I calculated a rectangle (that does not need to be axis aligned) in an image and now want to check which pixel are inside that rectangle.
I believe the best way is to rotate each point in the image around the center point of the rectangle, such that the bounding box is axis aligned in the "new image". Then, we can quite easily check if a point is inside the rectangle or not.
I tried achieving this, by calculating an oriented bounding box with the geom2d package for my rectangle, which also gives us "the orientation of the length axis. Orientation is counted in degrees, counter-clockwise."
Up to now, I have not been able to understand exactly how the angle is defined. If I do the following:
obox1 = orientedBox([1 3; 4 1; 12 3; 9 5]);
obox2 = orientedBox([12 8; 9 10; 20 2; 24 3]);
figure;
isInside = false(15,25);
imshow(isInside);hold on;
drawOrientedBox(obox1, 'green');
drawOrientedBox(obox2, 'red');
I get
with angle of green box = 14.03° and angle of red box = 154.98°. How do these values need to be interpreted?
Link to geom2d package:
http://www.mathworks.com/matlabcentral/fileexchange/7844-geom2d/content/geom2d/geom2d/orientedBox.m
Related
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.
I have created a synthetic image that consists of a circle at the centre of a box with the code below.
%# Create a logical image of a circle with image size specified as follows:
imageSizeY = 400;
imageSizeX = 300;
[ygv, xgv] = meshgrid(1:imageSizeY, 1:imageSizeX);
%# Next create a logical mask for the circle with specified radius and center
centerY = imageSizeY/2;
centerX = imageSizeX/2;
radius = 100;
Img = double( (ygv - centerY).^2 + (xgv - centerX).^2 <= radius.^2 );
%# change image labels from double to numeric
for ii = 1:numel(Img)
if Img(ii) == 0
Img(ii) = 2; %change label from 0 to 2
end
end
%# plot image
RI = imref2d(size(Img),[0 size(Img, 2)],[0 size(Img, 1)]);
figure, imshow(Img, RI, [], 'InitialMagnification','fit');
Now, i need to create a rectangular mask (with label == 3, and row/col dimensions: 1 by imageSizeX) across the image from top to bottom and at known angles with the edges of the circle (see attached figure). Also, how can i make the rectangle thicker than 1 by imageSizeX?. As another option, I would love to try having the rectangle stop at say column 350. Lastly, any ideas how I can improve on the resolution? I mean is it possible to keep the image size the same while increasing/decreasing the resolution.
I have no idea how to go about this. Please i need any help/advice/suggestions that i can get. Many thanks!.
You can use the cos function to find the x coordinate with the correct angle phi.
First notice that the angle between the radius that intersects the vertex of phi has angle with the x-axis given by:
and the x coordinate of that vertex is given by
so the mask simply needs to set that row to 3.
Example:
phi = 45; % Desired angle in degrees
width = 350; % Desired width in pixels
height = 50; % Desired height of bar in pixels
theta = pi-phi*pi/180; % The radius angle
x = centerX + round(radius*cos(theta)); % Find the nearest row
x0 = max(1, x-height); % Find where to start the bar
Img(x0:x,1:width)=3;
The resulting image looks like:
Note that the max function is used to deal with the case where the bar thickness would extend beyond the top of the image.
Regarding resolution, the image resolution is determined by the size of the matrix you create. In your example that is (400,300). If you want higher resolution simply increase those numbers. However, if you would like to link the resolution to a higher DPI (Dots per Inch) so there are more pixels in each physical inch you can use the "Export Setup" window in the figure File menu.
Shown here:
I have background subtracted images as input. The idea is to reduce search areas for person detection by using a smaller search area for the HOG algorithm. The output required is a bounding box around the person and the pixel positions of the box corners.
This is the input image:
This is the required output:
This is what I have tried so far:
x=imread('frame 0080.png');
y=im2bw(x);
s=regionprops(y);
imshow(y);
hold on
for i=1:numel(s)
rectangle('Position',s(i).BoundingBox,'edgecolor','y')
end
This was the output I got:
It looks like you have tried what I have suggested. However, you want the bounding box that encapsulates the entire object. This can easily be done by using the BoundingBox property, then calculating each of the four corners of each rectangle. You can then calculate the minimum spanning bounding box that encapsulates all of the rectangles which ultimately encapsulates the entire object.
I do notice that there is a thin white strip at the bottom of your image, and that will mess up the bounding box calculations. As such, I'm going to cut the last 10 rows of the image before we proceed calculating the minimum spanning bounding box. To calculate the minimum spanning bounding box, all you have to do is take all of the corners for all of the rectangles, then calculate the minimum and maximum x co-ordinates and minimum and maximum y co-ordinates. These will correspond to the top left of the minimum spanning bounding box and the bottom right of the minimum spanning bounding box.
When taking a look at the BoundingBox property using regionprops, each bounding box outputs a 4 element vector:
[x y w h]
x,y denote the top left co-ordinate of your bounding box. x would be the column and y would be the row of the top left corner. w,h denote the width and the height of the bounding box. We would use this and compute top left, top right, bottom left and bottom right of every rectangle that is detected. Once you complete this, stack all of these rectangle co-ordinates into a single 2D matrix, then calculate the minimum and maximum x and y co-ordinates. To calculate the rectangle, simply use the minimum x and y co-ordinates as the top left corner, then calculate the width and height by subtracting the maximum and minimum x and y co-ordinates respectively.
