I have an image with a white background and trying to extract the object within it using MATLAB. I tried different techniques (bounded box) but I am not getting desired results. I have tried this StackOverflow post but it isn't quite working: MATLAB Auto Crop
How I can extract the object? Any idea which technique is suitable?
The image is shown below:
I propose the following workflow:
Binarize the image
Get the centroid and orientation of the main object
Crop the image to set the center of the object at the center of the image
Rotate the object with respect to the center to compensate for the tilt
Crop again to get the object
In practice, step 1 turned out to be rather complicated to perform because of the shadow and of the not-so-well-defined boundaries between the white part of the object and the background. The trick I used relies on the fact that, because of the many details inside the object (and the jpg compression) there is a lot of "noise" all over it. So looking for edges combined with a BW-dilation produced a very decent binarized image. I then used regioprops to filter the small outliers and get the properties of the object.
Here is the code:
% --- Load image
Img = imread('Tilted.jpg');
% --- Binarisation
% Find edges and dilate to get a bloc
BW = edge(mean(Img,3),'sobel');
BW = imdilate(BW, strel('disk',5));
% Get BW image
RP = regionprops(BW, 'Area', 'PixelIdxList', 'Centroid', 'Orientation');
[~, i] = max([RP(:).Area]);
BW = Img(:,:,1)*0;
BW(RP(i).PixelIdxList) = 1;
x = RP(i).Centroid(1);
y = RP(i).Centroid(2);
theta = -RP(i).Orientation;
% imshow(BW)
% plot(x+[0 500*cos(theta*pi/180)], y+[0 500*sin(theta*pi/180)]);
% return
% --- First crop
w = min(x-1, size(Img,2)-x);
h = min(y-1, size(Img,1)-y);
C = Img(ceil(y-h):floor(y+h), ceil(x-w):floor(x+w),:);
% --- Rotate image
R = imrotate(C, theta);
R(R==0) = 255;
% --- Second crop
% Parameters (to adjust)
th = 70;
margin = 25;
% Remove grey levels
Rwg = std(double(R),0,3).^2;
% Get x-bounds
sx = sum(Rwg,1);
x1 = max(find(sx>prctile(sx,th), 1, 'first')-margin, 1);
x2 = min(find(sx>prctile(sx,30), 1, 'last')+margin, size(Rwg,1));
% Get y-bounds
sy = sum(Rwg,2);
y1 = max(find(sy>prctile(sy,th), 1, 'first')-margin, 1);
y2 = min(find(sy>prctile(sy,th), 1, 'last')+margin, size(Rwg,1));
Res = R(y1:y2, x1:x2, :);
% --- Display
imshow(Res)
Of course, if you have many different images to process this may not work every time. This code is only a starting point and you now have to modify it to meet your precise needs.
Best,
Related
I have an image (white background with 1-5 black dots) that is called main.jpg (main image).
I am trying to place another image (secondary.jpg) in every black dot that is found in main image.
In order to do that:
I found the black pixels in main image
resize the secondary image to specific size that I want
plot the image in every coordinate that I found in step one. (the black pixel should be the center coordinates of the secondary image)
Unfortunately, I don't know how to do the third step.
for example:
main image is:
secondary image is:
output:
(The dots are behind the chairs. They are the image center points)
This is my code:
mainImage=imread('main.jpg')
secondaryImage=imread('secondary.jpg')
secondaryImageResized = resizeImage(secondaryImage)
[m n]=size(mainImage)
for i=1:n
for j=1:m
% if it's black pixel
if (mainImage(i,j)==1)
outputImage = plotImageInCoordinates(secondaryImageResized, i, j)
% save this image
imwrite(outputImage,map,'clown.bmp')
end
end
end
% resize the image to (250,350) width, height
function [ Image ] = resizeImage(img)
image = imresize(img, [250 350]);
end
function [outputImage] = plotImageInCoordinates(image, x, y)
% Do something
end
Any help appreciated!
Here's an alternative without convolution. One intricacy that you must take into account is that if you want to place each image at the centre of each dot, you must determine where the top left corner is and index into your output image so that you draw the desired object from the top left corner to the bottom right corner. You can do this by taking each black dot location and subtracting by half the width horizontally and half the height vertically.
Now onto your actual problem. It's much more efficient if you loop through the set of points that are black, not the entire image. You can do this by using the find command to determine the row and column locations that are 0. Once you do this, loop through each pair of row and column coordinates, do the subtraction of the coordinates and then place it on the output image.
I will impose an additional requirement where the objects may overlap. To accommodate for this, I will accumulate pixels, then find the average of the non-zero locations.
Your code modified to accommodate for this is as follows. Take note that because you are using JPEG compression, you will have compression artifacts so regions that are 0 may not necessarily be 0. I will threshold with an intensity of 128 to ensure that zero regions are actually zero. You will also have the situation where objects may go outside the boundaries of the image. Therefore to accommodate for this, pad the image sufficiently with twice of half the width horizontally and twice of half the height vertically then crop it after you're done placing the objects.
mainImage=imread('https://i.stack.imgur.com/gbhWJ.png');
secondaryImage=imread('https://i.stack.imgur.com/P0meM.png');
secondaryImageResized = imresize(secondaryImage, [250 300]);
% Find half height and width
rows = size(secondaryImageResized, 1);
cols = size(secondaryImageResized, 2);
halfHeight = floor(rows / 2);
halfWidth = floor(cols / 2);
% Create a padded image that contains our main image. Pad with white
% pixels.
rowsMain = size(mainImage, 1);
colsMain = size(mainImage, 2);
outputImage = 255*ones([2*halfHeight + rowsMain, 2*halfWidth + colsMain, size(mainImage, 3)], class(mainImage));
outputImage(halfHeight + 1 : halfHeight + rowsMain, ...
halfWidth + 1 : halfWidth + colsMain, :) = mainImage;
% Find a mask of the black pixels
mask = outputImage(:,:,1) < 128;
% Obtain black pixel locations
[row, col] = find(mask);
% Reset the output image so that they're all zeros now. We use this
% to output our final image. Also cast to ensure accumulation is proper.
outputImage(:) = 0;
outputImage = double(outputImage);
% Keeps track of how many times each pixel was hit by the object
% This is so that we can find the average at each location.
counts = zeros([size(mask), size(mainImage, 3)]);
% For each row and column location in the image
for i = 1 : numel(row)
% Get the row and column locations
r = row(i); c = col(i);
% Offset to get the top left corner
r = r - halfHeight;
c = c - halfWidth;
% Place onto final image
outputImage(r:r+rows-1, c:c+cols-1, :) = outputImage(r:r+rows-1, c:c+cols-1, :) + double(secondaryImageResized);
% Accumulate the counts
counts(r:r+rows-1,c:c+cols-1,:) = counts(r:r+rows-1,c:c+cols-1,:) + 1;
end
% Find average - Any values that were not hit, change to white
outputImage = outputImage ./ counts;
outputImage(counts == 0) = 255;
outputImage = uint8(outputImage);
% Now crop and show
outputImage = outputImage(halfHeight + 1 : halfHeight + rowsMain, ...
halfWidth + 1 : halfWidth + colsMain, :);
close all; imshow(outputImage);
% Write the final output
imwrite(outputImage, 'finalimage.jpg', 'Quality', 100);
We get:
Edit
I wasn't told that your images had transparency. Therefore what you need to do is use imread but ensure that you read in the alpha channel. We then check to see if one exists and if one does, we will ensure that the background of any values with no transparency are set to white. You can do that with the following code. Ensure this gets placed at the very top of your code, replacing the images being loaded in:
mainImage=imread('https://i.stack.imgur.com/gbhWJ.png');
% Change - to accommodate for transparency
[secondaryImage, ~, alpha] = imread('https://i.imgur.com/qYJSzEZ.png');
if ~isempty(alpha)
m = alpha == 0;
for i = 1 : size(secondaryImage,3)
m2 = secondaryImage(:,:,i);
m2(m) = 255;
secondaryImage(:,:,i) = m2;
end
end
secondaryImageResized = imresize(secondaryImage, [250 300]);
% Rest of your code follows...
