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How do I detect defect/missing tablets in the tablet strips. Assuming there is one missing tablet in the tablet strip. I've tried stdfilt(), but the image contains alot noise .I ve also tried average and median filtring such as canny and prewitt. I also added noise such as salt and paper to the image.
Is there any othe segmentation method? Any coding will be helpful
I2=rgb2gray(I);
J = imnoise(I2,'salt & pepper',0.02);
figure
imshow(J)
Kaverage = filter2(fspecial('average',3),J)/255;
figure
imshow(Kaverage)
Kmedian = medfilt2(J);
imshowpair(Kaverage,Kmedian,'montage')
BW1 = edge(Kmedian,'Canny');
BW2 = edge(Kmedian,'Prewitt');`
Here's an approach based on my comment
% reduce to grayscale
d=rgb2gray(your_img);
% find edge and blur a bit so we can find circles
d2=conv2(edge(d),ones(9),'same');
d2=max(d2(:))-d2;
% find circles
Rmin = 71; Rmax = 80;
[center, radius] = imfindcircles(d2,[Rmin Rmax],'Sensitivity',0.98);
% Display what we found
imagesc(d);axis square
hold on;
viscircles(center,radius);
plot(center(:,1),center(:,2),'yx','LineWidth',2);
hold off;
% histogram of each circle content:
[x, y]=meshgrid(1:size(d,2),1:size(d,1));
for n=1:numel(radius)
circle_pixels{n}=find ((x-center(n,1)).^2+(y-center(n,2)).^2<=radius(n).^2);
h(:,n) = histcounts(d(circle_pixels{n}),0:max(d(:)) );
subplot(2,5,n); plot(h(:,n));title(['circle # ' num2str(n)]);
end
Now we can see how the intensity is distributed in each circle and choose a metric to discriminate the missing pills. We can see that for the missing pills (#5,#9,#10) we have a less simple distribution of intensities (more than one peak) , and in particular a saturation of the maximal intensity that happen because of the glare reflection of the foil probably. So you can choose now a threshold based on that, or any other statistical metric you want (# of peaks in the distributions etc)...
A similar question was asked before, unfortunately I cannot comment Samgaks answer so I open up a new post with this one. Here is the link to the old question:
How to calculate ray in real-world coordinate system from image using projection matrix?
My goal is to map from image coordinates to world coordinates. In fact I am trying to do this with the Camera Intrinsics Parameters of the HoloLens Camera.
Of course this mapping will only give me a ray connecting the Camera Optical Centre and all points, which can lie on that ray. For the mapping from image coordinates to world coordinates we can use the inverse camera matrix which is:
K^-1 = [1/fx 0 -cx/fx; 0 1/fy -cy/fy; 0 0 1]
Pcam = K^-1 * Ppix;
Pcam_x = P_pix_x/fx - cx/fx;
Pcam_y = P_pix_y/fy - cy/fy;
Pcam_z = 1
Orientation of Camera Coordinate System and Image Plane
In this specific case the image plane is probably at Z = -1 (However, I am a bit uncertain about this). The Section Pixel to Application-specified Coordinate System on page HoloLens CameraProjectionTransform describes how to go form pixel coordinates to world coordinates. To what I understand two signs in the K^-1 are flipped s.t. we calculate the coordinates as follows:
Pcam_x = (Ppix_x/fx) - (cx*(-1)/fx) = P_pix_x/fx + cx/fx;
Pcam_y = (Ppix_y/fy) - (cy*(-1)/fy) = P_pix_y/fy + cy/fy;
Pcam_z = -1
Pcam = (Pcam_x, Pcam_y, -1)
CameraOpticalCentre = (0,0,0)
Ray = Pcam - CameraOpticalCentre
I do not understand how to create the Camera Intrinsics for the case of the image plane being at a negative Z-coordinate. And I would like to have a mathematical explanation or intuitive understanding of why we have the sign flip (P_pix_x/fx + cx/fx instead of P_pix_x/fx - cx/fx).
Edit: I read in another post that the thirst column of the camera matrix has to be negated for the case that the camera is facing down the negative z-direction. This would explain the sign flip. However, why do we need to change the sign of the third column. I would like to have a intuitive understanding of this.
