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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 have a z-image from a ToF Camera (Kinect V2). I do not have the pixel size, but I know that the depth image has a resolution of 512x424. I also know that I have a fov of 70.6x60 degrees.
I asked how to get the Pixel size before here. In Matlab this code looks like the following.
The brighter the pixel, the closer the object.
close all
clear all
%Load image
depth = imread('depth_0_30_0_0.5.png');
frame_width = 512;
frame_height = 424;
horizontal_scaling = tan((70.6 / 2) * (pi/180));
vertical_scaling = tan((60 / 2) * (pi/180));
%pixel size
with_size = horizontal_scaling * 2 .* (double(depth)/frame_width);
height_size = vertical_scaling * 2 .* (double(depth)/frame_height);
The image itself is a cube rotated by 30 degree, and can be seen here: .
What I want to do now is calculate the horizontal angle of a pixel to the camera-plane and the vertical angle to the camera plane.
I tried to do this with triangulation, I calculate the z-distance from one pixel to another, first in the horizontal direction and then in the vertical direction. I do this with a convolution:
%get the horizontal errors
dx = abs(conv2(depth,[1 -1],'same'));
%get the vertical errors
dy = abs(conv2(depth,[1 -1]','same'));
After this I calculate it via the atan, like this:
horizontal_angle = rad2deg(atan(with_size ./ dx));
vertical_angle = rad2deg(atan(height_size ./ dy));
horizontal_angle(horizontal_angle == NaN) = 0;
vertical_angle(vertical_angle == NaN) = 0;
Which gives back promising results, like these:
However, using a little bit more complex image like this, which is turned by 60° and 30°.
Gives back the same angle images for horizontal and vertical angles, which look like this:
After subtracting both images from each other, I get the following image - which shows that there is a difference between those two.
So, I have the following questions: How can I proof this concept? Is the math correct, and the test case is just poorly chosen? Is the angle difference from horizontal to vertical angles in the two images too close? Are there any errors in the calculation ?
While my previous code may looks good, it had a flaw. I tested it with smaller images (5x5,3x3 and so on) and saw, that there is an offset created by the difference picture (dx,dy) made by the convolution. It is simple not possible to map the difference picture (which holds the difference between two pixels) to the pixels itself, since the difference picture is smaller than the original one.
For a fast fix, I do a downsampling. So I changed the filter mask to:
%get the horizontal differences
dx = abs(conv2(depth,[1 0 -1],'valid'));
%get the vertical differences
dy = abs(conv2(depth,[1 0 -1]','valid'));
And changed the angle function to:
%get the angles by the tangent
horizontal_angle = rad2deg(atan(with_size(2:end-1,2:end-1)...
./ dx(2:end-1,:)))
vertical_angle = rad2deg(atan(height_size(2:end-1,2:end-1)...
./ dy(:,2:end-1)))
Also I used a padding function to get the angle map to the same size as the original images.
horizontal_angle = padarray(horizontal_angle,[1 1],0);
vertical_angle = padarray(vertical_angle[1 1],0);
I have some damaged line segments in a binary image and I need to fix them (make them straight and at their original thick). In order to do that I have to find the middle points of the segment, so when I check the neighborhood to find the thickness of the lines I'll be able to find where the pixel stops being 1 and becomes 0.
Assuming your damaged line segments are straight, you can use regionprops in MATLAB to find the center of each bounding box. Because if a segment is straight, its is always the diagonal line of the bounding box, thus the center of the box is also the center of the semgent.
Let's call the points A and B to reduce ambiguity, A(Xa, Ya) and B(Xb, Yb)
Let C be the middle point.
C(Xc, Yc)
Xc = (Xa + Xb) / 2
Yc = (Ya + Yb) / 2
We have four interesting numbers, two for the X coordinates and two for the Y coordinates.
Xmin = floor(Xc)
Xmax = ceil(Xc)
Ymin = floor(Yc)
Ymax = ceil(Yc)
The X coordinate of your middle point is either Xmin or Xmax, the Y coordinate of your middle point is either Ymin or Ymax.
So we have four potential points: (Xmin, Ymin), (Xmin, Ymax), (Xmax, Ymin), (Xmax, Ymax).
So, finally, we must decide which point is nearest to C.
