Consider the following image:
On the left, let's say I have 3 labels. The background = 1, the black = 2, the orange = 3. What I want to do is be able to first identify all connected components that are orange, then if there are black objects touching the orange, then I'd like to convert this to orange as well.
I know the following steps:
Assume the labeled image is called labeled. I do orange = labeled == 3
CC = bwconncomp(orange);
But from here, I'm not sure how I can check if any of the black components are touching the orange. Once I know which black components are touching the orange, I can do the following: RP = regionprops(black, 'PixelIdxList'); labeled(RP(index).PixelIdxList) = 3;
To test code: here is an example input and output generation:
%input matrix
I = zeros([8 8]);
I(1:5, 1:5) = 1;
I(2:4, 3:7) = 2;
I(7:8, 1:2) = 1;
I(7:8, 7:8) = 2;
I
%output
O = zeros([8 8]);
O(1:5, 1:5) = 1;
O(2:4, 3:7) = 1;
O(7:8, 1:2) = 1;
O(7:8, 7:8) = 2;
O
You can do it as follows. Let target denote the target value, corresponding to orange.
Identify regions defined as connected components without regard to color, using bwlabel.
For each region, if it contains at least a pixel equal to target, set the whole region to target.
target = 1; % value corresponding to orange
O = I;
[regions, numRegions] = bwlabel(I);
for regionId = 1:numRegions
ind = regions==regionId;
if any(I(ind)==target)
O(ind) = target;
end
end
This can be quite simply solved using imreconstruct. This function, for binary (logical) inputs, does a flood fill. The idea is to find an mask image (the orange and the black regions together) and a marker (or seed) image (the orange regions).
As a demo, let's start with what OP already has:
orig = readim('https://i.stack.imgur.com/rH7yT.png');
labeled = uint8(round((255-orig{2}(0:866,:))/127));
orange = labeled==1; % for OP this is 3
black = labeled==2; % for OP this is 2 also
(Note that labeled is not identical to the one that OP has, the numbers for each label are different.)
Now we can apply imreconstruct:
black_or_orange = orange | black;
output = imreconstruct(orange,black_or_orange);
output now contains the orange regions, grown to encompass any touching black regions. We can create a new labeled image as follows:
new_labeled = uint8(output) + 2*uint8(black & ~output);
Related
after leaf segmentation i got the following binary image:
Is there a way to fill the gaps caused by the similiarity of the veins with the background? I've tried to use imclose, or imdilate etc but it affects teeth shape. I can't find out how to fill these gaps without affecting teeth shape.
You may try bwfill(I, 'hols'), with out without imclose:
I = imbinarize(rgb2gray(imread('leaf.jpg')));
I = I(3:end-4, 1:end-8); %Remove white frame
J = imclose(I, ones(2)); %Minor affect the teeth shape (result looks better with imclose).
K = bwfill(J, 'hols'); %Fill the black hols
Result:
In case you want to fill the "vein gaps", you can try the following approach:
I = imbinarize(rgb2gray(imread('leaf.jpg')));
I = I(3:end-4, 1:end-8); %Remove white frame
I = bwfill(I, 'hols'); %Fill small black hols.
J = imerode(imdilate(I, strel('disk',5)), strel('disk',10)); %Dilate with radius 5 and erode with 10
T = (I == 0) & (J == 1); %Create mask with 1 where I is black and J is white "vein mask".
K = I;
K(T) = 1; %Fill "vein mask" in I with white.
K = bwfill(K, 'hols'); %Fill small black hols (fill tiny holds left).
Result:
I have the following code:
close all;
star = imread('/Users/name/Desktop/folder/pics/OnTheBeach.png');
blrtype = fspecial('average',[3 3]);
blurred = imfilter(star, blrtype);
[rows,cols,planes] = size(star);
R = star(:,:,1); G = star(:,:,2); B = star(:,:,3);
starS = zeros(rows,cols);
ind = find(R > 190 & R < 240 & G > 100 & G < 170 & B > 20 & B < 160);
starS(ind) = 1;
K = imfill(starS,'holes');
stats = regionprops(logical(K), 'Area', 'Solidity');
ind = ([stats.Area] > 250 & [stats.Solidity] > 0.1);
L = bwlabel(K);
result = ismember(L,find(ind));
Up to this point I load an image, blur to filter out some noise, do colour segmentation to find the specific objects which fall in that range, then create a binary image that has value 1 for the object's colour, and 0 for all other stuff. Finally I do region filtering to remove any clutter that was left in the image so I'm only left with the objects I'm looking for.
Now I want to recolour the original image based on the segmentation mask to change the colour of the starfish. I want to create Red,Green,Blue channels, assign value to them then lay the mask over the image. (To have red starfishes for example)
red = star;
red(starS) = starS(:,:,255);
green = star;
green(starS) = starS(:,:,0);
blue = star;
blue(starS) = star(:,:,0);
out = cat(3, red, green, blue);
imshow(out);
This gives me an error: Index exceeds matrix dimensions.
Error in Project4 (line 28)
red(starS) = starS(:,:,255);
What is wrong with my current approach?
Your code is kinda confusing... I don't understand whether the mask you want to use is starS or result since both look like 2d indexers. In your second code snippet you used starS, but the mask you posted in your question is result.
Anyway, no matter what your desired mask is, all you have to do is to use the imoverlay function. Here is a small example based on your code:
out = imoverlay(star,result,[1 0 0]);
imshow(out);
and here is the output:
If the opaque mask of imoverlay suggested by Tommaso is not what you're after, you can modify the RGB values of the input to cast a hue over the selected pixels without saturating them. It is only slightly more involved.
I = find(result);
gives you an index of the pixels in the 2D image. However, star is 3D. Those indices will point at the same pixels, but only at the first 2D slice. That is, if I points at pixel (x,y), it is equivalently pointing to pixel (x,y,1). That is the red component of the pixel. To index (x,y,2) and (x,y,2), the green and blue components, you need to increment I by numel(result) and 2*numel(result). That is, star(I) accesses the red component of the selected pixels, star(I+numel(result)) accesses the green component, and star(I+2*numel(result)) accesses the blue component.
Now that we can access these values, how do we modify their color?
This is what imoverlay does:
I = find(result);
out = star;
out(I) = 255; % red channel
I = I + numel(result);
out(I) = 0; % green channel
I = I + numel(result);
out(I) = 0; % blue channel
Instead, you can increase the brightness of the red proportionally, and decrease the green and blue. This will change the hue, increase saturation, and preserve the changes in intensity within the stars. I suggest the gamma function, because it will not cause strong saturation artefacts:
I = find(result);
out = double(star)/255;
out(I) = out(I).^0.5; % red channel
I = I + numel(result);
out(I) = out(I).^1.5; % green channel
I = I + numel(result);
out(I) = out(I).^1.5; % blue channel
imshow(out)
By increasing the 1.5 and decreasing the 0.5 you can make the effect stronger.
This is the problem I have: I have an image as shown below. I want to detect the circular region which I have marked with a red line for display here (that particular bright ring).
Initially, this is what I do for now: (MATLAB)
binaryImage = imdilate(binaryImage,strel('disk',5));
binaryImage = imfill(binaryImage, 'holes'); % Fill holes.
binaryImage = bwareaopen(binaryImage, 20000); % Remove small blobs.
binaryImage = imerode(binaryImage,strel('disk',300));
out = binaryImage;
img_display = immultiply(binaryImage,rgb2gray(J1));
figure, imshow(img_display);
The output seems to be cut on one of the parts of the object (for a different image as input, not the one displayed above). I want an output in such a way that it is symmetric (its not always a perfect circle, when it is rotated).
I want to strictly avoid im2bw since as soon as I binarize, I lose a lot of information about the shape.
This is what I was thinking of:
I can detect the outer most circular (almost circular) contour of the image (shown in yellow). From this, I can find out the centroid and maybe find a circle which has a radius of 50% (to locate the region shown in red). But this won't be exactly symmetric since the object is slightly tilted. How can I tackle this issue?
I have attached another image where object is slightly tilted here
I'd try messing around with the 'log' filter. The region you want is essentially low values of the 2nd order derivative (i.e. where the slope is decreasing), and you can detect these regions by using a log filter and finding negative values. Here's a very basic outline of what you can do, and then tweak it to your needs.
img = im2double(rgb2gray(imread('wheel.png')));
img = imresize(img, 0.25, 'bicubic');
filt_img = imfilter(img, fspecial('log',31,5));
bin_img = filt_img < 0;
subplot(2,2,1);
imshow(filt_img,[]);
% Get regionprops
rp = regionprops(bin_img,'EulerNumber','Eccentricity','Area','PixelIdxList','PixelList');
rp = rp([rp.EulerNumber] == 0 & [rp.Eccentricity] < 0.5 & [rp.Area] > 2000);
bin_img(:) = false;
bin_img(vertcat(rp.PixelIdxList)) = true;
subplot(2,2,2);
imshow(bin_img,[]);
bin_img(:) = false;
bin_img(rp(1).PixelIdxList) = true;
bin_img = imfill(bin_img,'holes');
img_new = img;
img_new(~bin_img) = 0;
subplot(2,2,3);
imshow(img_new,[]);
bin_img(:) = false;
bin_img(rp(2).PixelIdxList) = true;
bin_img = imfill(bin_img,'holes');
img_new = img;
img_new(~bin_img) = 0;
subplot(2,2,4);
imshow(img_new,[]);
Output:
I need a script that "checks" if 4 given points form a square or a rhombus.
I am working in a QR code segmentation script in which I try to locate the vertex by looking for the non-negative values of a binary image traversing it by rows and columns.
There are some cases in which checking is not neccessary, like in this image:
It is a bit hard to see, but the vertex are labelled as 4 points in green, magenta, cyan and yellow. In this case the script should return the same input points, since no modification is needed.
On the other hand, there are cases in which the vertex are labeled as so:
It can be seen that the magenta and cyan labels rely on the top right corner of the image. This is obviously not correct, but it fullfills the specified condition: traverse each row of the image until you find a row satisfying sum(row)>1 (greater than 1 to avoid single, noisy pixels).
How can I locate the misplaced vertex and place it using the remaining vertex coordinates?
EDIT
Solved the problem. I'm posting the code of the function in case someone needs it:
function correctedCorners = square(corners)
correctedCorners = corners;
X = corners(:,1);
Y = corners(:,2);
sortedX = sort(corners(:,1));
sortedY = sort(corners(:,2));
%% DISTANCES BW POINTS
for i=1:4
for j=1:4
distances(i,j) = sqrt((corners(i,1)-corners(j,1))^2+ (corners(i,2)-corners(j,2))^2);
end
end
%% relationship bw distances
% check corner 1
d11 = distances(1,1);%0
d12 = distances(1,2);%x
d13 = distances(1,3);%sqrt(2)*x
d14 = distances(1,4);%x
bool1 = [(d12*0.8<=d14)&(d12*1.2>=d14) (d12*0.8*sqrt(2)<=d13)& (d12*1.2*sqrt(2)>=d13) (d14*0.8<=d12)&(d14*1.2>=d12) (d14*0.8*sqrt(2)<=d13)&(d14*1.2*sqrt(2)>=d13)];
% check corner 2