% ...
The code above has been modified to read in the basketball image. The rest of the code remains the same and we thus get:
You can use convolution to achieve the desired effect. This will place a copy of im everywhere there is a black dot in imz.
% load secondary image
im = double(imread('secondary.jpg'))/255.0;
% create some artificial image with black indicators
imz = ones(500,500,3);
imz(50,50,:) = 0;
imz(400,200,:) = 0;
imz(200,400,:) = 0;
% create output image
imout = zeros(size(imz));
imout(:,:,1) = conv2(1-imz(:,:,1),1-im(:,:,1),'same');
imout(:,:,2) = conv2(1-imz(:,:,2),1-im(:,:,2),'same');
imout(:,:,3) = conv2(1-imz(:,:,3),1-im(:,:,3),'same');
imout = 1-imout;
% output
imshow(imout);
Also, you probably want to avoid saving main.jpg as a .jpg since it results in lossy compression and will likely cause issues with any method that relies on exact pixel values. I would recommend using .png which is lossless and will also likely compress better than .jpg for synthetic images where the same colors repeat many times.
I have 2 images ("before" and "after"). I would like to show a final image where the left half is taken from the before image and the right half is taken from the after image.
The images should be separated by a white diagonal line of predefined width (2 or 3 pixels), where the diagonal is specified either by a certain angle or by 2 start and end coordinates. The diagonal should overwrite a part of the final image such that the size is the same as the sources'.
Example:
I know it can be done by looping over all pixels to recombine and create the final image, but is there an efficient way, or better yet, a built-in function that can do this?
Unfortunately I don't believe there is a built-in solution to your problem, but I've developed some code to help you do this but it will unfortunately require the image processing toolbox to play nicely with the code. As mentioned in your comments, you have this already so we should be fine.
The logic behind this is relatively simple. We will assume that your before and after pictures are the same size and also share the same number of channels. The first part is to declare a blank image and we draw a straight line down the middle of a certain thickness. The intricacy behind this is to declare an image that is slightly bigger than the original size of the image. The reason why is because I'm going to draw a line down the middle, then rotate this blank image by a certain angle to achieve the first part of what you desire. I'll be using imrotate to rotate an image by any angle you desire. The first instinct is to declare an image that's the same size as either the originals, draw a line down the middle and rotate it. However, if you do this you'll end up with the line being disconnected and not draw from the top to the bottom of the image. That makes sense because the line being drawn on an angle covers more pixels than if you were to draw this vertically.
Using Pythagorean's theorem, we know that the longest line that can ever be drawn on your image is the diagonal. Therefore we declare an image that is sqrt(rows*rows + cols*cols) in both the rows and columns where rows and cols are the rows and columns of the original image. After, we'll take the ceiling to make sure we've covered as much as possible and we add a bit of extra room to accommodate for the width of the line. We draw a line on this image, rotate it then we'll crop the image after so that it's the same size as the input images. This ensures that the line drawn at whatever angle you wish is fully drawn from top to bottom.
That logic is the hardest part. Once you do that, you declare two logical masks where you use imfill to fill the left side of the mask as one mask and we'll invert the mask to find the other mask. You will also need to use the line image that we created earlier with imrotate to index into the masks and set the values to false so that we ignore these pixels that are on the line.
Finally, you take each mask, index into your image and copy over each portion of the image you desire. You finally use the line image to index into the output and set the values to white.
