I have an intensity image where i marked with the impoly function a region of interest. I have a code that gives me back the vertices of the smallest rectangle still containing my polygon.
The rectangle is most of the times not aligned with the axes. I would like to receive the data inside the rectangle into a matrix form.I have been trying to find out the angle between the largest side of the rectangle and the x axis and then to rotate the image so the rectangle will be aligned with the axes.I will really appreciate any help or new ideas on how to extract the values inside my rectangle into a matrix.
Here is part of my code:
filename = uigetfile; %get the file name
obj = VideoReader(filename);
nFrames=obj.NumberOfFrames;
thisfig = figure();
for k = 1 : nFrames
this_frame = read(obj, k);
thisax = axes('Parent', thisfig);
image(this_frame, 'Parent', thisax);
if k==nFrames
title(thisax, sprintf('Frame #%d', k));
end
if k==1
result=input('How many polygons would you like to draw? ');
for i=1:result
handle=impoly;
accepted_pos = wait(handle);
BW = createMask(handle);
sparse_image=sparse(BW);
[XX, YY] = find(sparse_image);
[rectx,recty]=minboundrect(XX,YY); %the function that returns the vertices of the rectangle
points=[rectx(1),rectx(2),rectx(3),rectx(4);recty(1),recty(2),recty(3),recty(4)]
distance1=((points(1,2)-points(1,1))^2+(points(2,2)-points(2,1))^2)^0.5;
distance2=((points(1,3)-points(1,2))^2+(points(2,3)-points(2,2))^2)^0.5;
if(distance1>distance2) %which side of the rectangle is the largest
vector=[points(1,2)-points(1,1),points(2,2)-points(2,1)];
else
vector=[points(1,3)-points(1,2),points(2,3)-points(2,2)];
end
angleInDegrees = atan2(vector(2), vector(1)) * 180 / pi; %supposed angle between the largest side and the x axis
end
end
To be clear: I get a video and split it into frames and i need to follow a certain area in all of the frames in the video. I'm dealing with a video but i split it into frames so i'm really dealing with images. The rectangle i get is slanted most of the times,i.e not aligned with the axes of my image.
Confusing question.
The code refers to videos, not images.
Is the question about rotating an image? Is the question about retrieving a sub-matrix from an image?
Is the posted code really relevant to the question?
What have you tried?
From the doc:
this_frame = read(obj, k);
video = read(obj) reads in all video frames from the file associated with obj. The read method returns a H-by-W-by-B-by-F matrix, video, where H is the image frame height, W is the image frame width, B is the number of bands in the image (for example, 3 for RGB), and F is the number of frames read.
It seems that you do have the coordinates:
[rectx,recty]=minboundrect(XX,YY);
So I'd say that the matrix you're looking for is
A = this_frame(rectx(1):rectx(2), recty(1):rectx(y), 1, 1); % Red band
Please edit your question for clarity.
Related
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;
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!
I have skeletonize binary image and the junction information. I want to draw circle at junction points as center and want to find the point of intersection of circle and binary image.
I have written the following code:
BW = imread('circles.png');
imshow(BW);
BW2 = bwmorph(BW,'remove');
figure, imshow(BW2)
BW3 = bwmorph(BW,'skel',Inf);
figure, imshow(BW3)
BW3t = bwmorph(BW3,'thin');
figure, imshow(BW3t)
[rj, cj, re, ce] = findendsjunctions(BW3t, 1);
hold on
plot(cj(1),rj(1),'ob')
hold on
circle([cj(1),rj(1)],4,50,':r');
findendsjunctions.m and dependent file show.m can downloaded from here: http://www.csse.uwa.edu.au/~pk/research/matlabfns/LineSegments/findendsjunctions.m and here http://www.csse.uwa.edu.au/~pk/research/matlabfns/Misc/show.m respectively.
And circle.m can be downloaded from here: http://www.mathworks.co.uk/matlabcentral/fileexchange/2876-draw-a-circle/content/circle.m
I want to find whether circle intersect 2, 3 or 4 vessels around it (marked as star in an image). Even if circle transverse many times a single vessel but output should be one intersection point per vessel.
Please suggest how can I find the intersection of circle and binary vessels.
Thanks
I have found the point of intersection of circle and Binary image and the coordinates of 3 points (marked as star in an image provided with my question). I have change the function circle.m (mentioned in my question above ) to give output of all X and Y coordinates of circumference of the circle and then I have written the following matlab code:
[H, X, Y]=circle([cj(1),rj(1)],4,50,':r');
c = improfile(BW3t,X,Y)
x=1:length(c)
figure
plot(x, c,'r')
[maxtab, mintab]=peakdet(c, 1)
[pks,locs] = findpeaks(c)
pt1=[X(locs(1)) Y(locs(1))]
pt2=[X(locs(2)) Y(locs(2))]
pt3=[X(locs(3)) Y(locs(3))]
hold on
plot(pt1(1),pt1(2),'om','LineWidth',2)
hold on
plot(pt2(1),pt2(2),'og','LineWidth',2)
hold on
plot(pt3(1),pt3(2),'ob','LineWidth',2)
pt1, pt2 pt3 are three points where circle cuts the binary image
Below is an arbitrary hand-drawn Intensity profile of a line in an image:
The task is to draw the line. The profile can be approximated to an arc of a circle or ellipse.
This I am doing for camera calibration. Since I do not have the actual industrial camera, I am trying to simulate the correction needed for calibration.
The question can be rephrased as I want pixel values which will follow a plot similar to the above. I want to do this using program (Preferably using opencv) and not manually enter these values because I have thousands of pixels in the line.