Here the link to the post Negation of third column
Thanks a lot in advance,
Lisa
why do we need to change the sign of the third column
To understand why we need to negate the third column of K (i.e. negate the principal points of the intrinsic matrix) let's first understand how to get the pixel coordinates of a 3D point already in the camera coordinates frame. After that, it is easier to understand why -z requires negating things.
let's imagine a Camera c, and one point B in the space (w.r.t. the camera coordinate frame), let's put the camera sensor (i.e. image) at E' as in the image below. Therefore f (in red) will be the focal length and ? (in blue) will be the x coordinate in pixels of B (from the center of the image). To simplify things let's place B at the corner of the field of view (i.e. in the corner of the image)
We need to calculate the coordinates of B projected into the sensor d (which is the same as the 2d image). Because the triangles AEB and AE'B' are similar triangles then ?/f = X/Z therefore ? = X*f/Z. X*f is the first operation of the K matrix is. We can multiply K*B (with B as a column vector) to check.
This will give us coordinates in pixels w.r.t. the center of the image. Let's imagine the image is size 480x480. Therefore B' will look like this in the image below. Keep in mind that in image coordinates, the y-axis increases going down and the x-axis increases going right.
In images, the pixel at coordinates 0,0 is in the top left corner, therefore we need to add half of the width of the image to the point we have. then px = X*f/Z + cx. Where cx is the principal point in the x-axis, usually W/2. px = X*f/Z + cx is exactly as doing K * B / Z. So X*f/Z was -240, if we add cx (W/2 = 480/2 = 240) and therefore X*f/Z + cx = 0, same with the Y. The final pixel coordinates in the image are 0,0 (i.e. top left corner)
Now in the case where we use z as negative, when we divide X and Y by Z, because Z is negative, it will change the sign of X and Y, therefore it will be projected to B'' at the opposite quadrant as in the image below.
Now the second image will instead be:
Because of this, instead of adding the principal point, we need to subtract it. That is the same as negating the last column of K.
So we have 240 - 240 = 0 (where the second 240 is the principal point in x, cx) and the same for Y. The pixel coordinates are 0,0 as in the example when z was positive. If we do not negate the last column we will end up with 480,480 instead of 0,0.
Hope this helped a little bit
I have three sections (top, mid, bot) of grayscale images (3D). In each section, I have a point with coordinates (x,y) and intensity values [0-255]. The distance between each section is 20 pixels.
I created an illustration to show how those images were generated using a microscope:
Illustration
Illustration (side view): red line is the object of interest. Blue stars represents the dots which are visible in top, mid, bot section. The (x,y) coordinates of these dots are known. The length of the object remains the same but it can rotate in space - 'out of focus' (illustration shows a rotating line at time point 5). At time point 1, the red line is resting (in 2D image: 2 dots with a distance equal to the length of the object).
I want to estimate the x,y,z-coordinate of the end points (represents as stars) by using the changes in intensity, the knowledge about the length of the object and the information in the sections I have. Any help would be appreciated.
Here is an example of images:
Bot section
Mid section
Top section
My 3D PSF data:
https://drive.google.com/file/d/1qoyhWtLDD2fUy2zThYUgkYM3vMXxNh64/view?usp=sharing
Attempt so far:
enter image description here
I guess the correct approach would be to record three images with slightly different z-coordinates for your bot and your top frame, then do a 3D-deconvolution (using Richardson-Lucy or whatever algorithm).
However, a more simple approach would be as I have outlined in my comment. If you use the data for a publication, I strongly recommend to emphasize that this is just an estimation and to include the steps how you have done it.
I'd suggest the following procedure:
Since I do not have your PSF-data, I fake some by estimating the PSF as a 3D-Gaussiamn. Of course, this is a strong simplification, but you should be able to get the idea behind it.
First, fit a Gaussian to the PSF along z:
[xg, yg, zg] = meshgrid(-32:32, -32:32, -32:32);
rg = sqrt(xg.^2+yg.^2);
psf = exp(-(rg/8).^2) .* exp(-(zg/16).^2);
% add some noise to make it a bit more realistic
psf = psf + randn(size(psf)) * 0.05;
% view psf:
%
subplot(1,3,1);
s = slice(xg,yg,zg, psf, 0,0,[]);
title('faked PSF');
for i=1:2
s(i).EdgeColor = 'none';
end
% data along z through PSF's center
z = reshape(psf(33,33,:),[65,1]);
subplot(1,3,2);
plot(-32:32, z);
title('PSF along z');
% Fit the data
% Generate a function for a gaussian distibution plus some background
gauss_d = #(x0, sigma, bg, x)exp(-1*((x-x0)/(sigma)).^2)+bg;
ft = fit ((-32:32)', z, gauss_d, ...
