I have images of a leaf,now I want to find the diameter of the leaf using matlab, for feature extraction
]1
An help would be appreciated.
In your image you want scan the pixels top to bottom and bottom to top,and get the first pixel values,using this pixels you can get diameter.The same method using in left to right and right to left manner.then you find the maximum value as your diameter.
This is code for mark first pixels,try this.
%Mark horizontal pixel
[row1, column1] = find(thinImg, 1, 'last');
hold on;
plot(column1, row1, 'yX', 'MarkerSize', 15);
%Mark vertical pixel
[row2, column2] = find(thinImg, 1, 'first');
hold on;
plot(row2, column2, 'yX', 'MarkerSize', 15);
Related
I'm representing a surface using "surf" function, with a colorbar. I would like to keep the default ticks of the colorbar, but add a custom tick on this colorbar, at a specific value (that I could make red to distinguish it from other ticks for example). Any idea on how to add a custom tick like that with keeping existing ticks on the colorbar ?
Thanks
As Luis mentioned in the comments, you can add an additional tick mark like so
h = colorbar;
newTick = 0.75;
h.Ticks = sort([h.Ticks newTick]);
If you want to add a line to the bar, the easiest thing (I think) is to use an annotation which is positioned relative to the figure (the same as the colorbar), so we can overlay it
pos = h.Position;
r = (newTick - min(h.Ticks))/(max(h.Ticks)-min(h.Ticks));
annotation( 'line', pos(1)+[0, pos(3)], [1, 1]*(pos(2)+pos(4)*r), ...
'color', [1,0,0], 'linewidth', 2 );
I'm setting the x position of the annotation to match the left and right sides of the colorbar, and the y position to match the bottom plus the relative % of the height according to the tick value.
Result:
Similarly, you could use annotatation exclusively to just get a red label, it's a little more convoluted to get everything lined up correctly, you have to make sure the text box is wide enough to be on a single line and vertically aligned to the middle to get the position right:
h = colorbar;
newTick = 0.75;
pos = h.Position;
r = (newTick - min(h.Ticks))/(max(h.Ticks)-min(h.Ticks));
h = 0.2;
annotation( 'textbox', [pos(1)+pos(3)/2, (pos(2)+pos(4)*r)-(h/2), pos(3)*2, h], ...
'color', [1,0,0], 'string', ['- ' num2str(newTick)], 'linestyle', 'none', ...
'VerticalAlignment', 'middle' );
so I am working for my disertation thesis and I have to detect the pupil from images using Hough Transform. So far I wrote a code that identifies 2 circles on my image, but right now I have to keep the black circle from the pupil.
When I run the code, it identifies me the pupil, but also a random circle on the cheek. My professor said that I should calculate the pixels mean and, considering the fact that the pupil is black, to keep the pixels from only that region. I don't know how to do this.
I will let my code here to have a look and if someone has an ideea on how should I write this and keep only the black pixels would be great. I also attached to this the final image to see what I obtained.
close all
clear all
path='C:\Users\Ioana PMEC\OneDrive\Ioana personal\Disertatie\test.jpg';
%Citire imagine initiala
xx = imread(path);
figure
imshow(xx)
title('Imagine initiala');% Binarizarea imaginii initiale
yy = rgb2gray(xx);
figure
imshow(yy);
title('Imagine binarizata');
e = edge(yy, 'canny');
imshow(e);
radii = 11:1:30;
h = circle_hough(e, radii, 'same', 'normalise');
peaks = circle_houghpeaks(h, radii, 'nhoodxy', 15, 'nhoodr', 21, 'npeaks', 2);
imshow(yy);
hold on;
for peak = peaks
[x, y]=circlepoints(peak(3));
plot(x+peak(1), y+peak(2), 'r-');
end
hold off
testimage
finalimage
I implemented something which should get the task done for you. The example is done with the image you provided.
Step 1: Read the file(s) and convert them to grayscale.
path = %user input;
RGB = imread(path);
lab = rgb2lab(RGB);
grayscale_image = rgb2gray(RGB);
Step 2: Do the Hough transformation with given parameters.
These, and also the sensitivity can be adapted according to your task. Hint: Play around in the Image Segmenter toolbox for fast parameter finding. Next, the inferred circles are converted to integer values, since these are required for indexing.
min_radius = 10;
max_radius = 50;
% Find circles
[centers,radii,~] = imfindcircles(RGB,[min_radius max_radius],'ObjectPolarity','dark','Sensitivity',0.95);
centers = uint16(centers);
radii = uint16(radii);
The annotated image appears as follows:
Step 3: Get the brightness values of the circles.
From the circle center and radius values, we infer their respective brightness. It is sufficient to only check for the x and y pixel values left/right and above/below the center. (-1 is just a safety margin to completely stay within the circles.)
brightness_checker = zeros(2, max(radii), 2);
for i=1:size(centers,1)
current_radii = radii(i)-1;
for j=1:current_radii
% X-center minus radius, step along x-axis
brightness_checker(i, j, 1) = grayscale_image((centers(i,2) - current_radii/2) + j,...
(centers(i,1) - radii(i)/2) + j);
% Y-center minus radius, step along y-axis
brightness_checker(i, j, 2) = grayscale_image((centers(i,2) - current_radii/2) + j,...
