I need to draw vertical lines at regular intervals in a rectangular box. this is what i have used so far:
xmin=000;
xmax=70000;
ymin=0;
ymax=1000;
line1Val=900;
line2Val=600;
line3Val=300;
xlim([xmin xmax])
ylim([ymin ymax])
xl=get(gca,'XLim');
line(xl,[line1Val line1Val],'Color','y');
line(xl,[line2Val line2Val],'Color','y');
line(xl,[line3Val line3Val],'Color','y');
hold on ;
rectangle('Position',[120000,900,(280000-120000),37],'faceColor','k')
so the width of the rectangle is 160000 units i want to divide this into 4 , where the vertical line is of a different color(say red) and the height of the line is 37 units.
any ideas on how i can draw this without drawing 4 rectangles whose edges are red and are filled with black color.
You could use the parameters xstart,ystart,width and height for drawing your rectangle:
rectangle('Position',[xstart,ystart,width,height],'faceColor','k');
After that, you could determine the line positions in a loop and simply draw these lines:
for i = 1:3
x = xstart+i*width/4;
line([x x],[ystart ystart+height],'Color','r');
end
If you want a red line at the start and end of the rectangle, let i = 0:4.
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.
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);
Is there any way to constraint the the imline to be always perpendicular to the other imline drawn on the same object. for ex. I draw a first line using "imline" now I want to draw second line across the first line to be perpendicular to it. if there is a way to force the second imline to be perpendicular to the first line keeping the flexibility of extending the length it will solve my problem some extent.
I want something like a flexible cross hair(which can rotate along the axis and have flexible sides) on my image to measure the height and width of the certain object.
Code:
function perpline()
imshow(rand(200),[]);
line1 = imline(gca,[50 50; 150 150]);
setColor(line1,'r');
line2 = imline(gca,[50 150; 150 50]);
setColor(line2,'g');
addNewPositionCallback(line2,#(pos)callback_line(pos));
function callback_line(pos)
% Must update line1 based on line2's position
pos_line1 = getPosition(line1);
pos_line2 = getPosition(line2);
% Get middle
pos_center = [(pos_line2(1,1)+pos_line2(2,1))/2 (pos_line2(1,2)+pos_line2(2,2))/2];
% Find displacement
vec_disp = [pos_line2(2,1)-pos_line2(1,1) pos_line2(2,2)-pos_line2(1,2)];
% Get normal unit vector
vec_perp = [-vec_disp(2) vec_disp(1)]/norm(vec_disp);
% Preserve length of line2
length_line1 = norm([pos_line1(2,1)-pos_line1(1,1) pos_line1(2,2)-pos_line1(1,2)]);
pos_line1_update = [-vec_perp*length_line1/2+pos_center;
vec_perp*length_line1/2+pos_center];
% Set position
setPosition(line1,pos_line1_update);
end
end
Save it as a function then call it. You can drag the green line around and the red line remains perpendicular. Note that you have to define how you want it to preserve the perpendicularity. I chose to preserve the length of the red line and keep it in the center of the green line.
I'm writing code in which I use MATLAB's fill command to plot 2D shapes. I can specify the fill color of the shape. However, the border line color is always black. I want the border line color the same as the fill color. How can I also specify the border line color?
See this thread:
To set the edgecolor to white do the following.
h = fill([-1 -1 1 1],[-1 1 1 -1],'w');
axis([-2 2 -2 2]);
set(h,'edgecolor','white');
that should take care of the border.
In addition to schnaader's answer, you can also set the edge color in the initial call to FILL:
hPatch = fill(xData,yData,'r','EdgeColor','r'); %# Red patch with red edges
Or stop the edges from being drawn altogether:
hPatch = fill(xData,yData,'r','EdgeColor','none'); %# Red patch with no edges