'Start', [0 16 0] ... % You may find proper start points by looking at your data
);
subplot(1,3,3);
plot(-32:32, z, '.');
hold on;
plot(-32:.1:32, feval(ft, -32:.1:32), 'r-');
title('fit to z-profile');
The function that relates the intensity I to the z-coordinate is
gauss_d = #(x0, sigma, bg, x)exp(-1*((x-x0)/(sigma)).^2)+bg;
You can re-arrange this formula for x. Due to the square root, there are two possibilities:
% now make a function that returns the z-coordinate from the intensity
% value:
zfromI = #(I)ft.sigma * sqrt(-1*log(I-ft.bg))+ft.x0;
zfromI2= #(I)ft.sigma * -sqrt(-1*log(I-ft.bg))+ft.x0;
Note that the PSF I have faked is normalized to have one as its maximum value. If your PSF data is not normalized, you can divide the data by its maximum.
Now, you can use zfromI or zfromI2 to get the z-coordinate for your intensity. Again, I should be normalized, that is the fraction of the intensity to the intensity of your reference spot:
zfromI(.7)
ans =
9.5469
>> zfromI2(.7)
ans =
-9.4644
Note that due to the random noise I have added, your results might look slightly different.
I've been using a function file [ret]=drawellipse(x,y,a,b,angle,steps,color,img). Calling the function through a script file to draw random ellipses in image. But once i set the random center point(x,y), and random a, b, there is high possibility that the ellipses intersection would occur. How can i prevent the intersection? (I'm supposed to draw the ellipses that are all separate from each other)
Well, over here i have a function file which is to check whether the ellipses got overlap or not,overlap = overlap_ellipses(x0,y0,a0,b0,angle0,x1,y1,a1,b1,angle1). If the two ellipses are overlap, then the 'overlap=1', otherwise 'overlap=0'.
Based on all these, i tested in the command window:
x=rand(4,1)*400; % x and y are the random coodinates for the center of ellipses
y=rand(4,1)*400;
a=[50 69 30 60]; % major axis for a and b, i intend to use random also in the future
b=[20 40 10 40]; % minor axis
angle=[30 90 45 0]; % angle of ellipse
steps=10000;
color=[255 0 0]; % inputs for another function file to draw the ellipse
img=zeros(500,500,3);
The following i want to dispaly the ellipses if overlap==0, and 'if overlap==1', decrease the a and b, till there is no intersection. Lastly, to imshow the img.
for i=1:length(x)
img=drawellipse(x(i),y(i),a(i),b(i),angle(i),steps,color,img);
end
For me now, i have difficulty in coding the middle part. How can i use the if statement to get the value of overlap and how to make the index corresponding to the ellipse i need to draw.
i tested a bit like
for k=1:(length(x)-1)
overlap = overlap_ellipses(x(1),y(1),a(1),b(1),angle(1),x(1+k),y(1+k),a(1+k),b(1+k),angle(1+k))
end
it returns
overlap=0
overlap=0
overlap=1
it is not [0 0 1]. I can't figure it out, thus stuck in the process.
The final image shoule look like the picture in this voronoi diagram of ellipses.
(There is no intersection between any two ellipses)
Assuming you are drawing the ellipses into a raster graphics image, you could calculate the pixels you would have to draw for an ellipse, check whether these pixels in the image are still of the background color, and draw the ellipse only if the answer is yes, otherwise reject it (because something else, i.e. another ellipse, is in the way) and try other x,y,a and b.
Alternatively, you could split your image into rectangles (not neccessarily of equal size) and place one ellipse in each of those, picking x,y,a,b such that no ellipse exceeds its rectangle - then the ellipses cannot overlap either, but it depends on how much "randomness" your ellipse placing should have whether this suffices.
The mathematically rigorous way would be to store x,y,a,b of each drawn ellipse and for each new ellipse, do pairwise checks with each of those whether they have common points by solving a system of two quadratic equations. However, this might be a bit complicated, especially once the angle is not 0.
Edit in response to the added code: Instead of fixing all x's and y's before the loop, you can determine them inside the loop. Since you know how many ellipses you want, but not how many you have to sample, you need a while loop. The test loop you give may come in handy, but you need to compare all previous ellipses to the one created in the loop iteration, not the first one.
i=1;
while (i<=4) %# or length(a), or, more elegantly, some pre-defined max
x(i) = rand*400; y(i) = rand*400; %# or take x and y as givren and decrease a and b
%# now, check overlap for given center
overlap = false;
for k=1:(i-1)
overlap = overlap || overlap_ellipses(x(i),y(i),a(i),b(i),angle(i),x(k),y(k),a(k),b(k),angle(k))
end
if (~overlap)
img = drawellipse(x(i),y(i),a(i),b(i),angle(i),steps,color,img);
i = i+1; %# determine next ellipse
end %# else x(i) and y(i) will be overwritten in next while loop iteration
end
Of course, if a and b are fixed, it may happen that no ellipse fits the image dimensions if the already present ones are unfortunately placed, resulting in an infinite loop.
