Rotating triangle in image - MATLAB - matlab

I'm writing a program for detecting basic shape in MATLAB.
when I detect the shape I evaluate its Orientation then I rotate the shape to make its Orientation Zero, then I can evaluate its Projection and specify what it is.
The problem is the MATLAB function: regionprops() doesn't evaluate Orientation of Triangle correctly.
I = zeros(256,256);
pos_triangle = [64 64 128 192 128 128];
Is = insertShape(I, 'FilledPolygon', pos_triangle);
imshow(Is)
original = Is;
originalBW = im2bw(original);
figure; imshow(originalBW);
S = regionprops(originalBW,'All');
bwr=imrotate(originalBW,S.Orientation);
S2 = regionprops(bwr,'Centroid','Orientation');
figure;imshow(bwr);
I use imrotate fnc for rotating image, I don't have problem with rotation, imrotate is working well. the problem is in calculating the Orientation [using 'regionprps()' fnc] of the image! for example:I want to turn the triangle from this position
http://postimg.org/image/4un4sc7pn/
Orentation value: -28.9621
So I rotate it 28.9621 degree to change its postion to this
http://postimg.org/image/x68opdrm3/
but the output is this:
http://postimg.org/image/yf15or8y3/
using their Orientation (or other possible property of image)
an other example: changing the position from Up-Left 2nd triangle to Up-Left 1st triangle

Here is a way to work it out. Notice that:
1) I don't have the Computer Vision System Toolbox so I can't use insertShape; instead I used the fill function to create the triangle in the image and then getframe to obtain the actual image. I guess it comes down to the same than using insertShape.
2) I used the 'FilledArea' property of regionprops to detect the triangular shape. That's easy when there is a single shape, but if there were many of them you would have to modify a bit the code.
3) I performed a rotation equal to -1* the orientation given by regionprops to bring it back to 0. You can of course change it. The rotated image is bigger to account for the rotation.
Here is the code:
clear
clc
close all
I = zeros(256,256);
pos_triangle = [64 64 128 192 128 128];
xt = [64 128 128];
yt = [64 192 128];
%// Since I don't have the Computer Vision System Toolbox I use 'fill'.
%//Is = insertShape(I, 'FilledPolygon', pos_triangle);
imshow(I)
hold on
fill(xt,yt,'w')
hold off
The original triangle:
%// Create image using getframe.
hFrame = getframe(gca);
original = hFrame.cdata;
originalBW = im2bw(original(:,:,1));
%// Remove border of the image.
originalBW = imclearborder(originalBW);
figure;
imshow(originalBW);
S = regionprops(originalBW,'FilledArea','Orientation');
%// Find region filled with the most pixels: that's the shape.
[a,b] =max([S.FilledArea]);
%// Get corresponding orientation
Orientation = S(b).Orientation;
%// Rotate by the inverse of the orientation; I'm not sur that's what you
%// want but it looks OK.
bwr=imrotate(originalBW,-1*Orientation);
S2 = regionprops(bwr,'Centroid','Orientation');
figure;imshow(bwr);
The rotated triangle, which looks like its orientation is 0.:
Hope that helps!

I'm not an expert in image processing but to achieve rotation is not hard. The only thing you need to know the center of your image. In Matlab, the center of an image in the upper left corner. To rotate an object around the center, you need to multiply every point by the rotation matrix. Let's rotate the triangle around the center of the image.
clear
clc
close all
I = zeros(256,256);
pos_triangle = [64 64 128 192 128 128];
xt = [64 128 128];
yt = [64 192 128];
a = deg2rad(20);
R = [cos(a) -sin(a);
sin(a) cos(a)];
p1 = R*[64; 64]
p2 = R*[128; 192]
p3 = R*[128; 128]
% the rotated points
x1 = [p1(1) p2(1) p3(1)];
y1 = [p1(2) p2(2) p3(2)];
imshow(I)
hold on
fill(x1,y1,'r')
fill(xt,yt,'w')
hold off
The result is now
If you would like to rotate it around itself, then you need to translate the center of the image to the center of the triangle. This is not a complete solution but I hope it helps.