Without further ado, here's the code:
% Load some example data
load mandrill;
% im is the image before
% im2 is the image after
% Before image is a colour image
im = im2uint8(ind2rgb(X, map));
% After image is a grayscale image
im2 = rgb2gray(im);
im2 = cat(3, im2, im2, im2);
% Declare line image
rows = size(im, 1); cols = size(im, 2);
width = 5;
m = ceil(sqrt(rows*rows + cols*cols + width*width));
ln = false([m m]);
mhalf = floor(m / 2); % Find halfway point width wise and draw the line
ln(:,mhalf - floor(width/2) : mhalf + floor(width/2)) = true;
% Rotate the line image
ang = 20; % 20 degrees
lnrotate = imrotate(ln, ang, 'crop');
% Crop the image so that it's the same dimensions as the originals
mrowstart = mhalf - floor(rows/2);
mcolstart = mhalf - floor(cols/2);
lnfinal = lnrotate(mrowstart : mrowstart + rows - 1, mcolstart : mcolstart + cols - 1);
% Make the masks
mask1 = imfill(lnfinal, [1 1]);
mask2 = ~mask1;
mask1(lnfinal) = false;
mask2(lnfinal) = false;
% Make sure the masks have as many channels as the original
mask1 = repmat(mask1, [1 1 size(im,3)]);
mask2 = repmat(mask2, [1 1 size(im,3)]);
% Do the same for the line
lnfinal = repmat(lnfinal, [1 1 size(im, 3)]);
% Specify output image
out = zeros(size(im), class(im));
out(mask1) = im(mask1);
out(mask2) = im2(mask2);
out(lnfinal) = 255;
% Show the image
figure;
imshow(out);
We get:
If you want the line to go in the other direction, simply make the angle ang negative. In the example script above, I've made the angle 20 degrees counter-clockwise (i.e. positive). To reproduce the example you gave, specify -20 degrees instead. I now get this image:
Here's a solution using polygons:
function q44310306
% Load some image:
I = imread('peppers.png');
B = rgb2gray(I);
lt = I; rt = B;
% Specify the boundaries of the white line:
width = 2; % [px]
offset = 13; % [px]
sz = size(I);
wlb = [floor(sz(2)/2)-offset+[0,width]; ceil(sz(2)/2)+offset-[width,0]];
% [top-left, top-right; bottom-left, bottom-right]
% Configure two polygons:
leftPoly = struct('x',[1 wlb(1,2) wlb(2,2) 1], 'y',[1 1 sz(1) sz(1)]);
rightPoly = struct('x',[sz(2) wlb(1,1) wlb(2,1) sz(2)],'y',[1 1 sz(1) sz(1)]);
% Define a helper grid:
[XX,YY] = meshgrid(1:sz(2),1:sz(1));
rt(inpolygon(XX,YY,leftPoly.x,leftPoly.y)) = intmin('uint8');
lt(repmat(inpolygon(XX,YY,rightPoly.x,rightPoly.y),1,1,3)) = intmin('uint8');
rt(inpolygon(XX,YY,leftPoly.x,leftPoly.y) & ...
inpolygon(XX,YY,rightPoly.x,rightPoly.y)) = intmax('uint8');
final = bsxfun(#plus,lt,rt);
% Plot:
figure(); imshow(final);
The result:
One solution:
im1 = imread('peppers.png');
im2 = repmat(rgb2gray(im1),1,1,3);
imgsplitter(im1,im2,80) %imgsplitter(image1,image2,angle [0-100])
function imgsplitter(im1,im2,p)
s1 = size(im1,1); s2 = size(im1,2);
pix = floor(p*size(im1,2)/100);
val = abs(pix -(s2-pix));
dia = imresize(tril(ones(s1)),[s1 val]);
len = min(abs([0-pix,s2-pix]));
if p>50
ind = [ones(s1,len) fliplr(~dia) zeros(s1,len)];
else
ind = [ones(s1,len) dia zeros(s1,len)];
end
ind = uint8(ind);
imshow(ind.*im1+uint8(~ind).*im2)
hold on
plot([pix,s2-pix],[0,s1],'w','LineWidth',1)
end
OUTPUT:
If I have an image, in which there is a page of text shot on a uniform background, how can I auto detect the boundaries between the paper and the background?