Regarding your plan of leaving the center fixed and decreasing the ellipse's size until it fits: where does your overlap_ellipses method come from? Maybe itcan be adapted to return a factor by which one ellipse needs to be shrinked to fit next to the other (and 1 if it fits already)?
The solution proposed by #arne.b (the first one) is a good way to rasterize non-overlapping ellipses.
Let me illustrate that idea with an example. I will be extending my previous answer:
%# color image
I = imread('pears.png');
sz = size(I);
%# parameters of ellipses
num = 7;
h = zeros(1,num);
clr = lines(num); %# color of each ellipse
x = rand(num,1) .* sz(2); %# center x-coords
y = rand(num,1) .* sz(1); %# center y-coords
a = rand(num,1) .* 200; %# major axis length
b = rand(num,1) .* 200; %# minor axis length
angle = rand(num,1) .* 360; %# angle of rotation
%# label image, used to hold rasterized ellipses
BW = zeros(sz(1),sz(2));
%# randomly place ellipses one-at-a-time, skip if overlaps previous ones
figure, imshow(I)
axis on, hold on
for i=1:num
%# ellipse we would like to draw directly on image matrix
[ex,ey] = calculateEllipse(x(i),y(i), a(i),b(i), angle(i), 100);
%# lets plot the ellipse (overlayed)
h(i) = plot(ex,ey, 'LineWidth',2, 'Color',clr(i,:));
%# create mask for image pixels inside the ellipse polygon
mask = poly2mask(ex,ey,sz(1),sz(2));
%# get the perimter of this mask
mask = bwperim(mask,8);
%# skip if there is an existing overlapping ellipse
if any( BW(mask)~=0 ), continue, end
%# use the mask to place the ellipse in the label image
BW(mask) = i;
end
hold off
legend(h, cellstr(num2str((1:num)','Line%d')), 'Location','BestOutside') %'
%# set pixels corresponding to ellipses using specified colors
clr = im2uint8(clr);
II = I;
for i=1:num
BW_ind = bsxfun(#plus, find(BW==i), prod(sz(1:2)).*(0:2));
II(BW_ind) = repmat(clr(i,:), [size(BW_ind,1) 1]);
end
figure, imshow(II, 'InitialMagnification',100, 'Border','tight')
Note how the overlap test is performed in the order the ellipses are added, thus after Line1 (blue) and Line2 (green) are drawn, Line3 (red) will be skipped because it overlaps one of the previous ones, and so on for the rest...
One option is to keep track of all the ellipses already drawn, and to make sure the next set of [x,y,a,b] does not produce a new ellipse which intersects with the existing ones. You can either invoke random numbers until you come up with a set that fulfills the condition, or once you have a set which violates the condition, decrease the values of a and/or b until no intersection occurs.
I have an image in MATLAB:
im = rgb2gray(imread('some_image.jpg');
% normalize the image to be between 0 and 1
im = im/max(max(im));
And I've done some processing that resulted in a number of points that I want to highlight:
points = some_processing(im);
Where points is a matrix the same size as im with ones in the interesting points.
Now I want to draw a circle on the image in all the places where points is 1.
Is there any function in MATLAB that does this? The best I can come up with is:
[x_p, y_p] = find (points);
[x, y] = meshgrid(1:size(im,1), 1:size(im,2))
r = 5;
circles = zeros(size(im));
for k = 1:length(x_p)
circles = circles + (floor((x - x_p(k)).^2 + (y - y_p(k)).^2) == r);
end
% normalize circles
circles = circles/max(max(circles));
output = im + circles;
imshow(output)
This seems more than somewhat inelegant. Is there a way to draw circles similar to the line function?
You could use the normal PLOT command with a circular marker point:
[x_p,y_p] = find(points);
imshow(im); %# Display your image
hold on; %# Add subsequent plots to the image
plot(y_p,x_p,'o'); %# NOTE: x_p and y_p are switched (see note below)!
hold off; %# Any subsequent plotting will overwrite the image!
You can also adjust these other properties of the plot marker: MarkerEdgeColor, MarkerFaceColor, MarkerSize.
If you then want to save the new image with the markers plotted on it, you can look at this answer I gave to a question about maintaining image dimensions when saving images from figures.
NOTE: When plotting image data with IMSHOW (or IMAGE, etc.), the normal interpretation of rows and columns essentially becomes flipped. Normally the first dimension of data (i.e. rows) is thought of as the data that would lie on the x-axis, and is probably why you use x_p as the first set of values returned by the FIND function. However, IMSHOW displays the first dimension of the image data along the y-axis, so the first value returned by FIND ends up being the y-coordinate value in this case.
