image. This binary image is what we will operate on to determine those pixels we want to set to 0 to get the outer border.
Next, we declare a structuring element that is a disk of a radius of 7, then erode the mask. Once I'm done, I duplicate this mask in 3D so that it has the same number of channels as the pseudo-coloured image, then use the locations of the mask to set the values that are internal to the object to 0. The result would be the original image, but having the outer contours of all of the objects remain.
The result I get is:
I was working on my image processing problem with detecting coins.
I have some images like this one here:
and wanted to separate the falsely connected coins.
We already tried the watershed method as stated on the MATLAB-Homepage:
the-watershed-transform-strategies-for-image-segmentation.html
especially since the first example is exactly our problem.
But instead we get a somehow very messed up separation as you can see here:
We already extracted the area of the coin using the regionprops Extrema parameter and casting the watershed only on the needed area.
I'd appreciate any help with the problem or even another method of getting it separated.
If you have the Image Processing Toolbox, I can also suggest the Circular Hough Transform through imfindcircles. However, this requires at least version R2012a, so if you don't have it, this won't work.
For the sake of completeness, I'll assume you have it. This is a good method if you want to leave the image untouched. If you don't know what the Hough Transform is, it is a method for finding straight lines in an image. The circular Hough Transform is a special case that aims to find circles in the image.
The added advantage of the circular Hough Transform is that it is able to detect partial circles in an image. This means that those regions in your image that are connected, we can detect them as separate circles. How you'd call imfindcircles is in the following fashion:
[centers,radii] = imfindcircles(A, radiusRange);
A would be your binary image of objects, and radiusRange is a two-element array that specifies the minimum and maximum radii of the circles you want to detect in your image. The outputs are:
centers: A N x 2 array that tells you the (x,y) co-ordinates of each centre of a circle that is detected in the image - x being the column and y being the row.
radii: For each corresponding centre detected, this also gives the radius of each circle detected. This is a N x 1 array.
There are additional parameters to imfindcircles that you may find useful, such as the Sensitivity. A higher sensitivity means that it is able to detect circular shapes that are more non-uniform, such as what you are showing in your image. They aren't perfect circles, but they are round shapes. The default sensitivity is 0.85. I set it to 0.9 to get good results. Also, playing around with your image, I found that the radii ranged from 50 pixels to 150 pixels. Therefore, I did this:
im = im2bw(imread('http://dennlinger.bplaced.net/t06-4.jpg'));
[centers,radii] = imfindcircles(im, [50 150], 'Sensitivity', 0.9);
The first line of code reads in your image directly from StackOverflow. I also convert this to logical or true black and white as the image you uploaded is of type uint8. This image is stored in im. Next, we call imfindcircles in the method that we described.
Now, if we want to visualize the detected circles, simply use imshow to show your image, then use the viscircles to draw the circles in the image.
imshow(im);
viscircles(centers, radii, 'DrawBackgroundCircle', false);
viscircles by default draws the circles with a white background over the contour. I want to disable this because your image has white circles and I don't want to show false contouring. This is what I get with the above code:
Therefore, what you can take away from this is the centers and radii variables. centers will give you the centre of each detected circle while radii will tell you what the radii is for each circle.
Now, if you want to simulate what regionprops is doing, we can iterate through all of the detected circles and physically draw them onto a 2D map where each circle would be labeled by an ID number. As such, we can do something like this:
[X,Y] = meshgrid(1:size(im,2), 1:size(im,1));
IDs = zeros(size(im));
for idx = 1 : numel(radii)
r = radii(idx);
cen = centers(idx,:);
loc = (X - cen(1)).^2 + (Y - cen(2)).^2 <= r^2;
IDs(loc) = idx;
end
We first define a rectangular grid of points using meshgrid and initialize an IDs array of all zeroes that is the same size as the image. Next, for each pair of radii and centres for each circle, we define a circle that is centered at this point that extends out for the given radius. We then use these as locations into the IDs array and set it to a unique ID for that particular circle. The result of IDs will be that which resembles the output of bwlabel. As such, if you want to extract the locations of where the idx circle is, you would do:
cir = IDs == idx;
For demonstration purposes, this is what the IDs array looks like once we scale the IDs such that it fits within a [0-255] range for visibility:
imshow(IDs, []);
Therefore, each shaded circle of a different shade of gray denotes a unique circle that was detected with imfindcircles.
