Related
I = imread('Sub1.png');
figure, imshow(I);
I = imcomplement(I);
I = double(I)/255;
I = adapthisteq(I,'clipLimit',0.0003,'Distribution','exponential');
k = 12;
beta = 2;
maxIter = 100;
for i=1:length(beta)
[seg,prob,mu,sigma,it(i)] = ICM(I, k, beta(i), maxIter,5);
pr(i) = prob(end);
hold on;
end
figure, imshow(seg,[]);
and ICM function is defined as
function [segmented_image,prob,mu,sigma,iter] = ICM(image, k, beta, max_iterations, neigh)
[width, height, bands] = size(image);
image = imstack2vectors(image);
segmented_image = init(image,k,1);
clear c;
iter = 0;
seg_old = segmented_image;
while(iter < max_iterations)
[mu, sigma] = stats(image, segmented_image, k);
E1 = energy1(image,mu,sigma,k);
E2 = energy2(segmented_image, beta, width, height, k);
E = E1 + E2;
[p2,~] = min(E2,[],2);
[p1,~] = min(E1,[],2);
[p,segmented_image] = min(E,[],2);
prob(iter+1) = sum(p);
%find mismatch with previous step
[c,~] = find(seg_old~=segmented_image);
mismatch = (numel(c)/numel(segmented_image))*100;
if mismatch<0.1
iter
break;
end
iter = iter + 1;
seg_old = segmented_image;
end
segmented_image = reshape(segmented_image,[width height]);
end
Output of my algorithm is a logical matrix (seg) of size 305-by-305. When I use
imshow(seg,[]);
I am able to display the image. It shows different component with varying gray value. But bwlabel returns 1. I want to display the connected components. I think bwlabel thresholds the image to 1. unique(seg) returns values 1 to 10 since number of classes used in k-means is 10. I used
[label n] = bwlabel(seg);
RGB = label2rgb(label);
figure, imshow(RGB);
I need all the ellipse-like structures which are in between the two squares close to the middle of the image. I don't know the number of classes present in it.
Input image:
Ground truth:
My output:
If you want to explode the label image to different connected components you need to use a loop to extract labels for each class and sum label images to get the out label image.
u = unique(seg(:));
out = zeros(size(seg));
num_objs = 0;
for k = 1: numel(u)
mask = seg==u(k);
[L,N] = bwlabel(mask);
L(mask) = L(mask) + num_objs;
out = out + L;
num_objs = num_objs + N ;
end
mp = jet(num_objs);
figure,imshow(out,mp)
Something like this is produced:
I have tried to do everything out of scratch. I wish it is of some help.
I have a treatment chain that get at first contours with parameters tuned on a trial-and-error basis, I confess. The last "image" is given at the bottom ; with it, you can easily select the connected components and do for example a reconstruction by markers using "imreconstruct" operator.
clear all;close all;
I = imread('C:\Users\jean-marie.becker\Desktop\imagesJPG10\spinalchord.jpg');
figure,imshow(I);
J = I(:,:,1);% select the blue channel because jpg image
J=double(J<50);% I haven't inverted the image
figure, imshow(J);
se = strel('disk',5);
J=J-imopen(J,se);
figure, imshow(J);
J=imopen(J,ones(1,15));% privilegizes long horizontal strokes
figure, imshow(J);
K=imdilate(J,ones(20,1),'same');
% connects verticaly not-to-far horizontal "segments"
figure, imshow(K);
I want to find the corners of objects.
I tried the following code:
Vstats = regionprops(BW2,'Centroid','MajorAxisLength','MinorAxisLength',...
'Orientation');
u = [Vstats.Centroid];
VcX = u(1:2:end);
VcY = u(2:2:end);
[VcY id] = sort(VcY); % sorting regions by vertical position
VcX = VcX(id);
Vstats = Vstats(id); % permute according sort
Bv = Bv(id);
Vori = [Vstats.Orientation];
VRmaj = [Vstats.MajorAxisLength]/2;
VRmin = [Vstats.MinorAxisLength]/2;
% find corners of vertebrae
figure,imshow(BW2)
hold on
% C = corner(VER);
% plot(C(:,1), C(:,2), 'or');
C = cell(size(Bv));
Anterior = zeros(2*length(C),2);
Posterior = zeros(2*length(C),2);
for i = 1:length(C) % for each region
cx = VcX(i); % centroid coordinates
cy = VcY(i);
bx = Bv{i}(:,2); % edge points coordinates
by = Bv{i}(:,1);
ux = bx-cx; % move to the origin
uy = by-cy;
[t, r] = cart2pol(ux,uy); % translate in polar coodinates
t = t - deg2rad(Vori(i)); % unrotate
for k = 1:4 % find corners (look each quadrant)
fi = t( (t>=(k-3)*pi/2) & (t<=(k-2)*pi/2) );
ri = r( (t>=(k-3)*pi/2) & (t<=(k-2)*pi/2) );
[rp, ip] = max(ri); % find farthest point
tc(k) = fi(ip); % save coordinates
rc(k) = rp;
end
[xc,yc] = pol2cart(tc+1*deg2rad(Vori(i)) ,rc); % de-rotate, translate in cartesian
C{i}(:,1) = xc + cx; % return to previous place
C{i}(:,2) = yc + cy;
plot(C{i}([1,4],1),C{i}([1,4],2),'or',C{i}([2,3],1),C{i}([2,3],2),'og')
% save coordinates :
Anterior([2*i-1,2*i],:) = [C{i}([1,4],1), C{i}([1,4],2)];
Posterior([2*i-1,2*i],:) = [C{i}([2,3],1), C{i}([2,3],2)];
end
My input image is :
I got the following output image
The bottommost object in the image is not detected properly. How can I correct the code? It fails to work for a rotated image.
You can get all the points from the image, and use kmeans clustering and partition the points into 8 groups. Once partition is done, you have the points in and and you can pick what ever the points you want.
rgbImage = imread('your image') ;
%% crop out the unwanted white background from the image
grayImage = min(rgbImage, [], 3);
binaryImage = grayImage < 200;
binaryImage = bwareafilt(binaryImage, 1);
[rows, columns] = find(binaryImage);
row1 = min(rows);
row2 = max(rows);
col1 = min(columns);
col2 = max(columns);
% Crop
croppedImage = rgbImage(row1:row2, col1:col2, :);
I = rgb2gray(croppedImage) ;
%% Get the white regions
[y,x,val] = find(I) ;
%5 use kmeans clustering
[idx,C] = kmeans([x,y],8) ;
%%
figure
imshow(I) ;
hold on
for i = 1:8
xi = x(idx==i) ; yi = y(idx==i) ;
id1=convhull(xi,yi) ;
coor = [xi(id1) yi(id1)] ;
[id,c] = kmeans(coor,4) ;
plot(coor(:,1),coor(:,2),'r','linewidth',3) ;
plot(c(:,1),c(:,2),'*b')
end
Now we are able to capture the regions..the boundary/convex hull points are in hand. You can do what ever math you want with the points.
Did you solve the problem? I Looked into it and it seems that the rotation given by 'regionprops' seems to be off. To fix that I've prepared a quick solution: I've dilated the image to close the gaps, found 4 most distant peaks of each spine, and then validated if a peak is on the left, or on the right of the centerline (that I have obtained by extrapolating form sorted centroids). This method seems to work for this particular problem.
BW2 = rgb2gray(Image);
BW2 = imbinarize(BW2);
%dilate and erode will help to remove extra features of the vertebra
se = strel('disk',4,4);
BW2_dilate = imdilate(BW2,se);
BW2_erode = imerode(BW2_dilate,se);
sb = bwboundaries(BW2_erode);
figure
imshow(BW2)
hold on
centerLine = [];
corners = [];
for bone = 1:length(sb)
x0 = sb{bone}(:,2) - mean(sb{bone}(:,2));
y0 = sb{bone}(:,1) - mean(sb{bone}(:,1));
%save the position of the centroid
centerLine = [centerLine; [mean(sb{bone}(:,1)) mean(sb{bone}(:,2))]];
[th0,rho0] = cart2pol(x0,y0);
%make sure that the indexing starts at the dip, not at the corner
lowest_val = find(rho0==min(rho0));
rho1 = [rho0(lowest_val:end); rho0(1:lowest_val-1)];
th00 = [th0(lowest_val:end); th0(1:lowest_val-1)];
y1 = [y0(lowest_val:end); y0(1:lowest_val-1)];
x1 = [x0(lowest_val:end); x0(1:lowest_val-1)];
%detect corners, using smooth data to remove noise
[pks,locs] = findpeaks(smooth(rho1));
[pksS,idS] = sort(pks,'descend');
%4 most pronounced peaks are where the corners are
edgesFndCx = x1(locs(idS(1:4)));
edgesFndCy = y1(locs(idS(1:4)));
edgesFndCx = edgesFndCx + mean(sb{bone}(:,2));
edgesFndCy = edgesFndCy + mean(sb{bone}(:,1));
corners{bone} = [edgesFndCy edgesFndCx];
end
[~,idCL] = sort(centerLine(:,1),'descend');
centerLine = centerLine(idCL,:);
%extrapolate the spine centerline
yDatExt= 1:size(BW2_erode,1);
extrpLine = interp1(centerLine(:,1),centerLine(:,2),yDatExt,'spline','extrap');
plot(centerLine(:,2),centerLine(:,1),'r')
plot(extrpLine,yDatExt,'r')
%find edges to the left, and to the right of the centerline
for bone = 1:length(corners)
x0 = corners{bone}(:,2);
y0 = corners{bone}(:,1);
for crn = 1:4
xCompare = extrpLine(y0(crn));
if x0(crn) < xCompare
plot(x0(crn),y0(crn),'go','LineWidth',2)
else
plot(x0(crn),y0(crn),'ro','LineWidth',2)
end
end
end
Solution
I'm writing the code in Matlab to find interest point using DoG in the image.
Here is the main.m:
imTest1 = rgb2gray(imread('1.jpg'));
imTest1 = double(imTest1);
sigma = 0.6;
k = 5;
thresh = 3;
[x1,y1,r1] = DoG(k,sigma,thresh,imTest1);
%get the interest points and show it on the image with its scale
figure(1);
imshow(imTest1,[]), hold on, scatter(y1,x1,r1,'r');
And the function DoG is:
function [x,y,r] = DoG(k,sigma,thresh,imTest)
x = []; y = []; r = [];
%suppose 5 levels of gaussian blur
for i = 1:k
g{i} = fspecial('gaussian',size(imTest),i*sigma);
end
%so 4 levels of DoG
for i = 1:k-1
d{i} = imfilter(imTest,g{i+1}-g{i});
end
%compare the current pixel in the image to the surrounding pixels (26 points),if it is the maxima/minima, this pixel will be a interest point
for i = 2:k-2
for m = 2:size(imTest,1)-1
for n = 2:size(imTest,2)-1
id = 1;
compare = zeros(1,27);
for ii = i-1:i+1
for mm = m-1:m+1
for nn = n-1:n+1
compare(id) = d{ii}(mm,nn);
id = id+1;
end
end
end
compare_max = max(compare);
compare_min = min(compare);
if (compare_max == d{i}(m,n) || compare_min == d{i}(m,n))
if (compare_min < -thresh || compare_max > thresh)
x = [x;m];
y = [y;n];
r = [r;abs(d{i}(m,n))];
end
end
end
end
end
end
So there's a gaussian function and the sigma i set is 0.6. After running the code, I find the position is not correct and the scales looks almost the same for all interest points. I think my code should work but actually the result is not. Anybody know what's the problem?
I'm trying to produce some computer generated holograms by using MATLAB. I used equally spaced mesh grid to initialize the spatial grid, and I got the following image
This pattern is sort of what I need except the center region. The fringe should be sharp but blurred. I think it might be the problem of the mesh grid. I tried generate a grid in polar coordinates and the map it into Cartesian coordinates by using MATLAB's pol2cart function. Unfortunately, it doesn't work as well. One may suggest that using fine grids. It doesn't work too. I think if I can generate a spiral mesh grid, perhaps the problem is solvable. In addition, the number of the spiral arms could, in general, be arbitrary, could anyone give me a hint on this?
