I need to transform my tilted image in a way I can find coins on an A4 paper. So far, I have been getting four coordinates of edges of my paper by manually selecting them with ginput.
targetImageData = imread('coin1.jpg');
imshow(targetImageData);
fprintf('Corner selection must be clockwise or anti-clockwise.\n');
[X,Y] = ginput(4);
Is there a way to automate this process, say, apply some edge detector and then find coordinates of each vertex and then pass them as the coordinates needed for transformation?
Manual selection:
Result:
You can try using detectHarrisFeatures on the S color channel of HSV color space:
I was looking for a color space that gets maximum contrast of the paper.
It looks like the saturation color channel of HSV makes a good contrast between the paper and the background.
Image is resized the image by a factor of 0.25, for removing noise.
detectHarrisFeatures finds the 4 corners of the paper, but it might not be robust enough.
You may need to find more features, and find the 4 correct features, using some logic.
Here is a code sample:
%Read input image
I = imread('im.png');
%Remove the margins, and replace them using padding (just because the image is a MATLAB figure)
I = padarray(I(11:end-10, 18:end-17, :), [10, 17], 'both', 'replicate');
HSV = rgb2hsv(I);
%H = HSV(:, :, 1);%figure;imshow(H);title('H');
S = HSV(:, :, 2);%figure;imshow(S);title('S');
%V = HSV(:, :, 3);%figure;imshow(V);title('V');
%Reduce image size by a factor of 0.25 in each axis
S = imresize(S, 0.25);
%S = imclose(S, ones(3)); %May be requiered
%Detect corners
corners = detectHarrisFeatures(S);
imshow(S); hold on;
plot(corners.selectStrongest(4));
Result:
Different approach you may try:
Take a photo without the coins.
Mark the corners manually, and extract features of the 4 corners.
Use image matching techniques to match the image with the coins with the image without the coins (mach basted on the 4 corners).
Related
I try to remove the white annotations of this image (the numbers and arrows), as well as the black grid, with MATLAB:
I tried to compute, for each pixel, the mode of neighbors, but this process is very slow and I get poor results.
How can I obtain an image like this one?
Thank you for your time.
The general name for such a task is inpainting. If you search for that you will find better methods than what I'm showing here. This is no more than a proof of concept. I'm using DIPimage 3 (because I'm an author and it's easy for me to use).
First we need to create a mask for the regions that we want to remove (inpaint). It is easy to find pixels where all three channels have a high value (white) or a low value (black):
img = readim('https://i.stack.imgur.com/16r9N.png');
% Find a mask for the areas to remove
whitemask = min(img,'tensor') > 50;
blackmask = max(img,'tensor') < 30;
mask = whitemask | blackmask;
This mask doesn't capture all of the black grid, if we increase the threshold we will also remove the dark region of sea off the coast of Spain. And it also captures the white outline of the coasts. We can do a little bit better than this with some additional filtering:
% Find a mask for the areas to remove
whitemask = min(img,'tensor') > 50;
whitemask = whitemask - pathopening(whitemask,50);
blackmask = max(img,'tensor');
blackmask2 = blackmask < 80;
blackmask2 = blackmask2 - areaopening(blackmask2,6);
blackmask = blackmask < 30 | blackmask2;
mask = whitemask | blackmask;
This produces the following mask:
Still far from perfect, but a good start for our proof of concept.
One simple inpainting method uses normalized convolution: using the inverse of the mask we made, convolve the image multiplied by the mask, and convolve the mask separately. The ratio of these two results is a smoothed image that doesn't take the masked pixels into account. Finally, we replace the pixels in the original image under the mask with the values from this normalized convolution:
% Solution 1: normalized convolution
smooth = gaussf(img * ~mask, 2) / gaussf(~mask, 2);
img(mask) = smooth(mask);
An alternative solution applies a closing on the image multiplied by the mask (note that this multiplication makes the pixels we don't want completely black; the closing will spread the surrounding colors over the black areas):
% Solution 2: morphology
smooth = iterate('closing',img * ~mask, 13);
img(mask) = smooth(mask);
I am calling some images inside a for loop and then doing some processing on those images. After that, I am using the step function to display those frames and their masks inside a video player. How can I add a boundary to an object inside the mask image? Also, how can I make the boundary thicker and plot the centroids of each blob in the mask in the mask image? Below is the rough sketch of the code.
videoPlayer = vision.VideoPlayer();
maskPlayer = vision.VideoPlayer();
for ii = 1:nfiles
filenameii = [............]
frame= imread(filenameii);
mask = dOB(frame,BackgroundImg);
% some processing on the images
mask= bwareaopen(mask,27);
boundaries = bwboundaries(mask,'noholes');
B=boundaries{1};
Centroid = regionprops(mask,'centroid');
Centroids = cat(1, Centroid.Centroid);
plot(B(:,2),B(:,1),'g','LineWidth',3);
plot(Centroids(:,1), Centroids(:,2), 'r+', 'MarkerSize', 10); step(videoPlayer,frame);
step(maskPlayer, mask);
P.S: I know how to display it on a figure using hold on but I would like this done directly on the image before displaying it in the video player. Any guidance would be appreciated.