However, the shades of gray are probably a bit ambiguous for certain coins as this blends into the background. Another way that we could visualize this is to apply a different colour map to the IDs array. We can try using the cool colour map, with the total number of colours to be the number of unique circles + 1 for the background. Therefore, we can do something like this:
cmap = cool(numel(radii) + 1);
RGB = ind2rgb(IDs, cmap);
imshow(RGB);
The above code will create a colour map such that each circle gets mapped to a unique colour in the cool colour map. The next line applies a mapping where each ID gets associated with a colour with ind2rgb and we finally show the image.
This is what we get:
Edit: the following solution is more adequate to scenarios where one does not require fitting the exact circumferences, although simple heuristics could be used to approximate the radii of the coins in the original image based on the centers found in the eroded one.
Assuming you have access to the Image Processing toolbox, try imerode on your original black and white image. It will apply an erosion morphological operator to your image. In fact, the Matlab webpage with the documentation of that function has an example strikingly similar to your problem/image and they use a disk structure.
Run the following code (based on the example linked above) assuming the image you submitted is called ima.jpg and is local to the code:
ima=imread('ima.jpg');
se = strel('disk',50);
eroded = imerode(ima,se);
imshow(eroded)
and you will see the image that follows as output. After you do this, you can use bwlabel to label the connected components and compute whatever properties you may want, for example, count the number of coins or detect their centers.
Simple rounded corner rectangle code in Matlab can be written as follows.
rectangle('Position',[0,-1.37/2,3.75,1.37],...
'Curvature',[1],...
'LineWidth',1,'LineStyle','-')
daspect([1,1,1])
How to get the x and y coordinates arrays of this figure?
To get the axes units boundaries, do:
axisUnits = axis(axesHandle) % axesHandle could be gca
axisUnits will be an four elements array, with the following syntax: [xlowlim xhighlim ylowlim yhighlim], it will also contain the zlow and zhigh for 3-D plots.
But I think that is not what you need to know. Checking the matlab documentation for the rectangle properties, we find:
Position four-element vector [x,y,width,height]
Location and size of rectangle. Specifies the location and size of the
rectangle in the data units of the axes. The point defined by x, y
specifies one corner of the rectangle, and width and height define the
size in units along the x- and y-axes respectively.
It is also documented on the rectangle documentation:
rectangle('Position',[x,y,w,h]) draws the rectangle from the point x,y
and having a width of w and a height of h. Specify values in axes data
units.
See if this illustrate what you want. You have an x axis that goes from −100 to 100 and y axis that goes from 5 to 15. Suppose you want to put a rectangle from −30 to −20 in x and 8 to 10 in y.
rectangle('Position',[-30,8,10,2]);
As explained by the comments there appears to be no direct way to query the figure created by rectangle and extract x/y coordinates. On the other hand, I can think of two simple strategies to arrive at coordinates that will closely reproduce the curve generated with rectangle:
(1) Save the figure as an image (say .png) and process the image to extract points corresponding to the curve. Some degree of massaging is necessary but this is relatively straightforward if blunt and I expect the code to be somewhat slow at execution compared to getting data from an axes object.
(2) Write your own code to draw a rectangle with curved edges. While recreating precisely what matlab draws may not be so simple, you may be satisfied with your own version.
Whether you choose one of these approaches boils down to (a) what speed of execution you consider acceptable (b) how closely you need to replicate what rectangle draws on screen (c) whether you have image processing routines, say for reading an image file.
Edit
If you have the image processing toolbox you can arrive at a set of points representing the rectangle as follows:
h=rectangle('Position',[0,-1.37/2,3.75,1.37],...
