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)
Related
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 felt confused to crop an image along any angle, i.e. pi/4, how can I make it?
The following example demonstrates how to crop rotated bounding box.
The example doesn't show how to select the bounding box, and the angle (user interface is excluded).
For keeping the example simple, the bounding box parameters are: center, size and angle (instead of 4 corners).
Center is kept in place, when rotating.
Size is the width and height of destination crop area (size after rotation).
In case you need to calculate corners transformation, you can use rotation matrix: https://en.wikipedia.org/wiki/Rotation_matrix
Code sample:
%Bounding box parameters: center, width, height and angle.
%Center is selected, because center is kept the same when rotating.
center_x = 256; %Center column index (applied center is 256.5)
center_y = 192; %Center row index (applied center is 192.5)
width = 300; %Number of columns of destination (cropped) area (even integer)
height = 200; %Number of rows of destination (cropped) area (even integer)
phi = 120; %Rotation angle in degrees
%Read sample image.
I = imread('peppers.png');
%Add center cross, for verifying center is kept.
I(center_y-1:center_y, center_x-30:center_x+30, :) = 255;
I(center_y-30:center_y+30, center_x-1:center_x, :) = 255;
%Rotate the entire input image, dimensions of J will be the same as I.
J = imrotate(I, phi, 'bicubic', 'crop');
%Crop rectangular are with size width by height around center_x, center_y from J.
C = J(center_y-height/2+1:center_y+height/2, center_x-width/2+1:center_x+width/2, :);
%Display result:
figure;imshow(C);
Result:
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 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);
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: