Finding the area of the black spots in a circle MATLAB - matlab

Is it possible to find the area of the black pixelation of an area within a circle? in other words I want to find the number of pixels (the area) of the RGB 0,0,0 (black pixels) within the circle. I do not want the areas of the white pixels (1,1,1) within the circle. I also have a radius of the circle if that helps. Here is the image:
Here is the code:
BW2= H(:,:) <0.45 ;%& V(:,:)<0.1;
aa=strel('disk',5);
closeBW = imclose(BW2,aa);
figure, imshow(closeBW)
imshow(closeBW)
viscircles([MYY1 MYX1], round(MYR2/2))
MYY1,MYX2, and the other values are calculated by my program. How can I find the area of the black pixelation in my circle?

Here is an idea:
1) Calculate the total # of black pixels in your original image (let's call it A).
2) Duplicate that image (let's call it B) and replace all pixels inside the circle with white. To do that, create a binary mask. (see below)
3) Calculate the total # of black pixels in that image (i.e. B).
4) Subtract both values. That should give you the number of black pixels within the circle.
Sample code: I used a dummy image I had on my computer and created a logical mask with the createMask method from imellipse. That seems complicated but in your case since you have the center position and radius of the circle you can create directly your mask like I did or by looking at this question/answer.
Once the mask is created, use find to get the linear indices of the white pixels of the mask (i.e. all of it) to replace the pixels in the circle of your original image with white pixels, which you use to calculate the difference in black pixels.
clc;clear;close all
A = im2bw(imread('TestCircle.png'));
imshow(A)
Center = [160 120];
Radius = 60;
%// In your case:
% Center = [MYY1 MYX1];
% Radius = round(MYR2/2);
%// Get sum in original image
TotalBlack_A = sum(sum(~A))
e = imellipse(gca, [Center(1) Center(2) Radius Radius]);
%// Create the mask
ROI = createMask(e);
%// Find white pixels
white_id = find(ROI);
%// Duplicate original image
B = A;
%// Replace only those pixels in the ROI with white
B(white_id) = 1;
%// Get new sum
NewBlack_B = sum(sum(~B))
%// Result!
BlackInRoi = TotalBlack_A - NewBlack_B
In this case I get this output:
TotalBlack_A =
158852
NewBlack_B =
156799
BlackInRoi =
2053
For this input image:

Related

Quantifying pixels from a list of coordinates

I have a list of coordinates, which are generated from another program, and I have an image.
I'd like to load those coordinates (making circular regions of interest (ROIs) with a diameter of 3 pixels) onto my image, and extract the intensity of those pixels.
I can load/impose the coordinates on to the image by using;
imshow(file);
hold on
scatter(xCoords, yCoords, 'g')
But can not extract the intensity.
Can you guys point me in the right direction?
I am not sure what you mean by a circle with 3 pixels diameter since you are in a square grid (as mentioned by Ander Biguri). But you could use fspecial to create a disk filter and then normalize. Something like this:
r = 1.5; % for diameter = 3
h = fspecial('disk', r);
h = h/h(ceil(r),ceil(r));
You can use it as a mask to get the intensities at the given region of the image.
im = imread(file);
ROI = im(xCoord-1:xCoord+1; yCoord-1:yCoord+1);
I = ROI.*h;