An example of the image I want to detect is shown below. The images that I will be dealing with consist of a single page on a uniform background and they can be rotated at any angle.
One simple method would be to threshold the image by some known value once you convert the image to grayscale. The problem with that approach is that we are applying a global threshold and so some of the paper at the bottom of the image will be lost if you make the threshold too high. If you make the threshold too low, then you'll certainly get the paper, but you'll include a lot of the background pixels too and it will probably be difficult to remove those pixels with post-processing.
One thing I can suggest is to use an adaptive threshold algorithm. An algorithm that has worked for me in the past is the Bradley-Roth adaptive thresholding algorithm. You can read up about it here on a post I commented on a while back:
Bradley Adaptive Thresholding -- Confused (questions)
However, if you want the gist of it, an integral image of the grayscale version of the image is taken first. The integral image is important because it allows you to calculate the sum of pixels within a window in O(1) complexity. However, the calculation of the integral image is usually O(n^2), but you only have to do that once. With the integral image, you scan neighbourhoods of pixels of size s x s and you check to see if the average intensity is less than t% of the actual average within this s x s window then this is pixel classified as the background. If it's larger, then it's classified as being part of the foreground. This is adaptive because the thresholding is done using local pixel neighbourhoods rather than using a global threshold.
I've coded an implementation of the Bradley-Roth algorithm here for you. The default parameters for the algorithm are s being 1/8th of the width of the image and t being 15%. Therefore, you can just call it this way to invoke the default parameters:
out = adaptiveThreshold(im);
im is the input image and out is a binary image that denotes what belongs to foreground (logical true) or background (logical false). You can play around with the second and third input parameters: s being the size of the thresholding window and t the percentage we talked about above and can call the function like so:
out = adaptiveThreshold(im, s, t);
Therefore, the code for the algorithm looks like this:
function [out] = adaptiveThreshold(im, s, t)
%// Error checking of the input
%// Default value for s is 1/8th the width of the image
%// Must make sure that this is a whole number
if nargin <= 1, s = round(size(im,2) / 8); end
%// Default value for t is 15
%// t is used to determine whether the current pixel is t% lower than the
%// average in the particular neighbourhood
if nargin <= 2, t = 15; end
%// Too few or too many arguments?
if nargin == 0, error('Too few arguments'); end
if nargin >= 4, error('Too many arguments'); end
%// Convert to grayscale if necessary then cast to double to ensure no
%// saturation
if size(im, 3) == 3
im = double(rgb2gray(im));
elseif size(im, 3) == 1
im = double(im);
else
error('Incompatible image: Must be a colour or grayscale image');
end
%// Compute integral image
intImage = cumsum(cumsum(im, 2), 1);
%// Define grid of points
[rows, cols] = size(im);
[X,Y] = meshgrid(1:cols, 1:rows);
%// Ensure s is even so that we are able to index the image properly
s = s + mod(s,2);
%// Access the four corners of each neighbourhood
x1 = X - s/2; x2 = X + s/2;
y1 = Y - s/2; y2 = Y + s/2;
%// Ensure no co-ordinates are out of bounds
x1(x1 < 1) = 1;
x2(x2 > cols) = cols;
y1(y1 < 1) = 1;
y2(y2 > rows) = rows;
%// Count how many pixels there are in each neighbourhood
count = (x2 - x1) .* (y2 - y1);
%// Compute row and column co-ordinates to access each corner of the
%// neighbourhood for the integral image
f1_x = x2; f1_y = y2;
f2_x = x2; f2_y = y1 - 1; f2_y(f2_y < 1) = 1;
f3_x = x1 - 1; f3_x(f3_x < 1) = 1; f3_y = y2;
f4_x = f3_x; f4_y = f2_y;
%// Compute 1D linear indices for each of the corners
ind_f1 = sub2ind([rows cols], f1_y, f1_x);
ind_f2 = sub2ind([rows cols], f2_y, f2_x);
ind_f3 = sub2ind([rows cols], f3_y, f3_x);
ind_f4 = sub2ind([rows cols], f4_y, f4_x);
%// Calculate the areas for each of the neighbourhoods
sums = intImage(ind_f1) - intImage(ind_f2) - intImage(ind_f3) + ...
intImage(ind_f4);
%// Determine whether the summed area surpasses a threshold
%// Set this output to 0 if it doesn't
locs = (im .* count) <= (sums * (100 - t) / 100);
out = true(size(im));
out(locs) = false;
end
When I use your image and I set s = 500 and t = 5, here's the code and this is the image I get:
im = imread('http://i.stack.imgur.com/MEcaz.jpg');
out = adaptiveThreshold(im, 500, 5);
imshow(out);
You can see that there are some spurious white pixels at the bottom white of the image, and there are some holes we need to fill in inside the paper. As such, let's use some morphology and declare a structuring element that's a 15 x 15 square, perform an opening to remove the noisy pixels, then fill in the holes when we're done:
se = strel('square', 15);
out = imopen(out, se);
out = imfill(out, 'holes');
imshow(out);
This is what I get after all of that:
Not bad eh? Now if you really want to see what the image looks like with the paper segmented, we can use this mask and multiply it with the original image. This way, any pixels that belong to the paper are kept while those that belong to the background go away:
out_colour = bsxfun(#times, im, uint8(out));
imshow(out_colour);
We get this:
You'll have to play around with the parameters until it works for you, but the above parameters were the ones I used to get it working for the particular page you showed us. Image processing is all about trial and error, and putting processing steps in the right sequence until you get something good enough for your purposes.
Happy image filtering!
After I did a 'imclearborder', there are still a bit of unwanted object around the barcode. How can I remove those objects to isolate the barcode? I have pasted my code for your reference.
rgb = imread('barcode2.jpg');
% Resize Image
rgb = imresize(rgb,0.33);
figure(),imshow(rgb);
% Convert from RGB to Gray
Igray = double(rgb2gray(rgb));
% Calculate the Gradients
[dIx, dIy] = gradient(Igray);
B = abs(dIx) - abs(dIy);
% Low-Pass Filtering
H = fspecial('gaussian', 20, 10);
C = imfilter(B, H);
C = imclearborder(C);
figure(),imagesc(C);colorbar;
Well, i have already explained it in your previous question How to find the location of red region in an image using MATLAB? , but with a opencv code and output images.
Instead of asking for code, try to implement it yourself.
Below is what to do next.
1) convert image 'C' in your code to binary.
2) Apply some erosion to remove small noises.( this time, barcode region also shrinks)
3) Apply dilation to compensate previous erosion.(most of noise will have removed in previous erosion. So they won't come back)
4) Find contours in the image.
5) Find their area. Most probably, contour which has maximum area will be the barcode, because other things like letters, words etc will be small ( you can understand it in the grayscale image you have provided)
6) Select contour with max. area. Draw a bounding rectangle for it.
Its result is already provided in your previous question. It works very nice. Try to implement it yourself with help of MATLAB documentation. Come here only when you get an error which you don't understand.
