i want to convert RGB values to HSV values . But if I devide 9 by 28, octave calculate 0. Can anyone explain me the reason??
function [hsv] = RGBtoHSV()
im = imread('picture.png');
R = im(:,:,1);
G = im(:,:,2);
B = im(:,:,3);
len = length(R); % R, G, B should have the same length
for i = 1:len
MAX = max([R(i),G(i),B(i)]);
MIN = min([R(i),G(i),B(i)]);
S = 0;
if MAX == MIN
H = 0;
elseif MAX == R(i)
disp(G(i) - B(i)); % 9
disp(MAX - MIN); % 28
H = 0.6 * ( 0 + ( (G(i) - B(i)) / MAX - MIN) ); % 0
disp(H) % why i get 0 if try to calculate ( 0 + ( (G(i) - B(i)) / MAX - MIN)?
....
end
return;
end
endfunction
RGBtoHSV()
Chris :D
You must cast the image into Double by doing:
im = double(imread('picture.png'));
This will solve your issues which happens since the image is type UINT8.
You can also use Octave's builtin rgb2hsv function instead of writing your own.
im_rgb = imread ("picture.png");
im_hsv = rgb2hsv (im_rgb);
If this is an exercise, then I'd suggest you look at its source, enter type rgb2hsv at the Octave prompt, and see how its implemented.
Related
While using Matlab for image processing (exactly improving img by Fuzzy Logic) I found a really strange thing. My fuzzy function is correct, I tested it on random values and they are basically simple linear functions.
function f = Udark(z)
if z < 50
f = 1;
elseif z > 125
f = 0;
elseif (z >= 50) && (z <= 125)
f = -z/75 + 125/75;
end
end
where z is a value of a pixel (in grayscale). Now there is a really strange thing going on.
f = -z/75 + 125/75;, where a is an image. However, it is giving really different results if used as an input. I.e. if I use a variable p = 99, the output of the function is 0.3467 as it should be, when if I use A(i,j) it is giving me result f=2. Since it is clearly impossible, I do not know where is the problem. I thought that maybe there is a case with the type of the variable but if I change it to uint8 it stays the same... If you know what's going on, please, let me know :)
1.Changed line:
f = (125/75) - (z/75);
After editing the third condition the resultant/transformed image has no pixel values of 2. Not sure if you intend to work with decimals. If decimals are necessary using the im2double() function to convert the image and scaling it up by a factor of 255 might suffice your needs. See heading 3 for rounding details.
2.Reading in Image and Testing:
%Reading in the image and applying the function%
Image = imread("RGB_Image.png");
Greyscale_Image = rgb2gray(Image);
[Image_Height,Image_Width] = size(Greyscale_Image);
Transformed_Image = zeros(Image_Height,Image_Width);
for Row = 1: +1: Image_Height
for Column = 1: +1: Image_Width
Pixel_Value = Greyscale_Image(Row,Column);
[Transformed_Pixel_Value] = Udark(Pixel_Value);
Transformed_Image(Row,Column) = Transformed_Pixel_Value;
end
end
subplot(1,2,1); imshow(Greyscale_Image);
subplot(1,2,2); imshow(Transformed_Image);
%Checking that no transformed pixels falls in this impossible range%
Check = (Transformed_Image > (125/75)) & (Transformed_Image ~= 1);
Check_Flag = any(Check,'all');
%Function to transform pixel values%
function f = Udark(z)
if z < 50
f = 1;
elseif z > 125
f = 0;
elseif (z >= 50) && (z <= 125)
f = (125/75) - (z/75);
end
end
3.Evaluating the Specifics of the Third Condition
Working with integers (uint8) will force the values to be rounded to the nearest integer. Any number that falls between the range (50,125] will evaluate to 1 or 0.
f = -z/75 + 125/75;
If z = 50.1,
-50.1/75 + 125/75 = 74.9/75 ≈ 0.9987 → rounds to 1
Using MATLAB version: R2019b
As introduced in this work (code), the following code sould refine an image (i.e. a gray scale saliency map).
Refinement Function:
function sal = Refinement(y, dim)
th_2 = graythresh(y);
if dim == 1
sal = y;
sal(y<th_2) = 10*(y(y < th_2))/th_2 - 10;
sal(y>=th_2) = 10*(y(y >= th_2) - th_2)/(1-th_2);
sal = 1 ./ (1 + exp(-sal)) + y;
sal = normalization(sal, 0);
elseif dim == 2
[r, c] = size(y);
y_col = reshape(y,[1 r*c]);
sal_col = y_col;
sal_col(y_col<th_2) = 10*(y_col(y_col < th_2))/th_2 - 10;
sal_col(y_col>=th_2) = 10*(y_col(y_col >= th_2) - th_2)/(1-th_2);
sal_col = 1 ./ (1 + exp(-sal_col)) + y_col;
sal = reshape(sal_col, [r c]);
end
end
normalization function:
function matrix = normalization(mat, flag)
% INPUT :
% flag: 1 denotes that the mat is a 3-d matrix;
% 0 denotes that the mat is a matrix;
%
if flag ~= 0
dim = size(mat,3);
matrix = mat;
for i = 1:dim
matrix(:,:,i) = ( mat(:,:,i) - min(min(mat(:,:,i)))) / ( max(max(mat(:,:,i))) - min(min( mat(:,:,i))) + eps);
end
else
matrix = ( mat - min(min(mat)))/( max(max(mat)) - min(min(mat)) + eps);
end
However, after applying the function values of the image matrix will be changed, the result remains the same as the image before the refinement.