d21 = distances(2,1);%x
d22 = distances(2,2);%0
d23 = distances(2,3);%x
d24 = distances(2,4);%sqrt(2)*x
bool2 = [(d21*0.8<=d23)&(d21*1.2>=d23) (d21*0.8*sqrt(2)<=d24)&(d21*1.2*sqrt(2)>=d24) (d23*0.8<=d21)&(d23*1.2>=d21) (d23*0.8*sqrt(2)<=d24)&(d23*1.2*sqrt(2)>=d24)];
% check corner 3
d31 = distances(3,1);%sqrt(2)*x
d32 = distances(3,2);%x
d33 = distances(3,3);%0
d34 = distances(3,4);%x
bool3 = [(d32*0.8<=d34)&(d32*1.2>=d34) (d32*0.8*sqrt(2)<=d31)&(d32*1.2*sqrt(2)>=d31) (d34*0.8<=d32)&(d34*1.2>=d32) (d34*0.8*sqrt(2)<=d31)&(d34*1.2*sqrt(2)>=d31)];
% check corner 4
d41 = distances(4,1);%x
d42 = distances(4,2);%sqrt(2)*x
d43 = distances(4,3);%x
d44 = distances(4,4);%0
bool4 = [(d41*0.8<=d43)&(d41*1.2>=d43) (d41*0.8*sqrt(2)<=d42)&(d41*1.2*sqrt(2)>=d42) (d43*0.8<=d41)&(d43*1.2>=d41) (d43*0.8*sqrt(2)<=d42)&(d43*1.2*sqrt(2)>=d42)];
bool = [bool1; bool2;bool3;bool4];
idx = 0;
for i=1:4
if (sum(bool(i,:))==0)
idx = [idx i];
end
end
if (length(idx)>=2)
for i=2:length(idx)
switch idx(i)
case 1
correctedCorners(1,:) = abs(corners(4,:)-(corners(3,:)-corners(2,:)));
case 2
correctedCorners(2,:) = abs(corners(3,:)-(corners(4,:)-corners(1,:)));
case 3
correctedCorners(3,:) = abs(corners(2,:)+(corners(1,:)-corners(1,:)));
case 4
correctedCorners(4,:) = abs(corners(1,:)+(corners(3,:)-corners(2,:)));
end
end
end
From basic geometry about squares:
TopLeft distance to BotLeft = x
TopLeft distance to TopRight= x
TopLeft distance to BotRight= sqrt(2)*x
Use the same logic for BotLeft to other points, etc.
Allow yourself something like 10-20% margin of error to declare an incorrect point. That is, if TopLeft distance to 2 points outside of the range (80%;120%)*x , and its distance to the third point is outside of the range (80%;120%)*sqrt(2)*x, you can declare the point as placed incorrectly.
In your case, the TopLeft point fails on all distance tests:
0 instead of x to TopRight (about 100% error)
sqrt(2)*x vs x to BotLeft (about 44% error)
x vs sqrt(2)*x to BotRight) (about 31% error)
As long as the rhombus is very similar to a square, a 20% margin of error while treating it as a square should still work.
Here is the code I used:
x = linspace(0,2);
e = exp(1);
lin = e;
quad = e-e.*x.*x/2;
cub = e-e.*x.*x/2;
quart = e-e.*x.*x/2+e.*x.*x.*x.*x/24;
act = e.^cos(x);
mplot = plot(x,act,x,lin,x,quad,x,cub,x,quart);
legend('actual','linear','quadratic','cubic','quartic')
This produces a legend matching the right colors to actual and linear, then after that it seems to skip over red on the graph, but not on the legend, i.e. the legend says quadratic should be red, but the graph shows it as green, the legend says cubic should be green, but the graph shows it as purple etc.
Any help is appreciated.
The lin curve needs to be fixed --- now you just have a bunch of points instead of a line. quad and cub need to be fixed as well (see below).
x = linspace(0,2);
e = exp(1);
lin = ones(size(x))*e; %#Now it's a vector with the same size as x
quad = e-e.*x.*x/2;
cub = e-e.*x.*x/2;
quart = e-e.*x.*x/2+e.*x.*x.*x.*x/24;
act = e.^cos(x);
mplot = plot(x,act,x,lin,x,quad,x,cub,x,quart);
legend('actual','linear','quadratic','cubic','quartic')
Are quad and cub meant to be the same? Maybe it should be:
quad = e-e.*x.*x/2;
cub = e-e.*x.*x.*x/2;