Distance from P(Xp, Yp) to C(Xc, Yc) is:
sqrt(sqr(Xp - Xc) + sqr(Yp - Yc))
Calculate the four distance from the four points to C, choose the minimum and that will be the best possible middle point.
Suppose
A = [xa ya];
B = [xb yb];
then
C = round( mean([A;B]) );
Matlab's round rounds numbers towards their nearest integer, so this minimizes the (city-block) distance from the analytical center (mean([A;B])) to the nearest pixel.
If you want to keep sub-pixel precision (which is actually advisable for most calculations until an explicit map from a result to pixel indices is required), just drop the round and use only the mean part.
I have a bit of a problem categorizing points based on relative normals.
What I would like to do is use the information I got below to fit a simplified polygon to the points, with a bias towards 90 degree angles to an extent.
I have the rough (although not very accurate) normal lines for each point, but I'm not sure how to separate the data base on closeness of points and closeness of the normals. I plan to do a linear regression after chunking the points for each face, as the normal lines sometimes does not fit well with the actual faces (although they are close to each other for each face)
Example:
alt text http://a.imageshack.us/img842/8439/ptnormals.png
Ideally, I would like to be able to fit a rectangle around this data. However, the polygon does not need to be convex, nor does it have to be aligned with the axis.
Any hints as to how to achieve something like this would be awesome.
Thanks in advance
I am not sure if this is what you are looking for, but here's my attempt at solving the problem as I understood it:
I am using the angles of the normal vectors to find points belonging to each side of the rectangle (left, right, up, down), then simply fit a line to each.
%# create random data (replace those with your actual data)
num = randi([10 20]);
pT = zeros(num,2);
pT(:,1) = rand(num,1);
pT(:,2) = ones(num,1) + 0.01*randn(num,1);
aT = 90 + 10*randn(num,1);
num = randi([10 20]);
pB = zeros(num,2);
pB(:,1) = rand(num,1);
pB(:,2) = zeros(num,1) + 0.01*randn(num,1);
aB = 270 + 10*randn(num,1);
num = randi([10 20]);
pR = zeros(num,2);
pR(:,1) = ones(num,1) + 0.01*randn(num,1);
pR(:,2) = rand(num,1);
aR = 0 + 10*randn(num,1);
num = randi([10 20]);
pL = zeros(num,2);
pL(:,1) = zeros(num,1) + 0.01*randn(num,1);
pL(:,2) = rand(num,1);
aL = 180 + 10*randn(num,1);
pts = [pT;pR;pB;pL]; %# x/y coords
angle = mod([aT;aR;aB;aL],360); %# angle in degrees [0,360]
%# plot points and normals
plot(pts(:,1), pts(:,2), 'o'), hold on
theta = angle * pi / 180;
quiver(pts(:,1), pts(:,2), cos(theta), sin(theta), 0.4, 'Color','g')
hold off
%# divide points based on angle
[~,bin] = histc(angle,[0 45 135 225 315 360]);
bin(bin==5) = 1; %# combine last and first bin
%# fit line to each segment
hold on
for i=1:4
%# indices of points in this segment
idx = ( bin == i );
%# x/y or y/x
if i==2||i==4, xx=1; yy=2; else xx=2; yy=1; end
%# fit line
coeff = polyfit(pts(idx,xx), pts(idx,yy), 1);
fit(:,1) = 0:0.05:1;
fit(:,2) = polyval(coeff, fit(:,1));
%# plot fitted line
plot(fit(:,xx), fit(:,yy), 'Color','r', 'LineWidth',2)
end
hold off
I'd try the following
Cluster the points based on proximity and similar angle. I'd use single-linkage hierarchical clustering (LINKAGE in Matlab), since you don't know a priori how many edges there will be. Single linkage favors linear structures, which is exactly what you're looking for. As the distance criterion between two points you can use the euclidean distance between point coordinates multiplied by a function of the angle that increases very steeply as soon as the angle differs more than, say, 20 or 30 degrees.
Do (robust) linear regression into the data. Using the normals may or may not help. My guess is that they won't help too much. For simplicity, you may want to disregard the normals initially.
Find the intersections between the lines.
If you have to, you can always try and improve the fit, for example by constraining opposite lines to be parallel.
If that fails, you could try and implement the approach in THIS PAPER, which allows fitting multiple straight lines at once.
You could get the mean value for the X and Y coordinates for each side and then just make lines based on that.
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)