What is the shortest way to find coordinates of figure left/right/top/bottom edges? 4 coordinates (2 horizontal, 2 vertical lines) are enough.
Tried to flip, transpose, etc. My mind is gonna blow :/.
[EDIT]: Image is binary. Figure is represented by 1's.
You can try to get its bounding box with the regionprops() function.
regionprops(img,'BoundingBox')
The result is (x,y) upper left coordinates x_width, y_width, size of the box.
I get [45.5000000000000 45.5000000000000 174 107] in your image.
The shortest solution i made:
% I - image array
V = sum(I,2);
edge_top = find(V,1,'first');
edge_bot = find(V,1,'last');
H = sum(I,1);
edge_left = find(H,1,'first');
edge_right = find(H,1,'last');
Related
My binary image has rectangular rotated objects of known size on it. I'd like to get the object inclination using axis-aligned bounding box that MATLAB's regionprops returns. What are my suggestions:
Let bounding box width be W, side of rectangle be C and inclination alpha
Then
Using Weierstrass substitution
After some simplification:
Solving the equation for tan(alpha/2) with
For any nonzero inclination discriminant is positive.
Logic seems to be OK, so as math. Could you please point where I make a mistake, or what is a better way to get inclination?
Here is corresponding MATLAB code:
img = false(25,25);
img(5:16,5:16) = true;
rot_img = imrotate(img, 30, 'crop');
props = regionprops(bwlabel(rot_img),'BoundingBox');
bbox = cat(1,props.BoundingBox);
w = bbox(3);
h = 12;
a = -1*(1+w/h); b = 2; c = 1 - w/h;
D = b^2 - 4*a*c;
alpha = 2*atand((-b + sqrt(D))/(2*a));
%alpha = 25.5288
EDIT Thank you for trigonometry hints. They significantly simplify the calculations, but they give wrong answer. As I now understand, the question is asked in wrong way. The thing I really need is finding inclination of short lines (10-50 pixels) with high accuracy (+/- 0.5 deg), the lines' position is out of interest.
The approach used in the question and answers show better accuracy for long lines, for c = 100 error is less than 0.1 degree. That means we're into rasterization error here, and need subpixel accuracy. At the moment I have only one algorithm that solves the problem - Radon transform, but I hope you can recommend something else.
p = bwperim(rot_img);
theta=0:0.1:179.9;
[R,xp] = radon(p,theta); %Radon transform of contours
a=imregionalmax(R,true(3,3)); %Regional maxima of the transform
[r,c]=find(a); idx=sub2ind(size(a),r,c); maxvals=R(idx);
[val,midx]=sort(maxvals,'descend'); %Choose 4 highest maxima
mean(rem(theta(c(midx(1:4))),90)) %And average corresponding angles
%29.85
If rectangle is square:
w/c=sin(a)+cos(a)
(w/c)^2=1+sin(2a)
sin(2a)=(w/c)^2-1
a=0.5*arcsin((w/c)^2-1)
May be use regionprops function with 'Orientation' option...
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 am trying to draw the lines or the edges of a cone using plot3 in matlab. Any help please? I do not need the surface. I need the edges only. SO that I can patch something on it. A useful link. But i need the circle at the bottom:
https://patentimages.storage.googleapis.com/US8514658B2/US08514658-20130820-D00021.png
Few horizontal lines are fine. But no tilted line as i need to patch something inside.
cylinder is your friend here...
You just need to pass it a vector of radii* and transpose the output*...
* negative radii tending to zero will flip the order so the apex is on top...
* so it draws rings not lines from the base to the apex
numRings = 10;
numPointsAround = 100;
[x,y,z] = cylinder(linspace(-1,0,nlines),numPointsAround);
plot3(y.',x.',z.','-k')
I think this is what you want. Most of the answer is directly taken from the above answer by #RTL.
numRings = 2;
numPointsAround = 100;
[x,y,z] = cylinder(linspace(-1,0,numRings),numPointsAround);
plot3(y.',x.',z.','-k')
hold on;line([-0.5878;0], [0.809;0],[0;1]);
hold on;line([0.9511;0], [-0.309;0],[0;1]);
axis square
I have a video of moving hose in an experiment and I need to detect certain points in that hose and calculate the amplitude of their movements, I am using the code below and I am able to extract the required point using detectSURFFeatures, the function get many unnecessary points so I am using cuba = ref_pts.selectStrongest(5); to choose only five points, the problem is I can not get a function to put a bounding box about this 5 points and get their pixel values through the video, Kindly advice what functions can be used, thanks :)
clear;
clc;
% Image aquisition from Video and converting into gray scale
vidIn = VideoReader('ItaS.mp4');
%% Load reference image, and compute surf features
ref_img = read(vidIn, 1);
ref_img_gray = rgb2gray(ref_img);
ref_pts = detectSURFFeatures(ref_img_gray);
[ref_features, ref_validPts] = extractFeatures(ref_img_gray, ref_pts);
figure; imshow(ref_img);
hold on; plot(ref_pts.selectStrongest(5));
cuba = ref_pts.selectStrongest(5);
stats1 = round(cuba.Location);
If you want to find the bounding box which covers all the five points you selected: stats1 now contains (x, y) coordinates of the selected 5 points. Find min and max for x and y coordinates. min values of x and y gives you the starting point of the rectangle. Width and height of the bounding box is now the difference of max and min in y and x directions.
If you want to extract the part of the original image inside the bounding box: just copy that part to another variable as you want. Consider the following example.
img2 = img1(y:h, x:w, :)
Here, x and y are the x and y coordinates of the top left corner of the bounding box. w and h are the width and height of the bounding box.
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!