How to remove horizontal and vertical lines

I need to remove horizontal and vertical lines in a binary image. Is there any method for filtering these lines? bwareaopen() is not good method to remove these lines and also Dilation and Erosion are not good for these cases.
Does any one know a solution?
Example image:
EDIT:(added more example images:
http://s1.upload7.ir/downloads/pPqTDnmsmjHUGTEpbwnksf3uUkzncDwr/example%202.png
source file of images:
https://www.dropbox.com/sh/tamcdqk244ktoyp/AAAuxkmYgBkB8erNS9SajkGVa?dl=0
www.directexe.com/9cg/pics.rar
Use regionprops and remove regions with high eccentricity (meaning the region is long and thin) and orientation near 0 or near 90 degrees (regions which are vertical or horizontal).
Code:
img = im2double(rgb2gray(imread('removelines.jpg')));
mask = ~im2bw(img);
rp = regionprops(mask, 'PixelIdxList', 'Eccentricity', 'Orientation');
% Get high eccentricity and orientations at 90 and 0 degrees
rp = rp([rp.Eccentricity] > 0.95 & (abs([rp.Orientation]) < 2 | abs([rp.Orientation]) > 88));
mask(vertcat(rp.PixelIdxList)) = false;
imshow(mask);
Output:
If all of your images are the same where the horizontal and vertical lines are touching the border, a simple call to imclearborder will do the trick. imclearborder removes any object pixels that are touching the borders of the image. You'll need to invert the image so that the characters are white and the background is dark, then reinvert back, but I'm assuming that isn't an issue. However, to be sure that none of the actual characters get removed because they may also be touching the border, it may be prudent to artificially pad the top border of the image with a single pixel thickness, clear the border, then recrop.
im = imread('http://i.stack.imgur.com/L1hUa.jpg'); %// Read image directly from StackOverflow
im = ~im2bw(im); %// Convert to black and white and invert
im_pad = zeros(size(im,1)+1, size(im,2)) == 1; %// Pad the image too with a single pixel border
im_pad(2:end,:) = im;
out = ~imclearborder(im_pad); %// Clear border pixels then reinvert
out = out(2:end,:); %// Crop out padded pixels
imshow(out); %// Show image
We get this:
You can firstly find the horizontal and vertical lines. Since, the edge map will also be binary so you can perform a logical subtraction operation in between the images. To find vertical lines, you can use (in MATLAB)
BW = edge(I,'sobel','vertical');
For horizontal lines, you can use
% Generate horizontal edge emphasis kernel
h = fspecial('sobel');
% invert kernel to detect vertical edges
h = h';
J = imfilter(I,h);

