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How to straighten a tilted square shape in an image?
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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.
Related
Having this coordinate system:
And this dominant vertical vanishing point:
I would like to rotate the image around x axis so the vanishing point is at infinity. That means that all vertical lines are parallel.
I am using matlab. I find the line segmentes using LSD and the vanishing point using homogeneous coordinates. I would like to use angle-axis representation, then convert it to a rotation matrix and pass this to imwarp and get the rotated image. Also would be good to know how to rotate the segments. The segments are as (x1,y1,x2,y2).
Image above example:
Vanishin point in homogenous coordinates:
(x,y,z) = 1.0e+05 * [0.4992 -2.2012 0.0026]
Vanishin point in cartesian coordinates (what you see in the image):
(x,y) = [190.1335 -838.3577]
Question: With this vanishing point how do I compute the rotation matrix in the world x axis as explained above?
If all you're doing is rotating the image so that the vector from the origin to the vanishing point, is instead pointing directly vertical, here's an example.
I = imread('cameraman.tif');
figure;imagesc(I);set(gcf,'colormap',gray);
vp=-[190.1335 -838.3577,0]; %3d version,just for cross-product use,-ve ?
y=[0,1,0]; %The vertical axis on the plot
u = cross(vp,y); %you know it's going to be the z-axis
theta = -acos(dot(vp/norm(vp),y)); %-ve ?
rotMat = vrrotvec2mat([u, theta]);
J=imwarp(I,affine2d (rotMat));
figure;imagesc(J);set(gcf,'colormap',gray); %tilted image
You can play with the negatives, and plotting, since I'm not sure about those parts applying to your situation. The negatives may come from plotting upside down, or from rotation of the world vs. camera coordinate system, but I don't have time to think about it right now.
EDIT
If you want to rotation about the X-axis, this might work (adapted from https://www.mathworks.com/matlabcentral/answers/113074-how-to-rotate-an-image-along-y-axis), or check out: Rotate image over X, Y and Z axis in Matlab
[rows, columns, numberOfColorChannels] = size(I);
newRows = rows * cos(theta);
rotatedImage = imresize(I, [newRows, columns]);
Is it possible to find the area of the black pixelation of an area within a circle? in other words I want to find the number of pixels (the area) of the RGB 0,0,0 (black pixels) within the circle. I do not want the areas of the white pixels (1,1,1) within the circle. I also have a radius of the circle if that helps. Here is the image:
Here is the code:
BW2= H(:,:) <0.45 ;%& V(:,:)<0.1;
aa=strel('disk',5);
closeBW = imclose(BW2,aa);
figure, imshow(closeBW)
imshow(closeBW)
viscircles([MYY1 MYX1], round(MYR2/2))
MYY1,MYX2, and the other values are calculated by my program. How can I find the area of the black pixelation in my circle?
Here is an idea:
1) Calculate the total # of black pixels in your original image (let's call it A).
2) Duplicate that image (let's call it B) and replace all pixels inside the circle with white. To do that, create a binary mask. (see below)
3) Calculate the total # of black pixels in that image (i.e. B).
4) Subtract both values. That should give you the number of black pixels within the circle.
Sample code: I used a dummy image I had on my computer and created a logical mask with the createMask method from imellipse. That seems complicated but in your case since you have the center position and radius of the circle you can create directly your mask like I did or by looking at this question/answer.
Once the mask is created, use find to get the linear indices of the white pixels of the mask (i.e. all of it) to replace the pixels in the circle of your original image with white pixels, which you use to calculate the difference in black pixels.
clc;clear;close all
A = im2bw(imread('TestCircle.png'));
imshow(A)
Center = [160 120];
Radius = 60;
%// In your case:
% Center = [MYY1 MYX1];
% Radius = round(MYR2/2);
%// Get sum in original image
TotalBlack_A = sum(sum(~A))
e = imellipse(gca, [Center(1) Center(2) Radius Radius]);
%// Create the mask
ROI = createMask(e);
%// Find white pixels
white_id = find(ROI);
%// Duplicate original image
B = A;
%// Replace only those pixels in the ROI with white
B(white_id) = 1;
%// Get new sum
NewBlack_B = sum(sum(~B))
%// Result!