%%hi, i am ading my code to yours at the end of your code%%%%
clear all;
rgb = imread('barcode.jpeg');
% Resize Image
rgb = imresize(rgb,0.33);
figure(),imshow(rgb);
% Convert from RGB to Gray
Igray = double(rgb2gray(rgb));
Igrayc = Igray;
% Calculate the Gradients
[dIx, dIy] = gradient(Igray);
B = abs(dIx) - abs(dIy);
% Low-Pass Filtering
H = fspecial('gaussian', 10, 5);
C = imfilter(B, H);
C = imclearborder(C);
imshow(Igray,[]);
figure(),imagesc(C);colorbar;
%%%%%%%%%%%%%%%%%%%%%%%%from here my code starts%%%%%%%%%%%%%%%%
bw = im2bw(C);%%%binarising the image
% imshow(bw);
%%%%if there are letters or any other noise is present around the barcode
%%Note: the size of the noise and letters should be smaller than the
%%barcode size
labelImage = bwlabel(bw,8);
len=0;labe=0;
for i=1:max(max(labelImage))
a = find(labelImage==i);
if(len<length(a))
len=length(a);
labe=i;
end
end
imag = zeros(size(l));
imag(find(labelImage==labe))=255;
% imtool(imag);
%%%if Necessary do errossion
% se2 = strel('line',10,0);
% imag= imerode(imag,se2);
% imag= imerode(imag,se2);
[r c]= find(imag==255);
minr = min(r);
maxc = max(c);
minc = min(c);
maxr = max(r);
imag1 = zeros(size(l));
for i=minr:maxr
for j=minc:maxc
imag1(i,j)=255;
end
end
% figure,imtool(imag1);
varit = find(imag1==0);
Igrayc(varit)=0;
%%%%%result image having only barcode
imshow(Igrayc,[]);
%%%%%original image
figure(),imshow(Igray,[]);
Hope it is useful
I have an image in MATLAB:
im = rgb2gray(imread('some_image.jpg');
% normalize the image to be between 0 and 1
im = im/max(max(im));
And I've done some processing that resulted in a number of points that I want to highlight:
points = some_processing(im);
Where points is a matrix the same size as im with ones in the interesting points.
Now I want to draw a circle on the image in all the places where points is 1.
Is there any function in MATLAB that does this? The best I can come up with is:
[x_p, y_p] = find (points);
[x, y] = meshgrid(1:size(im,1), 1:size(im,2))
r = 5;
circles = zeros(size(im));
for k = 1:length(x_p)
circles = circles + (floor((x - x_p(k)).^2 + (y - y_p(k)).^2) == r);
end
% normalize circles
circles = circles/max(max(circles));
output = im + circles;
imshow(output)
This seems more than somewhat inelegant. Is there a way to draw circles similar to the line function?
You could use the normal PLOT command with a circular marker point:
[x_p,y_p] = find(points);
imshow(im); %# Display your image
hold on; %# Add subsequent plots to the image
plot(y_p,x_p,'o'); %# NOTE: x_p and y_p are switched (see note below)!
hold off; %# Any subsequent plotting will overwrite the image!
You can also adjust these other properties of the plot marker: MarkerEdgeColor, MarkerFaceColor, MarkerSize.
If you then want to save the new image with the markers plotted on it, you can look at this answer I gave to a question about maintaining image dimensions when saving images from figures.
NOTE: When plotting image data with IMSHOW (or IMAGE, etc.), the normal interpretation of rows and columns essentially becomes flipped. Normally the first dimension of data (i.e. rows) is thought of as the data that would lie on the x-axis, and is probably why you use x_p as the first set of values returned by the FIND function. However, IMSHOW displays the first dimension of the image data along the y-axis, so the first value returned by FIND ends up being the y-coordinate value in this case.
This file by Zhenhai Wang from Matlab Central's File Exchange does the trick.
%----------------------------------------------------------------
% H=CIRCLE(CENTER,RADIUS,NOP,STYLE)
% This routine draws a circle with center defined as
% a vector CENTER, radius as a scaler RADIS. NOP is
% the number of points on the circle. As to STYLE,
% use it the same way as you use the rountine PLOT.
% Since the handle of the object is returned, you
% use routine SET to get the best result.