Complete partial circles in an image using MATLAB

I have binary images and they have semi or less circles. My aim is to find these circles, make them whole circles and remove all other objects . I found this but it is for MATLAB R2013a. I am using R2011b and it doesn't have the function centers = imfindcircles(A,radius).
How can I do that in MATLAB version R2011b?
Images:
Edit:
My aim is to get whole circle. I show this below for the last image.
Too bad about imfindcircles! One thing I can suggest is to invoke regionprops and specify the 'Area' and 'BoundingBox' flags. regionprops was available in MATLAB for as long as I can remember, so we can certainly use it here.
What this will do is that whatever distinct objects that are seen in the image that are connected, we will find both their areas and their bounding boxes that bound them. After you do this, threshold on the area so that any objects that have a very large area most likely contain circles of interest. Bear in mind that I'm only assuming that you have circles in your image. Should you have any objects that have a large area, this method will extract those out too.
As such, let's read in your image directly from Stack Overflow. When you uploaded the image, it's a RGB image, so I'll have to convert to binary:
im = imread('http://i.stack.imgur.com/wQLPi.jpg');
im_bw = im2bw(im);
Next, call regionprops:
s = regionprops(im_bw, 'Area', 'BoundingBox');
Now, collect all of the areas, and let's take a look at all of the unique areas of all objects seen in this image:
areas = [s.Area].';
unique(areas)
ans =
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
19
20
23
24
25
28
29
38
43
72
73
85
87
250
465
3127
If you take a look at the very end, you'll see that we have an object that has 3127 pixels in it. This probably contains our circle. As such, let's pick out that single element that contains this object:
s2 = s(areas == 3127);
In general, you'll probably have more than one circle in your image, so you should threshold the area to select those potential circles. Something like:
s2 = s(areas > 2000);
Now, let's create a new blank image that is the same size as the original image, then simply use the BoundingBox property to extract out the area that encompasses the circle in the original image and copy it over to the same location in the output image. The BoundingBox field is structured in the following way:
[x y w h]
x and y are the top-left corner of the bounding box. x would be the column and y would be the row. w and h are the width and height of the bounding box. As such, we can use this directly to access our image and copy those pixels over into the output image.
out = false(size(im_bw));
bb = floor(s2.BoundingBox); %// Could be floating point, so floor it
out(bb(2):bb(2)+bb(4)-1, bb(1):bb(1)+bb(3)-1) = im_bw(bb(2):bb(2)+bb(4)-1, bb(1):bb(1)+bb(3)-1);
This is what I get:
What you should probably do is loop over the circles in case we have more than one. The above code assumes that you detected just one circle. Therefore, do something like this:
out = false(size(im_bw));
for idx = 1 : numel(s2) %// For each potential circle we have...
bb = floor(s2(idx).BoundingBox); %// Could be floating point, so floor it
%// Copy over pixels from original bw image to output
out(bb(2):bb(2)+bb(4)-1, bb(1):bb(1)+bb(3)-1) = im_bw(bb(2):bb(2)+bb(4)-1, bb(1):bb(1)+bb(3)-1);
end
A small thing to note is that the bounding box encompasses the entire object, but there could also be some noisy pixels that are disconnected that are within that bounding box. You may have to apply some morphology to get rid of those pixels. A binary opening could suffice.
Here's what I get with your other images. I thresholded the area to search for those that have 2000 pixels or more (I did this above):
Just for self-containment and your copy-and-pasting pleasure, here's the code in one segment:
clear all;
close all;
%im = imread('http://i.stack.imgur.com/qychC.jpg');
%im = imread('http://i.stack.imgur.com/wQLPi.jpg');
im = imread('http://i.stack.imgur.com/mZMBA.jpg');
im_bw = im2bw(im);
s = regionprops(im_bw, 'Area', 'BoundingBox');
areas = [s.Area].';
s2 = s(areas > 2000);
out = false(size(im_bw));
for idx = 1 : numel(s2) %// For each potential circle we have...
bb = floor(s2(idx).BoundingBox); %// Could be floating point, so floor it
%// Copy over pixels from original bw image to output
out(bb(2):bb(2)+bb(4)-1, bb(1):bb(1)+bb(3)-1) = im_bw(bb(2):bb(2)+bb(4)-1, bb(1):bb(1)+bb(3)-1);
end
imshow(out);
All three images are there in the code. You just have to uncomment whichever one you want to use, comment out the rest, then run the code. It will display an image with all of your detected circles.
Edit
You would like to draw complete circles, instead of extracting the shape themselves. That isn't a problem to do. All you need to do is determine the best "radii" that can be enclosed inside each of the bounding boxes. This is simply the maximum of the width and height of each bounding box, then divide these quantities by 2.
After, create a 2D grid of co-ordinates through meshgrid that is the same size as the original image itself, then create a binary image such that the Euclidean distance between the centre of this bounding box with any point in this 2D grid less than the radius is set to logical true while the other positions are set to logical false.
In other words, do this:
clear all;
close all;
im = imread('http://i.stack.imgur.com/qychC.jpg');
%im = imread('http://i.stack.imgur.com/wQLPi.jpg');
%im = imread('http://i.stack.imgur.com/mZMBA.jpg');
im_bw = im2bw(im);
s = regionprops(im_bw, 'Area', 'BoundingBox');
areas = [s.Area].';
s2 = s(areas > 2000);
out = false(size(im_bw));
for idx = 1 : numel(s2) %// For each potential circle we have...
bb = floor(s2(idx).BoundingBox); %// Could be floating point, so floor it
%// Copy over pixels from original bw image to output
out(bb(2):bb(2)+bb(4)-1, bb(1):bb(1)+bb(3)-1) = im_bw(bb(2):bb(2)+bb(4)-1, bb(1):bb(1)+bb(3)-1);
end
figure;
imshow(out);
%// Image that contains all of our final circles
out2 = false(size(im_bw));
[X,Y] = meshgrid(1:size(im_bw,2), 1:size(im_bw,1)); %// Find a 2D grid of co-ordinates
for idx = 1 : numel(s2) %// For each circle we have...
bb = floor(s2(idx).BoundingBox); %// Could be floating point, so floor it
cenx = bb(1) + (bb(3) / 2.0); %// Get the centre of the bounding box
ceny = bb(2) + (bb(4) / 2.0);
radi = max(bb(3), bb(4)) / 2; %// Find the best radius
tmp = ((X - cenx).^2 + (Y - ceny).^2) <= radi^2; %// Draw our circle and place in a temp. image
out2 = out2 | tmp; %// Add this circle on top of our output image
end
figure;
imshow(out2);
This script now shows you the original extracted shapes, and the best "circles" that describes these shapes in two separate figures. Bear in mind that this is a bit different than what I showed you previously with one circle. What I have to do now is allocate a blank image, then incrementally add each circle to this new image. For each circle, I create a temporary binary image that has just a circle I'm looking for, then I add this on top of the new image. At the end, we will show all of the circles in the image that are fully drawn as you desire.
This is what I get for the best circle for each of your images:
Good luck!

How to remove outline circle in this image?