'Curvature',[1],...
'LineWidth',1,'LineStyle','-')
daspect([1,1,1])
axis off
saveas(gca,'test.png');
im = imread('test.png');
im = rgb2gray(im);
figure, imshow(im)
Note that you will still need to apply a threshold to pick the relevant points from the image and then transform the coordinate system and rearrange the points in order to display properly as a connected set. You'll probably also want to tinker with resolution of the initial image file or apply image processing functions to get a smooth curve.
I am having trouble achieving the correct segmentation of a grayscale image:
The ground truth, i.e. what I would like the segmentation to look like, is this:
I am most interested in the three components within the circle. Thus, as you can see, I would like to segment the top image into three components: two semi-circles, and a rectangle between them.
I have tried various combinations of dilation, erosion, and reconstruction, as well as various clustering algorithms, including k-means, isodata, and mixture of gaussians--all with varying degrees of success.
Any suggestions would be appreciated.
Edit: here is the best result I've been able to obtain. This was obtained using an active contour to segment the circular ROI, and then applying isodata clustering:
There are two problems with this:
The white halo around the bottom-right cluster, belonging to the top-left cluster
The gray halo around both the top-right and bottom-left cluster, belonging to the center cluster.
Here's a starter...
use circular Hough transform to find the circular part. For that I initially threshold the image locally.
im=rgb2gray(imread('Ly7C8.png'));
imbw = thresholdLocally(im,[2 2]); % thresold localy with a 2x2 window
% preparing to find the circle
props = regionprops(imbw,'Area','PixelIdxList','MajorAxisLength','MinorAxisLength');
[~,indexOfMax] = max([props.Area]);
approximateRadius = props(indexOfMax).MajorAxisLength/2;
radius=round(approximateRadius);%-1:approximateRadius+1);
%find the circle using Hough trans.
h = circle_hough(edge(imbw), radius,'same');
[~,maxIndex] = max(h(:));
[i,j,k] = ind2sub(size(h), maxIndex);
center.x = j; center.y = i;
figure;imagesc(im);imellipse(gca,[center.x-radius center.y-radius 2*radius 2*radius]);
title('Finding the circle using Hough Trans.');
select only what's inside the circle:
[y,x] = meshgrid(1:size(im,2),1:size(im,1));
z = (x-j).^2+(y-i).^2;
f = (z<=radius^2);
im=im.*uint8(f);
EDIT:
look for a place to start threshold the image to segment it by looking at the histogram, finding it's first local maxima, and iterating from there until 2 separate segments are found, using bwlabel:
p=hist(im(im>0),1:255);
p=smooth(p,5);
[pks,locs] = findpeaks(p);
bw=bwlabel(im>locs(1));
i=0;
while numel(unique(bw))<3
bw=bwlabel(im>locs(1)+i);
i=i+1;
end
imagesc(bw);
The middle part can now be obtained by taking out the two labeled parts from the circle, and what is left will be the middle part (+some of the halo)
bw2=(bw<1.*f);
but after some median filtering we get something more reasonble
bw2= medfilt2(medfilt2(bw2));
and together we get:
imagesc(bw+3*bw2);
The last part is a real "quick and dirty", I'm sure that with the tools you already used you'll get better results...
One can also obtain an approximate result using the watershed transformation. This is the watershed on the inverted image -> watershed(255-I) Here is an example result:
Another Simple method is to perform a morphological closing on the original image with a disc structuring element (one can perform multiscale closing for granulometries) and then obtain the full circle. After this extracting the circle is and components withing is easier.
se = strel('disk',3);
Iclo = imclose(I, se);% This closes open circular cells.
Ithresh = Iclo>170;% one can locate this threshold automatically by histogram modes (if you know apriori your cell structure.)
Icircle = bwareaopen(Ithresh, 50); %to remove small noise components in the bg
Ithresh2 = I>185; % This again needs a simple histogram.