Related

Matlab - rotate a card [duplicate]

This question already has answers here:
How to straighten a tilted square shape in an image?
(2 answers)
Closed 5 years ago.
I have a cropped image of a card:
The card is a rectangle with rounded corners, is brightly colored, and sits on a relatively dark background.
It is, therefore, easy to differentiate between pixels belonging to the card and pixels belonging to the background.
I want to use MATLAB to rotate the card so its sides are vertical and horizontal (and not diagonal) and create an image of nothing but the straightened card.
I need this to work for any reasonable card angle (say +45 to -45 degrees of initial card rotation).
What would be the best way of doing this?
Thanks!
You can do this by finding the lines made by the edges of the card. The angle of rotation is then the angle between one of the lines and the horizontal (or vertical).
In MATLAB, you can use the Hough line detector to find lines in a binary image.
0. Read the input image
I downloaded your image and renamed it card.png.
A = imread('card.png');
We don't need color information, so convert to grayscale.
I = rgb2gray(A);
1. Detect edges in the image
A simple way is to use the Canny edge detector. Adjust the threshold to reject noise and weak edges.
BW = edge(I, 'canny', 0.5);
Display the detected edges.
figure
imshow(BW)
title('Canny edges')
2. Use the Hough line detector
First, you need to use the Hough transform on the black and white image, with the hough function. Adjust the resolution so that you detect all lines you need later.
[H,T,R] = hough(BW, 'RhoResolution', 2);
Second, find the strongest lines in the image by finding peaks in the Hough transform with houghpeaks.
P = houghpeaks(H, 100); % detect a maximum of 100 lines
Third, detect lines with houghlines.
lines = houghlines(BW, T, R, P);
Display the detected lines to make sure you find at least one along the edge of the card. The white border around the black background in your image makes detecting the right edges a bit more difficult.
figure
imshow(A)
hold on
for k = 1:length(lines)
xy = [lines(k).point1; lines(k).point2];
plot(xy(:,1), xy(:,2), 'LineWidth', 2, 'Color', 'red');
end
title('Detected lines')
3. Calculate the angle of rotation
lines(3) is the left vertical edge of the card. lines(3).point2 is the end of the line that is at the bottom. We want this point to stay where it is, but we want to vector along the line to be aligned with the vector v = [0 -1]'. (The origin is the top-left corner of the image, x is horizontal to the right and y is vertical down.)
lines(3)
ans =
struct with fields:
point1: [179 50]
point2: [86 455]
theta: 13
rho: 184
Simply calculate the angle between the vector u = lines(3).point1 - lines(3).point2 and the vertical vector v.
u = lines(3).point1 - lines(3).point2; % vector along the vertical left edge.
v = [0 -1]; % vector along the vertical, oriented up.
theta = acos( u*v' / (norm(u) * norm(v)) );
The angle is in radians.
4. Rotate
The imrotate function lets you rotate an image by specifying an angle in degrees. You could also use imwarp with a rotation transform.
B = imrotate(A, theta * 180 / pi);
Display the rotated image.
figure
imshow(B)
title('Rotated image')
Then you would have to crop it.

Matlab: mask/create a circular roi knowing its origin point with a certain radius

Just a quick question. I've an image and I've extracted a certain point (feature), I know the coordinates of that point in every frame.
Say x1 and y1.
I need a circular ROI form that point on the image with a radius that I chose.
I tried impoly and roipoly - not sure how to use either of these when I know the point in the image.
Thanks
Since you know the coordinates of the center of the ROI along with the radius, you can modify a bit the code provided by #Jonas here to create a circular mask in a quite efficient way.
Example:
clc;clear
Im = imread('coins.png');
[rNum,cNum,~] = size(Im);
%// Define coordinates and radius
x1 = 60;
y1 = 100;
radius = 40;
%// Generate grid with binary mask representing the circle. Credit to Jonas for original code.
[xx,yy] = ndgrid((1:rNum)-y1,(1:cNum)-x1);
mask = (xx.^2 + yy.^2)<radius^2;
%// Mask the original image
Im(mask) = uint8(0);
imshow(Im)
Output:
EDIT
If you want to see only the outer edge of the ROI to see the center, add a logical condition with some tolerance for the radius of a smaller circle. Something like this:
mask = (xx.^2 + yy.^2)<radius^2 & (xx.^2 + yy.^2)>(radius-tol)^2;
With a tol of 2 it looks like this:

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!

OpenCV MATLAB: How to draw a line having a particular Intensity profile?

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:

How to measure the rotation of a image in MATLAB?

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.