I have an image Im of size 50x129 :
It is a part of big cercle in yellow, and a space with blue color.
I rotated this image Im using imrotate() with an angle of -45°, I get ImR :
I would like to compute the new size of the image ImR ?
One way to do this is to use the function axis of ImR :
axises = axis;
w1=axises(2)-axises(1);
h1=axises(4)-axises(3);
but I want to be independent from the resulting image ... I mean I would like to, knowing the size of original image and the angle of rotation, get the size of the rotated image.
Update
I intend to use this code which rotate the point A with an angle t around the Origin point.
function Af = rotate(A,t,Origin)
% Definition of the rotation matrix (rotation around origin)
R=[ ...
cosd(t) -sind(t)
sind(t) cosd(t)
];
% translation
At = A - Origin;
% rotation of the points A and B
Ar = R*At;
Compute the position of the corners of the rotated image using your rotation matrix. And put them all in a matrix as follows:
rotImgCorners = [
x_topLeft , y_topleft;
x_top_right , y_top_right;
%//... bottom left ...
%//... bottom right ...
]
Then the size of your new image would be:
%// [horizontal size, vertical size] - you can switch it around if you whish
newImgSize = max(rotImgCorners, [], 1) - min(rotImgCorners, [], 1);
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:
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:
As part of a circle recognition program, I have a background image with a geometrical circle with known coordinates and radius. I want the inside part of the circle filled by an image and the outside left alone. My best idea is some sort of circular mask, but I am not sure this is the best approach. Any suggestions?
X = imread('X.jpg'); % Background image jpg
Y = imread('Y.jpg'); % Filling image jpg
cent = [100,100]; % Center of circle
rad = 20; % Circle radius
% Fill circle ?
...
I have not provided the extended code, due to confidentiality.
I think the hard part was done by whomever authored this: http://matlab.wikia.com/wiki/FAQ#How_do_I_create_a_circle.3F
Assumptions:
I'm assuming you're not going to specify points that are out of range of the image (i.e. I'm not adding validation here).
I use the background image to relate the "center" of your circle to the coordinates.
I'm assuming radius is in pixels.
I didn't create a background image with a circle of known radius because I don't think that's necessary to create the filling effect you're looking for (unless I'm missing something).
Code:
X = imread('rdel_x.png'); % Background image jpg (I used a random image but this can be your blank + geometric circle)
Y = imread('rdel_y.png'); % Filling image jpg
cent = [100,150]; % Center of circle
rad = 70; % Circle radius
% make a mesh grid to provide coords for the circle (mask)
% taken from http://matlab.wikia.com/wiki/FAQ#How_do_I_create_a_circle.3F
[columnsInImage rowsInImage] = meshgrid(1:size(X,2), 1:size(X,1));
% circle points in pixels:
circlePixels = (rowsInImage - cent(1)).^2 ...
+ (columnsInImage - cent(2)).^2 <= rad.^2;
circlePixels3d=repmat(circlePixels,[1 1 3]); % turn into 3 channel (assuming X and Y are RGB)
X((circlePixels3d)) = Y((circlePixels3d)); % assign the filling image pixels to the background image for pixels where it's the desired circle
imagesc(X);
axis image off
Result: from left to right, background image, filling image, result from above code.
Edit: you might not even need the background image if everything is encapsulated in your coordinates. e.g. try appending this to the above code...
Z=zeros(size(X),'uint8'); % same size as your background
Z(circlePixels3d) = Y(circlePixels3d);
figure; % new fig
imagesc(Z);
axis image off
I have image which size was 600 * 600 and it was displayed on 800 * 800 pixel screen.
The x,y coordinate in which the user look on screen was recorded in an array:
x =[250,300,390,750,760];
y =[120,550,250,130,420];
In other program, I want to plot the x,y coordinate on the 600 * 600 image. The problem is that some of the x,y plot were out of the image (as shown on the picture below) since the coordinate was more that the maximum size of the image (600 * 600).
EDITED:
How to transform/adjust the coordinate of the bigger image (800*800) into the smaller image (600*600) so all x,y coordinate are inside the smaller image (600*600)?
Lets say for example, the coordinate of top left image of the 600*600 inside the image of the 800*800 image is e.g. x = -10, y = 3.
Thanks.
alt text http://img9.imageshack.us/img9/8836/e47184420f.jpg
To get the pixels in image coordinates, you need to know where the bottom left and top right corners of your image were placed on the screen. From that you can both calculate offset and zoom of the image.
%# define some parameters
imageSize = [600 600];
topLeftPixScreen = [200,100]; %# position of the top left image corner in screen pixels
bottomRightPixScreen = [800,750]; %# position of the bottom right image corner in screen pixels
%# transform coordinates
oldX =[250,300,390,750];
oldY =[120,550,250,130,420];
newX = (oldX - topLeftPixScreen(1))/(bottomRightPixScreen(1) - topLeftPixScreen(1) + 1);
newY = (oldY - topLeftPixScreen(2))/(bottomRightPixScreen(2) - topLeftPixScreen(2) + 1);
Having said that, I'd suggest using ginput to select the points with Matlab, since this function directly returns image pixels.
EDIT
If you only have the top left corner, you have to hope that there has not been any scaling - otherwise, there is no way you can transform the points.
With offset only, the above simplifies to
%# define some parameters
imageSize = [600 600];
topLeftPixScreen = [200,100]; %# position of the top left image corner in screen pixels
%# transform coordinates
oldX =[250,300,390,750];
oldY =[120,550,250,130,420];
newX = oldX - topLeftPixScreen(1);
newY = oldY - topLeftPixScreen(2);
It seems that just adjusting the coordinates by the ratio of the screen area and image size would do:
newX = x.*(600/800)
newY = y.*(600/800)