(centers(i,1) - current_radii/2) + j);
end
end
Step 4: Check which circle is a pupil.
The determined value of 30 could potentially be enhanced.
median_x = median(brightness_checker(:,:,1),2);
median_y = median(brightness_checker(:,:,2),2);
is_pupil = (median_x<30)&(median_y<30);
pupils_center = centers(is_pupil == true,:);
Step 5: Draw the pupils.
The marker can be changed. Refer to:
https://de.mathworks.com/help/matlab/ref/matlab.graphics.chart.primitive.line-properties.html
figure
imshow(grayscale_image);
hold on
plot(centers(:,1), centers(:,2), 'r+', 'MarkerSize', 20, 'LineWidth', 2);
hold on
plot(pupils_center(:,1), pupils_center(:,2), 'b+', 'MarkerSize', 20, 'LineWidth', 2);
This is the final output:
I want to detect the points shown in the image below:
I have done this so far:
[X,map] = rgb2ind(img,0.0);
img = ind2gray(X,map); % Convert indexed to grayscale
level = graythresh(img); % Compute an appropriate threshold
img_bw = im2bw(img,level);% Convert grayscale to binary
mask = zeros(size(img_bw));
mask(2:end-2,2:end-2) = 1;
img_bw(mask<1) = 1;
%invert image
img_inv =1-img_bw;
% find blobs
img_blobs = bwmorph(img_inv,'majority',10);
% figure, imshow(img_blobs);
[rows, columns] = size(img_blobs);
for col = 1 : columns
thisColumn = img_blobs(:, col);
topRow = find(thisColumn, 1, 'first');
bottomRow = find(thisColumn, 1, 'last');
img_blobs(topRow : bottomRow, col) = true;
end
inverted = imcomplement(img_blobs);
ed = edge(inverted,'canny');
figure, imshow(ed),title('inverted');
Now how to proceed to get the coordinates of the desired position?
The top point is obviously the white pixel with the highest ordinate, which is easily obtained.
The bottom point is not so well defined. What you can do is
follow the peak edges until you reach a local minimum, on the left and on the right. That gives you a line segment, which you can intersect with the vertical through the top point.
if you know a the peak width, try every pixel on the vertical through the top point, downward, and stop until it has no left nor right neighbors at a distance equal to the peak with.
as above, but stop when the distance between the left and right neighbors exceeds a threshold.
In this particular case, you could consider using houghlines in matlab. Setting the required Theta and MinLength parameter values, you should be able to get the two vertical lines parallel to your peak. You can use the end points of the vertical lines, to get the point at the bottom.
Here is a sample code.
[H,theta,rho] = hough(bw,'Theta',5:1:30);%This is the angle range
P = houghpeaks(H,500,'NHoodSize',[11 11]);
lines = houghlines(bw,theta,rho,P,'FillGap',10,'MinLength',300);
Here is a complete description of how houghlines actually works.
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 a satellite image and I want to show an example of multi-scale.
so what I did is, add zeros (black pixels) around the image and put the image in the centre. now how can I fill the zeros from the first left, right, top, and bottom pixel row/column?
Matlab code:
img=imread ('example.jpg');
padcam = padarray(img,[1000 1000],'both');
EDIT:
Maybe it is easy explain from this image. Is this image what I want is to repeat the sea part in such a ways at the black area that it looks like we have large empty sea and small ships at the centre. That's why I have made red lines that I want to repeat/make copy/extend first left, right, top, and bottom pixel row/column so that make image at centre and black will be converted to pixels values of first left, right, top, and bottom pixel row/column.
I don't know if this is what you have meant...
Instead of fill zeros, start from the highest resolution, and place lower resolution images at the center of each other.
Here is my code sample (you may use a for loop instead):
I0 = imread('peppers.png');
I0 = padarray(I0,[16 16],'both'); %I0 - full resolution.
I1 = imresize(I0, 0.5); %I1 - half resolution.
I2 = imresize(I0, 0.25); %I2 - quarter resolution.
J = I0;
%Place I1 at the center of J.
J(1+(end-size(I1,1))/2:(end+size(I1,1))/2, 1+(end-size(I1,2))/2:(end+size(I1,2))/2, :) = I1;
%Place I2 at the center of J.
J(1+(end-size(I2,1))/2:(end+size(I2,1))/2, 1+(end-size(I2,2))/2:(end+size(I2,2))/2, :) = I2;
figure;imshow(J);
Result:
Check the following:
The example is basted on Matlab documentation of imtransform
I = imresize(imread('peppers.png'), 0.5);
A = [1, 0, 0; 0, 1, 0; 0, 0, 1];
T = maketform('affine', A);
R = makeresampler({'cubic', 'nearest'}, 'replicate');
J = imtransform(I, T, R, 'XData', [-size(I,2), size(I,2)*2], 'YData', [-size(I,1), size(I,1)*2]);
figure;imshow(J);
Result:
I think you are looking for 'replicate' padding option:
padcam = padarray(img,[1000 1000],'both', 'replicate');
resulting with:
If you are looking for smoother result, consider using regionfill.