Without further ado, here's the code. Note that I want to extract all of the bounding box co-ordinates in a N x 4 matrix where N denotes the number of bounding boxes that are detected. You would have to use reshape to do this correctly:
% //Read in the image from StackOverflow
x=imread('http://i.stack.imgur.com/mRWId.png');
% //Threshold and remove last 10 rows
y=im2bw(x);
y = y(1:end-10,:);
% //Calculate all bounding boxes
s=regionprops(y, 'BoundingBox');
%// Obtain all of the bounding box co-ordinates
bboxCoords = reshape([s.BoundingBox], 4, []).';
% // Calculate top left corner
topLeftCoords = bboxCoords(:,1:2);
% // Calculate top right corner
topRightCoords = [topLeftCoords(:,1) + bboxCoords(:,3) topLeftCoords(:,2)];
% // Calculate bottom left corner
bottomLeftCoords = [topLeftCoords(:,1) topLeftCoords(:,2) + bboxCoords(:,4)];
% // Calculate bottom right corner
bottomRightCoords = [topLeftCoords(:,1) + bboxCoords(:,3) ...
topLeftCoords(:,2) + bboxCoords(:,4)];
% // Calculating the minimum and maximum X and Y values
finalCoords = [topLeftCoords; topRightCoords; bottomLeftCoords; bottomRightCoords];
minX = min(finalCoords(:,1));
maxX = max(finalCoords(:,1));
minY = min(finalCoords(:,2));
maxY = max(finalCoords(:,2));
% Draw the rectangle on the screen
width = (maxX - minX + 1);
height = (maxY - minY + 1);
rect = [minX minY width height];
% // Show the image
imshow(y);
hold on;
rectangle('Position', rect, 'EdgeColor', 'yellow');
This is the image I get:
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 background subtracted images as input. The idea is to reduce search areas for person detection by using a smaller search area for the HOG algorithm. The output required is a bounding box around the person and the pixel positions of the box corners.
This is the input image:
This is the required output:
This is what I have tried so far:
x=imread('frame 0080.png');
y=im2bw(x);
s=regionprops(y);
imshow(y);
hold on
for i=1:numel(s)
rectangle('Position',s(i).BoundingBox,'edgecolor','y')
end
This was the output I got:
It looks like you have tried what I have suggested. However, you want the bounding box that encapsulates the entire object. This can easily be done by using the BoundingBox property, then calculating each of the four corners of each rectangle. You can then calculate the minimum spanning bounding box that encapsulates all of the rectangles which ultimately encapsulates the entire object.
I do notice that there is a thin white strip at the bottom of your image, and that will mess up the bounding box calculations. As such, I'm going to cut the last 10 rows of the image before we proceed calculating the minimum spanning bounding box. To calculate the minimum spanning bounding box, all you have to do is take all of the corners for all of the rectangles, then calculate the minimum and maximum x co-ordinates and minimum and maximum y co-ordinates. These will correspond to the top left of the minimum spanning bounding box and the bottom right of the minimum spanning bounding box.
When taking a look at the BoundingBox property using regionprops, each bounding box outputs a 4 element vector:
[x y w h]
x,y denote the top left co-ordinate of your bounding box. x would be the column and y would be the row of the top left corner. w,h denote the width and the height of the bounding box. We would use this and compute top left, top right, bottom left and bottom right of every rectangle that is detected. Once you complete this, stack all of these rectangle co-ordinates into a single 2D matrix, then calculate the minimum and maximum x and y co-ordinates. To calculate the rectangle, simply use the minimum x and y co-ordinates as the top left corner, then calculate the width and height by subtracting the maximum and minimum x and y co-ordinates respectively.
Without further ado, here's the code. Note that I want to extract all of the bounding box co-ordinates in a N x 4 matrix where N denotes the number of bounding boxes that are detected. You would have to use reshape to do this correctly:
% //Read in the image from StackOverflow
x=imread('http://i.stack.imgur.com/mRWId.png');
% //Threshold and remove last 10 rows
y=im2bw(x);
y = y(1:end-10,:);
% //Calculate all bounding boxes
s=regionprops(y, 'BoundingBox');
%// Obtain all of the bounding box co-ordinates
bboxCoords = reshape([s.BoundingBox], 4, []).';
% // Calculate top left corner
topLeftCoords = bboxCoords(:,1:2);
% // Calculate top right corner
topRightCoords = [topLeftCoords(:,1) + bboxCoords(:,3) topLeftCoords(:,2)];
% // Calculate bottom left corner
bottomLeftCoords = [topLeftCoords(:,1) topLeftCoords(:,2) + bboxCoords(:,4)];
% // Calculate bottom right corner
bottomRightCoords = [topLeftCoords(:,1) + bboxCoords(:,3) ...
topLeftCoords(:,2) + bboxCoords(:,4)];
% // Calculating the minimum and maximum X and Y values
finalCoords = [topLeftCoords; topRightCoords; bottomLeftCoords; bottomRightCoords];
minX = min(finalCoords(:,1));
maxX = max(finalCoords(:,1));
minY = min(finalCoords(:,2));
maxY = max(finalCoords(:,2));
% Draw the rectangle on the screen
width = (maxX - minX + 1);
height = (maxY - minY + 1);
rect = [minX minY width height];
% // Show the image
imshow(y);
hold on;
rectangle('Position', rect, 'EdgeColor', 'yellow');
This is the image I get: