How to plot boundary and centroids on video frames called inside step() function - matlab

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

Related

Getting the coordinates of vertices of an A4 sheet with coins on it, for its further projective transformation and coin detection

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).

Can we rotate an image in MATLAB filled with background color of original image?

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..

How to remove outline circle in this image?

Can you guys suggest possible ways on how to remove the circle outline in this image? Imfindcircles doesnt work for me. Can you suggest other methods? http://i.stack.imgur.com/RuD7v.jpg
Assuming BW to be the binary image that has the outline circled and which is to be removed, you can use an approach based on regionprops -
perimtrs = regionprops(BW, 'Perimeter'); %// perimeters for each connected component
px = regionprops(BW, 'PixelIdxList'); %// pixel list for each connected component
[~,idx] = max(struct2array(perimtrs)); %// get the component with max perimeter
%// that represents the outline circle
BW(px(idx).PixelIdxList) = 0; %// Set all pixels of the outline circle to zero,
%// that is they are removed
If you would like to be on the safest side with the functionality, you can use BoundingBox properties from regionprops instead of 'Perimeter' as shown here -
%// Get the bounding box properties for each connected component
perimtrs = regionprops(BW, 'BoundingBox');
%// Get bounding box area for each component and get the ID for the largest
%// box that corresponds to the outline circle
bound_box = reshape(struct2array(perimtrs),4,[]);
bound_box_area = bound_box(3,:).*bound_box(4,:);
[~,idx] = max(bound_box_area);
%// Set the pixels corresponding to the outline circle to zeros
px = regionprops(BW, 'PixelIdxList');
BW(px(idx).PixelIdxList) = 0;
Alternatively, you can avoid the second use of regionprops to get the pixel list with a call to regionprops and that might be efficient with performance, but I haven't not tested, so can't guarantee that. The new approach would look something like this -
perimtrs = regionprops(BW, 'Perimeter');
[~,idx] = max(struct2array(perimtrs))
[L,num] = bwlabel( BW ); %// Label connected components
BW(L==idx)=0; %// Select all pixels corresponding to label idx and set those to zero
Similarly, you can mix this bwlabel approach with BoundingBox of regionprops.
OK so here goes one hypotheses that does not assume the interface to be a circle, neither to be a single region, or having the largest perimeter.
%Assume A as your original image (left image), and bin_A as your binary image (right image)
thres=graythresh(A)
mask_A=im2bw(A,thres);
mask_A=imerode(mask_A,ones(3));
bin_A=bin_A.*mask_A;

MATLAB: Display image inside circle

As part of a circle recognition program, I have a background image with a geometrical circle with known coordinates and radius. I want the inside part of the circle filled by an image and the outside left alone. My best idea is some sort of circular mask, but I am not sure this is the best approach. Any suggestions?
X = imread('X.jpg'); % Background image jpg
Y = imread('Y.jpg'); % Filling image jpg
cent = [100,100]; % Center of circle
rad = 20; % Circle radius
% Fill circle ?
...
I have not provided the extended code, due to confidentiality.
I think the hard part was done by whomever authored this: http://matlab.wikia.com/wiki/FAQ#How_do_I_create_a_circle.3F
Assumptions:
I'm assuming you're not going to specify points that are out of range of the image (i.e. I'm not adding validation here).
I use the background image to relate the "center" of your circle to the coordinates.
I'm assuming radius is in pixels.
I didn't create a background image with a circle of known radius because I don't think that's necessary to create the filling effect you're looking for (unless I'm missing something).
Code:
X = imread('rdel_x.png'); % Background image jpg (I used a random image but this can be your blank + geometric circle)
Y = imread('rdel_y.png'); % Filling image jpg
cent = [100,150]; % Center of circle
rad = 70; % Circle radius
% make a mesh grid to provide coords for the circle (mask)
% taken from http://matlab.wikia.com/wiki/FAQ#How_do_I_create_a_circle.3F
[columnsInImage rowsInImage] = meshgrid(1:size(X,2), 1:size(X,1));
% circle points in pixels:
circlePixels = (rowsInImage - cent(1)).^2 ...
+ (columnsInImage - cent(2)).^2 <= rad.^2;
circlePixels3d=repmat(circlePixels,[1 1 3]); % turn into 3 channel (assuming X and Y are RGB)
X((circlePixels3d)) = Y((circlePixels3d)); % assign the filling image pixels to the background image for pixels where it's the desired circle
imagesc(X);
axis image off
Result: from left to right, background image, filling image, result from above code.
Edit: you might not even need the background image if everything is encapsulated in your coordinates. e.g. try appending this to the above code...
Z=zeros(size(X),'uint8'); % same size as your background
Z(circlePixels3d) = Y(circlePixels3d);
figure; % new fig
imagesc(Z);
axis image off

Set pixels on screen efficiently

I am using WindowAPI (http://www.mathworks.com/matlabcentral/fileexchange/31437) to show a black full screen in matlab.
When drawing on screen, turns out drawing using line() and rectangle() functions is extremely slow.
How can I set values of pixels without going through matlab's mechanism? Getting the window's canvas for example would be great.
One way to imitate a "canvas" is by using a MATLAB image. The idea is to manually change its pixels and update the 'CData' of the plotted image.
Note that you can use an image with the same dimensions as your screen size (image pixels will correspond to screen pixels one-to-one), but updating it would be slower. The trick is to keep it small and let MATLAB map it to the entire fullscreen. That way the image can be thought of as having "fat" pixels. Of course the resolution of the image is going to affect the size of the marker you draw.
To illustrate, consider the following implementation:
function draw_buffer()
%# paramters (image width/height and the indexed colormap)
IMG_W = 50; %# preferably same aspect ratio as your screen resolution
IMG_H = 32;
CMAP = [0 0 0 ; lines(7)]; %# first color is black background
%# create buffer (image of super-pixels)
%# bigger matrix gives better resolution, but slower to update
%# indexed image is faster to update than truecolor
img = ones(IMG_H,IMG_W);
%# create fullscreen figure
hFig = figure('Menu','none', 'Pointer','crosshair', 'DoubleBuffer','on');
WindowAPI(hFig, 'Position','full');
%# setup axis, and set the colormap
hAx = axes('Color','k', 'XLim',[0 IMG_W]+0.5, 'YLim',[0 IMG_H]+0.5, ...
'Units','normalized', 'Position',[0 0 1 1]);
colormap(hAx, CMAP)
%# display image (pixels are centered around xdata/ydata)
hImg = image('XData',1:IMG_W, 'YData',1:IMG_H, ...
'CData',img, 'CDataMapping','direct');
%# hook-up mouse button-down event
set(hFig, 'WindowButtonDownFcn',#mouseDown)
function mouseDown(o,e)
%# convert point from axes coordinates to image pixel coordinates
p = get(hAx,'CurrentPoint');
x = round(p(1,1)); y = round(p(1,2));
%# random index in colormap
clr = randi([2 size(CMAP,1)]); %# skip first color (black)
%# mark point inside buffer with specified color
img(y,x) = clr;
%# update image
set(hImg, 'CData',img)
drawnow
end
end