BlackInRoi = TotalBlack_A - NewBlack_B
In this case I get this output:
TotalBlack_A =
158852
NewBlack_B =
156799
BlackInRoi =
2053
For this input image:
I need to remove horizontal and vertical lines in a binary image. Is there any method for filtering these lines? bwareaopen() is not good method to remove these lines and also Dilation and Erosion are not good for these cases.
Does any one know a solution?
Example image:
EDIT:(added more example images:
http://s1.upload7.ir/downloads/pPqTDnmsmjHUGTEpbwnksf3uUkzncDwr/example%202.png
source file of images:
https://www.dropbox.com/sh/tamcdqk244ktoyp/AAAuxkmYgBkB8erNS9SajkGVa?dl=0
www.directexe.com/9cg/pics.rar
Use regionprops and remove regions with high eccentricity (meaning the region is long and thin) and orientation near 0 or near 90 degrees (regions which are vertical or horizontal).
Code:
img = im2double(rgb2gray(imread('removelines.jpg')));
mask = ~im2bw(img);
rp = regionprops(mask, 'PixelIdxList', 'Eccentricity', 'Orientation');
% Get high eccentricity and orientations at 90 and 0 degrees
rp = rp([rp.Eccentricity] > 0.95 & (abs([rp.Orientation]) < 2 | abs([rp.Orientation]) > 88));
mask(vertcat(rp.PixelIdxList)) = false;
imshow(mask);
Output:
If all of your images are the same where the horizontal and vertical lines are touching the border, a simple call to imclearborder will do the trick. imclearborder removes any object pixels that are touching the borders of the image. You'll need to invert the image so that the characters are white and the background is dark, then reinvert back, but I'm assuming that isn't an issue. However, to be sure that none of the actual characters get removed because they may also be touching the border, it may be prudent to artificially pad the top border of the image with a single pixel thickness, clear the border, then recrop.
im = imread('http://i.stack.imgur.com/L1hUa.jpg'); %// Read image directly from StackOverflow
im = ~im2bw(im); %// Convert to black and white and invert
im_pad = zeros(size(im,1)+1, size(im,2)) == 1; %// Pad the image too with a single pixel border
im_pad(2:end,:) = im;
out = ~imclearborder(im_pad); %// Clear border pixels then reinvert
out = out(2:end,:); %// Crop out padded pixels
imshow(out); %// Show image
We get this:
You can firstly find the horizontal and vertical lines. Since, the edge map will also be binary so you can perform a logical subtraction operation in between the images. To find vertical lines, you can use (in MATLAB)
BW = edge(I,'sobel','vertical');
For horizontal lines, you can use
% Generate horizontal edge emphasis kernel
h = fspecial('sobel');
% invert kernel to detect vertical edges
h = h';
J = imfilter(I,h);
I'm creating a figure with many polygons which can overlap in Matlab. I do not want borders on any of the shapes, only a translucent fill. I was able to get the first shape to obey this request using:
fill(xv,yv,'blue','FaceAlpha',0.1,'EdgeColor','None','LineStyle','none');hold on;
However, each subsequent shape which is drawn in the loop ignores this style, instead cycling through colored borders. I was able to use the set command to override the color cycle, but this still draws a border...I do not want any border. The border cannot be drawn at all because we are using the overlapping properties of the shapes, and the presence of any borders would perturb the nature of our simulation.
Here is the complete code:
for count = 1:92
x=randn*clustering;
y=randn*clustering;
angle=randn*360;
rectangle(width,height,x,y,angle);
end
function rectangle(w,h,x,y,angle)
%Define the rectangle
xv=[x x+w x+w x x];
yv=[y y y+h y+h y];
%Define the rotation transformation matrix
transform=[cos(angle) -sin(angle);sin(angle) cos(angle)];
%Define the translation to origin transform
xcenter=x+.5*w;
ycenter=y+.5*h;
%Perform translation to origin
tx=xv-xcenter;
ty=yv-ycenter;
%Perform rotation
rotated=transform*[tx;ty];
%Perform translation back to original location
xv=rotated(1,:)+xcenter;
yv=rotated(2,:)+ycenter;
%Plot result
figure(1);
plot(xv,yv);
fill(xv,yv,'blue','FaceAlpha',0.1);hold on;
axis square;
axis([-30 30 -30 30]);
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: