I have a distribution of tube radius r, and I would like to plot tubes for all the sampled r in single figure as shown in the figure above. The tubes have following characteristics:
The length of all tubes is constant, but radius is varying.
The narrowest tube will be completely filled
with light gray color.
The length of the light gray color from bottom in all other tubes is
inversely proportional to the radius of the tube i.e.
length of light grey color from bottom = constant/r
The remaining length of the tube will be filled with dark gray color.
Magnitudes of r and total length of each tube is of the order of 1e-005m and 1e-002 m, respectively, so they need to be standardized compared to the X and Y axes units.
The white interspaces are just spaces and not tubes.
UPDATE (Based on the answer by Boris)
This is the code from Boris in which I made certain changes based on the characteristics of the tubes that I have described above. I am having scaling issues as I am not able to visualize my network of tubes as clearly as can be seen in the figure above.
function drawGrayTube (x, r, sigma_wn, theta, del_rho, rmin, rmax,L)
% sigma_wn is in N/m (=Kg/s^2), theta is in degrees, del_rho is in Kg/m^3
% and L is in m
h=((2*sigma_wn*cos((pi/180)*theta))./(del_rho*9.81.*r));
hmin=((2*sigma_wn*cos((pi/180)*theta))./(del_rho*9.81.*rmax));
hmax=((2*sigma_wn*cos((pi/180)*theta))./(del_rho*9.81.*rmin));
rectangle ('Position',[x,0,r/rmax,h], 'FaceColor',[0.7,0.7,0.7]);
ylim([0 1]);
if L>h
rectangle ('Position',[x,L,r/rmax,L-h], 'FaceColor',[0.3,0.3,0.3]);
ylim([0 1]);
else
rectangle ('Position',[x,L,r/rmax,L], 'FaceColor',[0.3,0.3,0.3]);
ylim([0 1]);
end
end
A simple function to draw the gray tubes could be for instance
function drawGrayTube (x, w, h)
rectangle ('Position',[x,0,w,h], 'FaceColor',[0.7,0.7,0.7]);
rectangle ('Position',[x,h,w,100-h], 'FaceColor',[0.3,0.3,0.3]);
end
Hereby, x is the x position of the tube, w denotes the width and h between 0 and 100 the height of the light gray part of the tube.
You can now use it in your example by calling
drawGrayTube (x, r, 100*constant/r)
where you have to adapt the constant such that constant/r is at most 1.
You can write a similar function for the white interspaces.
Assume that you have given a vector of radii (already scaled such that the values are between 0 and 1), e.g., r=[0.5, 0.7, 0.9, 0.1, 0.5, 0.01] then on possibility to draw the tubes is
interspace = 0.5;
for i=1:length(r)
drawGrayTube(sum(r(1:i-1))+i*interspace, 100*r(i)+1e-10, r(i)+1e-10);
end
You should use the function rectangle
Related
I want to plot 2D representation of a Gaussian wave function in MATLAB. I want the 2D plot to be in one color (green), which gets transparent away from the center of the Gaussian.
When I use imagesc (like in the next code) I get a Gaussian over a black square (as shown in the figure below).
I do not want the black backgrount, and I want the Gaussian to be in one color, but gets transparent away from the center, so that I won't get the black square, only green circle (over a white background) where the circle gets transparent away from its center.
How can I do that ???
close all;clc
figure
xlim_min=-4;
xlim_max=4;
ylim_min=-4;
ylim_max=4;
ylim([ylim_min ylim_max])
xlim([xlim_min xlim_max])
x=-1:0.001:1;
y=-1:0.001:1;
[X,Y]=meshgrid(x,y);
c1=10;
c1_new=c1*0.3;
x_offset=0;
y_offset=0;
w_function=0.5*0.25*exp(-c1_new*((X+x_offset).^2+...
(Y+y_offset).^2));
imagesc(x,y,w_function);
ylim([ylim_min ylim_max])
xlim([xlim_min xlim_max])
To obtain transparency, use the AlphaData and AlphaDataMapping properties of the image. In the following, I'm explicitly computing the transparency (alpha) as an affine function of the data values, in order to specify minimum and maximum transparency values; and I set AlphaDataMapping to none so those values are used without any modifications.
So, replace your line
imagesc(x,y,w_function);
by
min_alpha = .2; % desired minimum alpha
max_alpha = 1; % desired maximum alpha
alpha = min_alpha + (max_alpha-min_alpha)/max(w_function(:))*w_function; % compute alpha
imagesc(x,y,w_function,'AlphaData',alpha,'AlphaDataMapping','none'); % image with alpha
Note how the blue part is more transparent (smaller alpha) than the yellow part.
To have the color blend smoothly with the outer region, set a mininum alpha of 0 and define the function on a larger grid. You can also change the colormap if desired. Since the transparency already varies, you probably want constant color in the colormap.
close all;clc
figure
xlim_min=-4;
xlim_max=4;
ylim_min=-4;
ylim_max=4;
ylim([ylim_min ylim_max])
xlim([xlim_min xlim_max])
x=-2:0.001:2; % large enough that the function approximately ...
y=-2:0.001:2; % ... reaches 0 within this rectangle
[X,Y]=meshgrid(x,y);
c1=10;
c1_new=c1*0.3;
x_offset=0;
y_offset=0;
w_function=0.5*0.25*exp(-c1_new*((X+x_offset).^2+...
(Y+y_offset).^2));
min_alpha = 0; % desired minimum alpha: set to 0
max_alpha = 1; % desired maximum alpha
alpha = min_alpha + (max_alpha-min_alpha)/max(w_function(:))*w_function; % compute alpha
imagesc(x,y,w_function,'AlphaData',alpha,'AlphaDataMapping','none'); % image with alpha
cm = [0 .7 0]; % define colormap: single color, dark green
colormap(cm) % apply colormap
ylim([ylim_min ylim_max])
xlim([xlim_min xlim_max])
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'm very new to Matlab. I need to create squares on white canvases based off specs given to me in a file squares.txt. The text document contains a matrix of numbers with 6 columns and some rows. The first two numbers in each row of the matrix are the center coordinates of the square to be drawn. The third value is the width, and the last values are RGB values. I'm having a very difficult time finding a way to draw a square on a canvas given the center coordinates.
Drawing a colored square, given the centre coordinates, the width and the RGB values, looks like:
sq = rectangle('Position', [i-width/2, j-width/2, width, width],'EdgeColor',[r g b]);
depending on whether your color values are between 0 and 1 or 0 and 255 you may have to divide r, g and b by 255
Edit:
Minimal example without using the file:
width=100;
for count = 1:2
if (count == 1)
canvas = ones(400,400,3);
else
canvas = ones(300,700,3);
end
figure, imshow (canvas);
rectangle('Position', [100-width/2, 200-width/2, width, width],
EdgeColor', [0.5 0.5 0.2]);
end
I am doing a research for my higher studies in automation. I have done the automation part of the microscope but I need help in MATLAB. An example of what I would like to segment is shown here:
I need to extract the dark purple pixels from this image and only display that in a figure. It is almost like colour based segmentation but I just want to only take the dark purple pixel from the whole image.
What would I do in this case?
Here's something to get you started. Let's go with the theme of colour segmentation where you only want to extract pixels that are of a deep purple. I would like to point you to the HSV colour space before we get started. The HSV colour space is ideal for representing colours in a way that is most intuitive to humans. We tend to describe colours by their dominant colour, followed by attributes such as how washed out or how pure the colour is, and how bright or dark the colour is. The dominant colour is represented by the Hue, the appearance of how washed out or how pure the colour is is represented by the Saturation and the intensity of the colour is represented by the Value, and hence Hue-Saturation-Value, or the HSV colour space.
We can transform a RGB image so that it becomes HSV by rgb2hsv. This will return a 3D matrix that has the hue, saturation and value as 2D slices in a 3D matrix, much like a RGB image where each slices represents the red, green and blue channels. Let's see what each component looks like once we transform the image into HSV:
im = imread('https://www.cdc.gov/dpdx/images/malaria/ovale/Po_gametocyte_thickB.jpg');
hsv = rgb2hsv(im2double(im));
figure;
for idx = 1 : 3
subplot(1,3,idx);
imshow(hsv(:,:,idx));
end
The first line of code reads in an image from a URL. I'm going to use the one that Hoki referred you to, as it's the most simplest one to deal with. For self-containment, this is what the original image looks like:
Once we do this, we convert the image into the HSV colour space. It is important that you convert the image to double precision and you normalize each component to [0,1], and that is performed by im2double. Next, we spawn a new figure, and place each component in a single row over three columns. The first column represents the hue, next column the saturation and finally the last column being the value. This is the figure that we see:
With the first figure, it looks like the dominant colour is purple, whether it's a light shade or a dark shade of the colour, so the hue won't help us here. If you look at a HSV colour wheel:
(source: hobbitsandhobos.com)
Normalize the wheel so that it falls between [0,1] instead of 0 to 360 degrees. The hue is actually represented as degrees due to the nature of the colour space, but MATLAB normalizes this to [0,1]. You can see that purple falls within a hue of [0.6,0.8], which corresponds to the first figure I showed you that displays the hue for our image. If you examine the pixels around the image, they fluctuate between this range. Therefore, the hue won't help us much here.
What will certainly help us are the saturation and value components. If you take a look, the deep purple pixels have a higher saturation than the rest of the background, which makes sense because the deep purple has a much more pure version of purple than the rest of the background. For the value, you can see that the brightness of the dark purple is darker than the background.
We can use these two points as an exploit to segment out the purple colour in the image. The easiest thing to do would be to threshold the saturation and value planes so that any values that are within a certain range you keep while those that are outside you throw away. Therefore, you can do something like this:
sThresh = hsv(:,:,2) > 0.6 & hsv(:,:,2) < 0.9;
vThresh = hsv(:,:,3) > 0.4 & hsv(:,:,3) < 0.65;
I used impixelinfo and I hovered my mouse over the saturation and value components to examine what the values were for the deep purple regions. It looks like those pixels that are deep purple have a saturation value between 0.6 and 0.9, while the value component has values between 0.4 and 0.65. The above code will create two binary masks where true means that the pixel satisfies our criteria while false means it doesn't. Because I want to combine both things together and not leave any stone unturned, let's logical OR the masks together for the final result:
figure;
result = sThresh | vThresh;
imshow(result);
We will also show the result too. This is what we get:
As you can see, this does a pretty good job, but we have remnants of the red arrow that we don't want in the final result. To do a bit of cleanup, we can use morphology - specifically an opening filter of a small window so that we don't affect the pixels that we want as much. We can use imopen to perform our opening operation for us. A morphological opening removes isolated pixels that appear around your image. You use what is called a structuring element that is used to look at local neighbourhoods of your image. For the basics, any pixel regions that are as small as the shape that is contained within the structuring element get removed. Because we want to preserve the shape of the other objects, we can try using a 5 x 5 disk structuring element to clean these pixels up:
figure;
se = strel('disk', 2, 0);
final = imopen(result, se);
imshow(final);
This is what we get:
Not bad! There are some holes that we need to patch up, so let's fill in those holes with imfill:
figure;
final_noholes = imfill(final, 'holes');
imshow(final_noholes);
This is what we get:
OK! So we have our mask. The last thing we need to do is present the image so that you only show the deep purple colours from the original image, and nothing else. That can easily be achieved with bsxfun:
figure;
out = bsxfun(#times, im, uint8(final_noholes));
imshow(out);
The above operation takes your mask, and multiplies every pixel in your image by this mask. One small thing I'd like to point out is that the mask we found in the previous step needs to be cast to uint8, because bsxfun requires that the multiplication (or whatever operation you perform) need to be the same type. We replicate this mask in 3D so that you mask out the unwanted RGB pixels and only keep the ones you are looking for.
This is what we finally get:
As you can see, it isn't perfect, but it's certainly enough to get you started. Those thresholds are what are important, but with some very simple thresholding, I extracted most of the purple pixels out.
To make it easier for you, here's the code that I wrote above that can easily be copied and pasted into MATLAB for you to run:
clear all; close all; clc;
im = imread('https://www.cdc.gov/dpdx/images/malaria/ovale/Po_gametocyte_thickB.jpg');
hsv = rgb2hsv(im2double(im));
figure;
for idx = 1 : 3
subplot(1,3,idx);
imshow(hsv(:,:,idx));
end
sThresh = hsv(:,:,2) > 0.6 & hsv(:,:,2) < 0.9;
vThresh = hsv(:,:,3) > 0.4 & hsv(:,:,3) < 0.65;
figure;
result = sThresh | vThresh;
imshow(result);
figure;
se = strel('disk', 2, 0);
final = imopen(result, se);
imshow(final);
figure;
final_noholes = imfill(final, 'holes');
imshow(final_noholes);
figure;
out = bsxfun(#times, im, uint8(final_noholes));
imshow(out);
Good luck!
Try this:
function main
clc,clear
A = imread('https://www.cdc.gov/dpdx/images/malaria/ovale/Po_gametocyte_thickB.jpg');
subplot(1,2,1)
imshow(A)
RGB = [230 210 200]; % color you want
e = 40; % color shift
B = pix_in(A,RGB,e);
B = B + 255.*uint8(~B); % choosing white background
subplot(1,2,2)
imshow(B)
end
function B = pix_in(A,RGB,e)
% select specific pixels in image
% A - color image (3D matrix uint8)
% RGB - [R G B] - color to select
% e - color shift/deviation
A = double(A); % for same class operations (RGB - double)
[m, n, ~] = size(A);
RGB = reshape(RGB,1,1,3);
RGB = repmat(RGB,m,n,1); % creating 3D matrix
b = abs(A-RGB) < e; % logical 3D
b = sum(b,3) == 3; % if [R,G,B] of a pixel in range
B = A.*repmat(b,1,1,3); % selecting pixels those in range
B = uint8(B);
end
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]);