An algorithm/pseudo code will suffice. Also please note that I do not have any actual Intensity profile, otherwise I would have read those values.
When will you encounter such situation ?
Suppose you take a picture (assuming complete white) from a Camera, your object being placed on table, and camera just above it in vertical direction. The light coming on the center of the picture vertically downward from the camera will be stronger in intensity as compared to the light reflecting at the edges. You measure pixel values across any line in the Image, you will find intensity curve like shown above. Since I dont have camera for the time being, I want to emulate this situation. How to achieve this?
This is not exactly image processing, rather image generation... but anyways.
Since you want an arc, we still need three points on that arc, lets take the first, middle and last point (key characteristics in my opinion):
N = 100; % number of pixels
x1 = 1;
x2 = floor(N/2);
x3 = N;
y1 = 242;
y2 = 255;
y3 = 242;
and now draw a circle arc that contains these points.
This problem is already discussed here for matlab: http://www.mathworks.nl/matlabcentral/newsreader/view_thread/297070
x21 = x2-x1; y21 = y2-y1;
x31 = x3-x1; y31 = y3-y1;
h21 = x21^2+y21^2; h31 = x31^2+y31^2;
d = 2*(x21*y31-x31*y21);
a = x1+(h21*y31-h31*y21)/d; % circle center x
b = y1-(h21*x31-h31*x21)/d; % circle center y
r = sqrt(h21*h31*((x3-x2)^2+(y3-y2)^2))/abs(d); % circle radius
If you assume the middle value is always larger (and thus it's the upper part of the circle you'll have to plot), you can draw this with:
x = x1:x3;
y = b+sqrt(r^2-(x-a).^ 2);
plot(x,y);
you can adjust the visible window with
xlim([1 N]);
ylim([200 260]);
which gives me the following result:
I have two images. One is the original, and the another is rotated.
Now, I need to discover the angle that the image was rotated. Until now, I thought about discovering the centroids of each color (as every image I will use has squares with colors in it) and use it to discover how much the image was rotated, but I failed.
I'm using this to discover the centroids and the color in the higher square in the image:
i = rgb2gray(img);
bw = im2bw(i,0.01);
s = regionprops(bw,'Centroid');
centroids = cat(1, s.Centroid);
colors = impixel(img,centroids(1),centroids(2));
top = max(centroids);
topcolor = impixel(img,top(1),top(2));
You can detect the corners of one of the colored rectangles in both the image and the rotated version, and use these as control points to infer the transformation between the two images (like in image registration) using the CP2TFORM function. We can then compute the angle of rotation from the affine transformation matrix:
Here is an example code:
%# read first image (indexed color image)
[I1 map1] = imread('http://i.stack.imgur.com/LwuW3.png');
%# constructed rotated image
deg = -15;
I2 = imrotate(I1, deg, 'bilinear', 'crop');
%# find blue rectangle
BW1 = (I1==2);
BW2 = imrotate(BW1, deg, 'bilinear', 'crop');
%# detect corners in both
p1 = corner(BW1, 'QualityLevel',0.5);
p2 = corner(BW2, 'QualityLevel',0.5);
%# sort corners coordinates in a consistent way (counter-clockwise)
p1 = sortrows(p1,[2 1]);
p2 = sortrows(p2,[2 1]);
idx = convhull(p1(:,1), p1(:,2)); p1 = p1(idx(1:end-1),:);
idx = convhull(p2(:,1), p2(:,2)); p2 = p2(idx(1:end-1),:);
%# make sure we have the same number of corner points
sz = min(size(p1,1),size(p2,1));
p1 = p1(1:sz,:); p2 = p2(1:sz,:);
%# infer transformation from corner points
t = cp2tform(p2,p1,'nonreflective similarity'); %# 'affine'
%# rotate image to match the other
II2 = imtransform(I2, t, 'XData',[1 size(I1,2)], 'YData',[1 size(I1,1)]);
%# recover affine transformation params (translation, rotation, scale)
ss = t.tdata.Tinv(2,1);
sc = t.tdata.Tinv(1,1);
tx = t.tdata.Tinv(3,1);
ty = t.tdata.Tinv(3,2);
translation = [tx ty];
scale = sqrt(ss*ss + sc*sc);
rotation = atan2(ss,sc)*180/pi;
%# plot the results
subplot(311), imshow(I1,map1), title('I1')
hold on, plot(p1(:,1),p1(:,2),'go')
subplot(312), imshow(I2,map1), title('I2')
hold on, plot(p2(:,1),p2(:,2),'go')
subplot(313), imshow(II2,map1)
title(sprintf('recovered angle = %g',rotation))
If you can identify a color corresponding to only one component it is easier to:
Calculate the centroids for each image
Calculate the mean of the centroids (in x and y) for each image. This is the "center" of each image
Get the red component color centroid (in your example) for each image
Subtract the mean of the centroids for each image from the red component color centroid for each image
Calculate the ArcTan2 for each of the vectors calculated in 4), and subtract the angles. That is your result.
If you have more than one figure of each color, you need to calculate all possible combinations for the rotation and then select the one that is compatible with the other possible rotations.
I could post the code in Mathematica, if you think it is useful.
I would take a variant to the above mentioned approach:
% Crude binarization method to knock out background and retain foreground
% features. Note one looses the cube in the middle
im = im > 1
Then I would get the 2D autocorrelation:
acf = normxcorr2(im, im);
From this result, one can easily detect the peaks, and as rotation carries into the autocorrelation function (ACF) domain, one can ascertain the rotation by matching the peaks between the original ACF and the ACF from the rotated image, for example using the so-called Hungarian algorithm.