Simply paint the pixels on the mask first before displaying it in the video player. What you have does work, but it will plot the boundary inside the figure for the mask player. Therefore, take your boundaries that you detected from bwboundaries, create linear indices from these coordinates and set the values in your image to white. What may be even simpler is to take your mask that you detected and use bwperim to automatically produce a mask that contains the boundaries of the blobs. I also see that you are filling in the holes of the mask, so you can use imfill directly on the output of your post-processing so that it gives you an image instead of coordinates. You would then use this mask to directly index into your image and set the coordinates of the boundaries of the blob to your desired colour. If you desire to make the perimeter thicker, a simple image dilation with imdilate using the appropriately sized square structuring element will help. Simply define the size of the neighbourhood of this structuring element to be the thickness of the perimeter that you desire. Finally, if you want to insert the centroids into the mask and since you have the MATLAB Computer Vision System Toolbox, use the insertMarker function so that you can use a set of points for each centroid and put them directly in the image. However, you must be sure to change the mask from a logical to a data type more suitable for images. uint8 should work. Therefore, cast the image to this type then multiply all nonzero values by 255 to ensure the white colours are maintained in the mask. With insertMarker, you want to insert red pluses with a size of 10 so we need to make sure we call insertMarker to reflect that. Also, because you want to have a colour image you will have to make your mask artificially colour and to do this painting individually for each plane for the colour that you want. Since you want green, this corresponds to the RGB value of (0,255,0).
Therefore, I have modified your code so that it does this. In addition, I've calculated the centroids of the filled mask instead of the original. We wouldn't want to falsely report the centroids of objects with gaps... unless that's what you're aiming for, but let's assume you're not:
videoPlayer = vision.VideoPlayer();
maskPlayer = vision.VideoPlayer();
% New - Specify colour you want
clr = [0 255 0]; % Default is green
% New - Define thickness of the boundaries in pixels.
thickness = 3;
% New - Create structuring element
se = strel('square', thickness);
for ii = 1:nfiles
filenameii = [............]
frame = imread(filenameii);
mask = dOB(frame, BackgroundImg);
% some processing on the images
mask = bwareaopen(mask,27);
%boundaries = bwboundaries(mask,'noholes');
%B=boundaries{1};
% New code - fill in the holes
mask = imfill(mask, 'holes');
Centroid = regionprops(mask,'centroid');
% New code - Create a boundary mask
mask_p = bwperim(mask, 8);
% New code - Make the boundaries thicker
mask_p = imdilate(mask_p, se);
% New code - create a colour image out of the mask
[red, green, blue] = deal(255*uint8(mask));
% Paint the perimeter of the blobs in the desired colour
red(mask_p) = clr(1); green(mask_p) = clr(2); blue(mask_p) = clr(3);
Centroids = cat(1, Centroid.Centroid);
%plot(B(:,2),B(:,1),'g','LineWidth',3);
%plot(Centroids(:,1), Centroids(:,2), 'r+', 'MarkerSize', 10);
% New - Make mask into RGB image for marker painting and to
% show to the user
mask_p = cat(3, red, green, blue);
% New - Insert the centroids directly in the mask image
mask_p = insertMarker(mask_p, Centroids, '+', 'color', 'r', 'size', 10);
step(videoPlayer, frame);
% New - Show new mask in the player
step(maskPlayer, mask_p);
end
I have a list of coordinates, which are generated from another program, and I have an image.
I'd like to load those coordinates (making circular regions of interest (ROIs) with a diameter of 3 pixels) onto my image, and extract the intensity of those pixels.
I can load/impose the coordinates on to the image by using;
imshow(file);
hold on
scatter(xCoords, yCoords, 'g')
But can not extract the intensity.
Can you guys point me in the right direction?
I am not sure what you mean by a circle with 3 pixels diameter since you are in a square grid (as mentioned by Ander Biguri). But you could use fspecial to create a disk filter and then normalize. Something like this:
r = 1.5; % for diameter = 3
h = fspecial('disk', r);
h = h/h(ceil(r),ceil(r));
You can use it as a mask to get the intensities at the given region of the image.
im = imread(file);
ROI = im(xCoord-1:xCoord+1; yCoord-1:yCoord+1);
I = ROI.*h;
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
By default, MATLAB function imrotate rotate image with black color filled in rotated portion. See this, http://in.mathworks.com/help/examples/images_product/RotationFitgeotransExample_02.png
We can have rotated image with white background also.
Question is, Can we rotate an image (with or without using imrotate) filled with background of original image?
Specific to my problem: Colored image with very small angle of rotation (<=5 deg.)
Here's a naive approach, where we simply apply the same rotation to a mask and take only the parts of the rotated image, that correspond to the transformed mask. Then we just superimpose these pixels on the original image.
I ignore possible blending on the boundary.
A = imread('cameraman.tif');
angle = 10;
T = #(I) imrotate(I,angle,'bilinear','crop');
%// Apply transformation
TA = T(A);
mask = T(ones(size(A)))==1;
A(mask) = TA(mask);
%%// Show image
imshow(A);
You can use padarray() function with 'replicate' and 'both' option to interpolate your image. Then you can use imrotate() function.
In the code below, I've used ceil(size(im)/2) as pad size; but you may want bigger pad size to eliminate the black part. Also I've used s and S( writing imR(S(1)-s(1):S(1)+s(1), S(2)-s(2):S(2)+s(2), :)) to crop the image where you can extract bigger part of image just expanding boundary of index I used below for imR.
Try this:
im = imread('cameraman.tif'); %// You can also read a color image
s = ceil(size(im)/2);
imP = padarray(im, s(1:2), 'replicate', 'both');
imR = imrotate(imP, 45);
S = ceil(size(imR)/2);
imF = imR(S(1)-s(1):S(1)+s(1)-1, S(2)-s(2):S(2)+s(2)-1, :); %// Final form
figure,
subplot(1, 2, 1)
imshow(im);
title('Original Image')
subplot(1, 2, 2)
imshow(imF);
title('Rotated Image')
This gives the output below:
Not so good but better than black thing..