Can you guys suggest possible ways on how to remove the circle outline in this image? Imfindcircles doesnt work for me. Can you suggest other methods? http://i.stack.imgur.com/RuD7v.jpg
Assuming BW to be the binary image that has the outline circled and which is to be removed, you can use an approach based on regionprops -
perimtrs = regionprops(BW, 'Perimeter'); %// perimeters for each connected component
px = regionprops(BW, 'PixelIdxList'); %// pixel list for each connected component
[~,idx] = max(struct2array(perimtrs)); %// get the component with max perimeter
%// that represents the outline circle
BW(px(idx).PixelIdxList) = 0; %// Set all pixels of the outline circle to zero,
%// that is they are removed
If you would like to be on the safest side with the functionality, you can use BoundingBox properties from regionprops instead of 'Perimeter' as shown here -
%// Get the bounding box properties for each connected component
perimtrs = regionprops(BW, 'BoundingBox');
%// Get bounding box area for each component and get the ID for the largest
%// box that corresponds to the outline circle
bound_box = reshape(struct2array(perimtrs),4,[]);
bound_box_area = bound_box(3,:).*bound_box(4,:);
[~,idx] = max(bound_box_area);
%// Set the pixels corresponding to the outline circle to zeros
px = regionprops(BW, 'PixelIdxList');
BW(px(idx).PixelIdxList) = 0;
Alternatively, you can avoid the second use of regionprops to get the pixel list with a call to regionprops and that might be efficient with performance, but I haven't not tested, so can't guarantee that. The new approach would look something like this -
perimtrs = regionprops(BW, 'Perimeter');
[~,idx] = max(struct2array(perimtrs))
[L,num] = bwlabel( BW ); %// Label connected components
BW(L==idx)=0; %// Select all pixels corresponding to label idx and set those to zero
Similarly, you can mix this bwlabel approach with BoundingBox of regionprops.
OK so here goes one hypotheses that does not assume the interface to be a circle, neither to be a single region, or having the largest perimeter.
%Assume A as your original image (left image), and bin_A as your binary image (right image)
thres=graythresh(A)
mask_A=im2bw(A,thres);
mask_A=imerode(mask_A,ones(3));
bin_A=bin_A.*mask_A;

MATLAB: Display image inside circle

As part of a circle recognition program, I have a background image with a geometrical circle with known coordinates and radius. I want the inside part of the circle filled by an image and the outside left alone. My best idea is some sort of circular mask, but I am not sure this is the best approach. Any suggestions?
X = imread('X.jpg'); % Background image jpg
Y = imread('Y.jpg'); % Filling image jpg
cent = [100,100]; % Center of circle
rad = 20; % Circle radius
% Fill circle ?
...
I have not provided the extended code, due to confidentiality.
I think the hard part was done by whomever authored this: http://matlab.wikia.com/wiki/FAQ#How_do_I_create_a_circle.3F
Assumptions:
I'm assuming you're not going to specify points that are out of range of the image (i.e. I'm not adding validation here).
I use the background image to relate the "center" of your circle to the coordinates.
I'm assuming radius is in pixels.
I didn't create a background image with a circle of known radius because I don't think that's necessary to create the filling effect you're looking for (unless I'm missing something).
Code:
X = imread('rdel_x.png'); % Background image jpg (I used a random image but this can be your blank + geometric circle)
Y = imread('rdel_y.png'); % Filling image jpg
cent = [100,150]; % Center of circle
rad = 70; % Circle radius
% make a mesh grid to provide coords for the circle (mask)
% taken from http://matlab.wikia.com/wiki/FAQ#How_do_I_create_a_circle.3F
[columnsInImage rowsInImage] = meshgrid(1:size(X,2), 1:size(X,1));
% circle points in pixels:
circlePixels = (rowsInImage - cent(1)).^2 ...
+ (columnsInImage - cent(2)).^2 <= rad.^2;
circlePixels3d=repmat(circlePixels,[1 1 3]); % turn into 3 channel (assuming X and Y are RGB)
X((circlePixels3d)) = Y((circlePixels3d)); % assign the filling image pixels to the background image for pixels where it's the desired circle
imagesc(X);
axis image off
Result: from left to right, background image, filling image, result from above code.
Edit: you might not even need the background image if everything is encapsulated in your coordinates. e.g. try appending this to the above code...
Z=zeros(size(X),'uint8'); % same size as your background
Z(circlePixels3d) = Y(circlePixels3d);
figure; % new fig
imagesc(Z);
axis image off