%
% Usage Examples,
%
% circle([1,3],3,1000,':');
% circle([2,4],2,1000,'--');
%
% Zhenhai Wang <zhenhai#ieee.org>
% Version 1.00
% December, 2002
%----------------------------------------------------------------
Funny! There are 6 answers here, none give the obvious solution: the rectangle function.
From the documentation:
Draw a circle by setting the Curvature property to [1 1]. Draw the circle so that it fills the rectangular area between the points (2,4) and (4,6). The Position property defines the smallest rectangle that contains the circle.
pos = [2 4 2 2];
rectangle('Position',pos,'Curvature',[1 1])
axis equal
So in your case:
imshow(im)
hold on
[y, x] = find(points);
for ii=1:length(x)
pos = [x(ii),y(ii)];
pos = [pos-0.5,1,1];
rectangle('position',pos,'curvature',[1 1])
end
As opposed to the accepted answer, these circles will scale with the image, you can zoom in an they will always mark the whole pixel.
Hmm I had to re-switch them in this call:
k = convhull(x,y);
figure;
imshow(image); %# Display your image
hold on; %# Add subsequent plots to the image
plot(x,y,'o'); %# NOTE: x_p and y_p are switched (see note below)!
hold off; %# Any subsequent plotting will overwrite the image!
In reply to the comments:
x and y are created using the following code:
temp_hull = stats_single_object(k).ConvexHull;
for k2 = 1:length(temp_hull)
i = i+1;
[x(i,1)] = temp_hull(k2,1);
[y(i,1)] = temp_hull(k2,2);
end;
it might be that the ConvexHull is the other way around and therefore the plot is different. Or that I made a mistake and it should be
[x(i,1)] = temp_hull(k2,2);
[y(i,1)] = temp_hull(k2,1);
However the documentation is not clear about which colum = x OR y:
Quote: "Each row of the matrix contains the x- and y-coordinates of one vertex of the polygon. "
I read this as x is the first column and y is the second colum.
In newer versions of MATLAB (I have 2013b) the Computer Vision System Toolbox contains the vision.ShapeInserter System object which can be used to draw shapes on images. Here is an example of drawing yellow circles from the documentation:
yellow = uint8([255 255 0]); %// [R G B]; class of yellow must match class of I
shapeInserter = vision.ShapeInserter('Shape','Circles','BorderColor','Custom','CustomBorderColor',yellow);
I = imread('cameraman.tif');
circles = int32([30 30 20; 80 80 25]); %// [x1 y1 radius1;x2 y2 radius2]
RGB = repmat(I,[1,1,3]); %// convert I to an RGB image
J = step(shapeInserter, RGB, circles);
imshow(J);
With MATLAB and Image Processing Toolbox R2012a or newer, you can use the viscircles function to easily overlay circles over an image. Here is an example:
% Plot 5 circles at random locations
X = rand(5,1);
Y = rand(5,1);
% Keep the radius 0.1 for all of them
R = 0.1*ones(5,1);
% Make them blue
viscircles([X,Y],R,'EdgeColor','b');
Also, check out the imfindcircles function which implements the Hough circular transform. The online documentation for both functions (links above) have examples that show how to find circles in an image and how to display the detected circles over the image.
For example:
% Read the image into the workspace and display it.
A = imread('coins.png');
imshow(A)
% Find all the circles with radius r such that 15 ≤ r ≤ 30.
[centers, radii, metric] = imfindcircles(A,[15 30]);
% Retain the five strongest circles according to the metric values.
centersStrong5 = centers(1:5,:);
radiiStrong5 = radii(1:5);
metricStrong5 = metric(1:5);
% Draw the five strongest circle perimeters.
viscircles(centersStrong5, radiiStrong5,'EdgeColor','b');
Here's the method I think you need:
[x_p, y_p] = find (points);
% convert the subscripts to indicies, but transposed into a row vector
a = sub2ind(size(im), x_p, y_p)';
% assign all the values in the image that correspond to the points to a value of zero
im([a]) = 0;
% show the new image
imshow(im)