Is there a conceptual error with that or the implementation failed?
P.S. The input image (saliency map) for refinery is something like below. in the refined saliency map, the foreground (satrfish in this image) should stand out(becomes homogeneously white as possible) and the background noise should be removed (becomes homogenously black as possible):
I'm doing this exercise by Andrew NG about using k-means to reduce the number of colors in an image. It worked correctly but I'm afraid it's a little slow because of all the for loops in the code, so I'd like to vectorize them. But there are those loops that I just can't seem to vectorize effectively. Please help me, thank you very much!
Also if possible please give some feedback on my coding style :)
Here is the link of the exercise, and here is the dataset.
The correct result is given in the link of the exercise.
And here is my code:
function [] = KMeans()
Image = double(imread('bird_small.tiff'));
[rows,cols, RGB] = size(Image);
Points = reshape(Image,rows * cols, RGB);
K = 16;
Centroids = zeros(K,RGB);
s = RandStream('mt19937ar','Seed',0);
% Initialization :
% Pick out K random colours and make sure they are all different
% from each other! This prevents the situation where two of the means
% are assigned to the exact same colour, therefore we don't have to
% worry about division by zero in the E-step
% However, if K = 16 for example, and there are only 15 colours in the
% image, then this while loop will never exit!!! This needs to be
% addressed in the future :(
% TODO : Vectorize this part!
done = false;
while done == false
RowIndex = randperm(s,rows);
ColIndex = randperm(s,cols);
RowIndex = RowIndex(1:K);
ColIndex = ColIndex(1:K);
for i = 1 : K
for j = 1 : RGB
Centroids(i,j) = Image(RowIndex(i),ColIndex(i),j);
end
end
Centroids = sort(Centroids,2);
Centroids = unique(Centroids,'rows');
if size(Centroids,1) == K
done = true;
end
end;
% imshow(imread('bird_small.tiff'))
%
% for i = 1 : K
% hold on;
% plot(RowIndex(i),ColIndex(i),'r+','MarkerSize',50)
% end
eps = 0.01; % Epsilon
IterNum = 0;
while 1
% E-step: Estimate membership given parameters
% Membership: The centroid that each colour is assigned to
% Parameters: Location of centroids
Dist = pdist2(Points,Centroids,'euclidean');
[~, WhichCentroid] = min(Dist,[],2);
% M-step: Estimate parameters given membership
% Membership: The centroid that each colour is assigned to
% Parameters: Location of centroids
% TODO: Vectorize this part!
OldCentroids = Centroids;
for i = 1 : K
PointsInCentroid = Points((find(WhichCentroid == i))',:);
NumOfPoints = size(PointsInCentroid,1);
% Note that NumOfPoints is never equal to 0, as a result of
% the initialization. Or .... ???????
if NumOfPoints ~= 0
Centroids(i,:) = sum(PointsInCentroid , 1) / NumOfPoints ;
end
end
% Check for convergence: Here we use the L2 distance
IterNum = IterNum + 1;
Margins = sqrt(sum((Centroids - OldCentroids).^2, 2));
if sum(Margins > eps) == 0
break;
end
end
IterNum;
Centroids ;
% Load the larger image
[LargerImage,ColorMap] = imread('bird_large.tiff');
LargerImage = double(LargerImage);
[largeRows,largeCols,NewRGB] = size(LargerImage); % RGB is always 3
% TODO: Vectorize this part!
largeRows
largeCols
NewRGB
% Replace each of the pixel with the nearest centroid
NewPoints = reshape(LargerImage,largeRows * largeCols, NewRGB);
Dist = pdist2(NewPoints,Centroids,'euclidean');
[~,WhichCentroid] = min(Dist,[],2);
NewPoints = Centroids(WhichCentroid,:);
LargerImage = reshape(NewPoints,largeRows,largeCols,NewRGB);
% for i = 1 : largeRows
% for j = 1 : largeCols
% Dist = pdist2(Centroids,reshape(LargerImage(i,j,:),1,RGB),'euclidean');
% [~,WhichCentroid] = min(Dist);
% LargerImage(i,j,:) = Centroids(WhichCentroid,:);
% end
% end
% Display new image
imshow(uint8(round(LargerImage)),ColorMap)
UPDATE: Replaced
for i = 1 : K
for j = 1 : RGB
Centroids(i,j) = Image(RowIndex(i),ColIndex(i),j);
end
end
with
for i = 1 : K
Centroids(i,:) = Image(RowIndex(i),ColIndex(i),:);
end
I think this may be vectorized further by using linear indexing, but for now I should just focus on the while loop since it takes most of the time.