This file by Zhenhai Wang from Matlab Central's File Exchange does the trick.
%----------------------------------------------------------------
% H=CIRCLE(CENTER,RADIUS,NOP,STYLE)
% This routine draws a circle with center defined as
% a vector CENTER, radius as a scaler RADIS. NOP is
% the number of points on the circle. As to STYLE,
% use it the same way as you use the rountine PLOT.
% Since the handle of the object is returned, you
% use routine SET to get the best result.
%
% Usage Examples,
%
% circle([1,3],3,1000,':');
% circle([2,4],2,1000,'--');
%
% Zhenhai Wang <zhenhai#ieee.org>
% Version 1.00
% December, 2002
%----------------------------------------------------------------
Funny! There are 6 answers here, none give the obvious solution: the rectangle function.
From the documentation:
Draw a circle by setting the Curvature property to [1 1]. Draw the circle so that it fills the rectangular area between the points (2,4) and (4,6). The Position property defines the smallest rectangle that contains the circle.
pos = [2 4 2 2];
rectangle('Position',pos,'Curvature',[1 1])
axis equal
So in your case:
imshow(im)
hold on
[y, x] = find(points);
for ii=1:length(x)
pos = [x(ii),y(ii)];
pos = [pos-0.5,1,1];
rectangle('position',pos,'curvature',[1 1])
end
As opposed to the accepted answer, these circles will scale with the image, you can zoom in an they will always mark the whole pixel.
Hmm I had to re-switch them in this call:
k = convhull(x,y);
figure;
imshow(image); %# Display your image
hold on; %# Add subsequent plots to the image
plot(x,y,'o'); %# NOTE: x_p and y_p are switched (see note below)!
hold off; %# Any subsequent plotting will overwrite the image!
In reply to the comments:
x and y are created using the following code:
temp_hull = stats_single_object(k).ConvexHull;
for k2 = 1:length(temp_hull)
i = i+1;
[x(i,1)] = temp_hull(k2,1);
[y(i,1)] = temp_hull(k2,2);
end;
it might be that the ConvexHull is the other way around and therefore the plot is different. Or that I made a mistake and it should be
[x(i,1)] = temp_hull(k2,2);
[y(i,1)] = temp_hull(k2,1);
However the documentation is not clear about which colum = x OR y:
Quote: "Each row of the matrix contains the x- and y-coordinates of one vertex of the polygon. "
I read this as x is the first column and y is the second colum.
In newer versions of MATLAB (I have 2013b) the Computer Vision System Toolbox contains the vision.ShapeInserter System object which can be used to draw shapes on images. Here is an example of drawing yellow circles from the documentation:
yellow = uint8([255 255 0]); %// [R G B]; class of yellow must match class of I
shapeInserter = vision.ShapeInserter('Shape','Circles','BorderColor','Custom','CustomBorderColor',yellow);
I = imread('cameraman.tif');
circles = int32([30 30 20; 80 80 25]); %// [x1 y1 radius1;x2 y2 radius2]
RGB = repmat(I,[1,1,3]); %// convert I to an RGB image
J = step(shapeInserter, RGB, circles);
imshow(J);
With MATLAB and Image Processing Toolbox R2012a or newer, you can use the viscircles function to easily overlay circles over an image. Here is an example:
% Plot 5 circles at random locations
X = rand(5,1);
Y = rand(5,1);
% Keep the radius 0.1 for all of them
R = 0.1*ones(5,1);
% Make them blue
viscircles([X,Y],R,'EdgeColor','b');
Also, check out the imfindcircles function which implements the Hough circular transform. The online documentation for both functions (links above) have examples that show how to find circles in an image and how to display the detected circles over the image.
For example:
% Read the image into the workspace and display it.
A = imread('coins.png');
imshow(A)
% Find all the circles with radius r such that 15 ≤ r ≤ 30.
[centers, radii, metric] = imfindcircles(A,[15 30]);
% Retain the five strongest circles according to the metric values.
centersStrong5 = centers(1:5,:);
radiiStrong5 = radii(1:5);
metricStrong5 = metric(1:5);
% Draw the five strongest circle perimeters.
viscircles(centersStrong5, radiiStrong5,'EdgeColor','b');
Here's the method I think you need:
[x_p, y_p] = find (points);
% convert the subscripts to indicies, but transposed into a row vector
a = sub2ind(size(im), x_p, y_p)';
% assign all the values in the image that correspond to the points to a value of zero
im([a]) = 0;
% show the new image
imshow(im)