Also when I tried #Dev-iL's suggestion and replaced
for i = 1 : K
PointsInCentroid = Points((find(WhichCentroid == i))',:);
NumOfPoints = size(PointsInCentroid,1);
% Note that NumOfPoints is never equal to 0, as a result of
% the initialization. Or .... ???????
if NumOfPoints ~= 0
Centroids(i,:) = sum(PointsInCentroid , 1) / NumOfPoints ;
end
end
with
E = sparse(1:size(WhichCentroid), WhichCentroid' , 1, Num, K, Num);
Centroids = (E * spdiags(1./sum(E,1)',0,K,K))' * Points ;
the results were always worse: With K = 16, the first takes 2,414s , the second takes 2,455s ; K = 32, the first takes 4,529s , the second takes 5,022s. Seems like vectorization does not help, but maybe there's something wrong with my code :( .
Replaced
for i = 1 : K
for j = 1 : RGB
Centroids(i,j) = Image(RowIndex(i),ColIndex(i),j);
end
end
with
for i = 1 : K
Centroids(i,:) = Image(RowIndex(i),ColIndex(i),:);
end
I think this may be vectorized further by using linear indexing, but for now I should just focus on the while loop since it takes most of the time.
Also when I tried #Dev-iL's suggestion and replaced
for i = 1 : K
PointsInCentroid = Points((find(WhichCentroid == i))',:);
NumOfPoints = size(PointsInCentroid,1);
% Note that NumOfPoints is never equal to 0, as a result of
% the initialization. Or .... ???????
if NumOfPoints ~= 0
Centroids(i,:) = sum(PointsInCentroid , 1) / NumOfPoints ;
end
end
with
E = sparse(1:size(WhichCentroid), WhichCentroid' , 1, Num, K, Num);
Centroids = (E * spdiags(1./sum(E,1)',0,K,K))' * Points ;
the results were always worse: With K = 16, the first takes 2,414s , the second takes 2,455s ; K = 32, the first took 4,529s , the second took 5,022s. Seems like vectorization did not help in this case.
However, when I replaced
Dist = pdist2(Points,Centroids,'euclidean');
[~, WhichCentroid] = min(Dist,[],2);
(in the while loop) with
Dist = bsxfun(#minus,dot(Centroids',Centroids',1)' / 2 , Centroids * Points' );
[~, WhichCentroid] = min(Dist,[],1);
WhichCentroid = WhichCentroid';
the code ran much faster, especially when K is large (K=32)
Thank you everyone!
I am implementing a fast optimization algorithm using fixed point method in matlab. The goal of that method is that find optimal value of u. Denote u={u_i,i=1..2}. The optimal value of u can be obtained as following steps:
Sorry about my image because I cannot type mathematics equation in here.
To do that task, I tried to find u follows above steps. However, I don't know how to implement the term \sum_{j!=i} (u_j-1) in equation 25. This is my code. Please see it and could you give me some comment or suggestion about my implementation to correct them. Currently, I tried to run that code but it give an incorrect answer.
function u = compute_u_TV(Im0, N_class)
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% Initialization
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
theta=0.001;
gamma=0.01;
tau=0.1;
sigma=0.1;
N_class=2; % only have u1 and u2
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% Iterative segmentation process
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
for i=1:N_class
v(:,:,i) = Im0/max(Im0(:)); % u between 0 and 1.
qxv(:,:,i) = zeros(size(Im0));
qyv(:,:,i) = zeros(size(Im0));
u(:,:,i) = v(:,:,i);
for iteration=1:10000
u_temp=u;
% Update v
Divqi = ( BackwardX(qxv(:,:,i)) + BackwardY(qyv(:,:,i)) );
Term = Divqi - u(:,:,i)/ (theta*gamma);
TermX = ForwardX(Term);
TermY = ForwardY(Term);
Norm = sqrt(TermX.^2 + TermY.^2);
Denom = 1 + tau*Norm;
%Equation 24
qxv(:,:,i) = (qxv(:,:,i) + tau*TermX)./Denom;
qyv(:,:,i) = (qyv(:,:,i) + tau*TermY)./Denom;
v(:,:,i) = u(:,:,i) - theta*gamma* Divqi; %Equation 23
% Update u
u(:,:,i) = (v(:,:,i) - theta* gamma* Divqi -theta*gamma*sigma*(sum(u(:))-u(:,:,i)-1))./(1+theta* gamma*sigma);
u(:,:,i) = max(u(:,:,i),0);
u(:,:,i) = min(u(:,:,i),1);
check=u_temp(:,:,i)-u(:,:,i);
if(abs(sum(check(:)))<=0.1)
break;
end
end
end
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% Sub-functions- X.Berson
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
function [dx]=BackwardX(u);
[Ny,Nx] = size(u);
dx = u;
dx(2:Ny-1,2:Nx-1)=( u(2:Ny-1,2:Nx-1) - u(2:Ny-1,1:Nx-2) );
dx(:,Nx) = -u(:,Nx-1);
function [dy]=BackwardY(u);
[Ny,Nx] = size(u);
dy = u;
dy(2:Ny-1,2:Nx-1)=( u(2:Ny-1,2:Nx-1) - u(1:Ny-2,2:Nx-1) );
dy(Ny,:) = -u(Ny-1,:);
function [dx]=ForwardX(u);
[Ny,Nx] = size(u);
dx = zeros(Ny,Nx);
dx(1:Ny-1,1:Nx-1)=( u(1:Ny-1,2:Nx) - u(1:Ny-1,1:Nx-1) );
function [dy]=ForwardY(u);
[Ny,Nx] = size(u);
dy = zeros(Ny,Nx);
dy(1:Ny-1,1:Nx-1)=( u(2:Ny,1:Nx-1) - u(1:Ny-1,1:Nx-1) );
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% End of sub-function
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
You should do
u(:,:,i) = (v(:,:,i) - theta* gamma* Divqi -theta*gamma*sigma* ...
(sum(u(:,:,1:size(u,3) ~= i),3) -1))./(1+theta* gamma*sigma);
The part you were searching for is
sum(u(:,:,1:size(u,3) ~= i),3)
Let's decompose this :
1:size(u,3) ~= i
is a vector containing all values from 1 to the max size of u on the third dimension except i.
Then
u(:,:,1:size(u,3) ~= i)
is all the matrix of the third dimension of u except for j = i
Finally,
sum(...,3)
is the sum of all the matrix by the thrid dimension.
Let me know if it does help!
I am using for my project the "LucasKanade" code in matlab. It gives me as output 2 matrices (u and v). These are(i believe so) the velocities of the image in the x and y axes respectively. Now how can i convert these velocities to object velocities(eg in meters/second)?
Thanks in advance
"LucasKanade" code:
function [u, v] = LucasKanade(im1, im2, windowSize);
%LucasKanade lucas kanade algorithm, without pyramids (only 1 level);
%REVISION: NaN vals are replaced by zeros
[fx, fy, ft] = ComputeDerivatives(im1, im2);
u = zeros(size(im1));
v = zeros(size(im2));
halfWindow = floor(windowSize/2);
for i = halfWindow+1:size(fx,1)-halfWindow
for j = halfWindow+1:size(fx,2)-halfWindow
curFx = fx(i-halfWindow:i+halfWindow, j-halfWindow:j+halfWindow);
curFy = fy(i-halfWindow:i+halfWindow, j-halfWindow:j+halfWindow);
curFt = ft(i-halfWindow:i+halfWindow, j-halfWindow:j+halfWindow);
curFx = curFx';
curFy = curFy';
curFt = curFt';
curFx = curFx(:);
curFy = curFy(:);
curFt = -curFt(:);
A = [curFx curFy];
U = pinv(A'*A)*A'*curFt;
u(i,j)=U(1);
v(i,j)=U(2);
end;
end;
u(isnan(u))=0;
v(isnan(v))=0;
%u=u(2:size(u,1), 2:size(u,2));
%v=v(2:size(v,1), 2:size(v,2));
%%
function [fx, fy, ft] = ComputeDerivatives(im1, im2);
%ComputeDerivatives Compute horizontal, vertical and time derivative
% between two gray-level images.
if (size(im1,1) ~= size(im2,1)) | (size(im1,2) ~= size(im2,2))
error('input images are not the same size');
end;
if (size(im1,3)~=1) | (size(im2,3)~=1)
error('method only works for gray-level images');
end;
fx = conv2(im1,0.25* [-1 1; -1 1]) + conv2(im2, 0.25*[-1 1; -1 1]);
fy = conv2(im1, 0.25*[-1 -1; 1 1]) + conv2(im2, 0.25*[-1 -1; 1 1]);
ft = conv2(im1, 0.25*ones(2)) + conv2(im2, -0.25*ones(2));
% make same size as input
fx=fx(1:size(fx,1)-1, 1:size(fx,2)-1);
fy=fy(1:size(fy,1)-1, 1:size(fy,2)-1);
ft=ft(1:size(ft,1)-1, 1:size(ft,2)-1);