I've attached the code (My final projects are not exactly the same, but it has a similar problem).
clc; clear all; close all;
%% initialization
tic
lambda = 1.55e-6;
k0 = 2*pi/lambda;
c0 = 3e8;
eta0 = 377;
scale = 0.25e-6;
NELEMENTS = 1600;
GoldenRatio = (1+sqrt(5))/2;
g = 2*pi*(1-1/GoldenRatio);
pntsrc = zeros(NELEMENTS, 3);
phisrc = zeros(NELEMENTS, 1);
for idxe = 1:NELEMENTS
pntsrc(idxe, :) = scale*sqrt(idxe)*[cos(idxe*g), sin(idxe*g), 0];
phisrc(idxe) = angle(-sin(idxe*g)+1i*cos(idxe*g));
end
phisrc = 3*phisrc/2; % 3 arms (topological charge ell=3)
%% post processing
sigma = 1;
polfilter = [0, 0, 1i*sigma; 0, 0, -1; -1i*sigma, 1, 0]; % cp filter
xboundl = -100e-6; xboundu = 100e-6;
yboundl = -100e-6; yboundu = 100e-6;
xf = linspace(xboundl, xboundu, 100);
yf = linspace(yboundl, yboundu, 100);
zf = -400e-6;
[pntobsx, pntobsy] = meshgrid(xf, yf);
% how to generate a right mesh grid such that we can generate a decent result?
pntobs = [pntobsx(:), pntobsy(:), zf*ones(size(pntobsx(:)))];
% arbitrary mesh may result in "wrong" results
NPNTOBS = size(pntobs, 1);
nxp = length(xf);
nyp = length(yf);
%% observation
Eobs = zeros(NPNTOBS, 3);
matlabpool open local 12
parfor nobs = 1:NPNTOBS
rp = pntobs(nobs, :);
Erad = [0; 0; 0];
for idx = 1:NELEMENTS
rs = pntsrc(idx, :);
p = exp(sigma*1i*2*phisrc(idx))*[1 -sigma*1i 0]/2; % simplified here
u = rp - rs;
r = sqrt(u(1)^2+u(2)^2+u(3)^2); %norm(u);
u = u/r; % unit vector
ut = [u(2)*p(3)-u(3)*p(2),...
u(3)*p(1)-u(1)*p(3), ...
u(1)*p(2)-u(2)*p(1)]; % cross product: u cross p
Erad = Erad + ... % u cross p cross u, do not use the built-in func
c0*k0^2/4/pi*exp(1i*k0*r)/r*eta0*...
[ut(2)*u(3)-ut(3)*u(2);...
ut(3)*u(1)-ut(1)*u(3); ...
ut(1)*u(2)-ut(2)*u(1)];
end
Eobs(nobs, :) = Erad; % filter neglected here
end
matlabpool close
Eobs = Eobs/max(max(sum(abs(Eobs), 2))); % normailized
%% source, gaussian beam
E0 = 1;
w0 = 80e-6;
theta = 0; % may be titled
RotateX = [1, 0, 0; ...
0, cosd(theta), -sind(theta); ...
0, sind(theta), cosd(theta)];
Esrc = zeros(NPNTOBS, 3);
for nobs = 1:NPNTOBS
rp = RotateX*[pntobs(nobs, 1:2).'; 0];
z = rp(3);
r = sqrt(sum(abs(rp(1:2)).^2));
zR = pi*w0^2/lambda;
wz = w0*sqrt(1+z^2/zR^2);
Rz = z^2+zR^2;
zetaz = atan(z/zR);
gaussian = E0*w0/wz*exp(-r^2/wz^2-1i*k0*z-1i*k0*0*r^2/Rz/2+1i*zetaz);% ...
Esrc(nobs, :) = (polfilter*gaussian*[1; -1i; 0]).'/sqrt(2)/2;
end
Esrc = [Esrc(:, 2), Esrc(:, 3), Esrc(:, 1)];
Esrc = Esrc/max(max(sum(abs(Esrc), 2))); % normailized
toc
%% visualization
fringe = Eobs + Esrc; % I'll have a different formula in my code
normEsrc = reshape(sum(abs(Esrc).^2, 2), [nyp nxp]);
normEobs = reshape(sum(abs(Eobs).^2, 2), [nyp nxp]);
normFringe = reshape(sum(abs(fringe).^2, 2), [nyp nxp]);
close all;
xf0 = linspace(xboundl, xboundu, 500);
yf0 = linspace(yboundl, yboundu, 500);
[xfi, yfi] = meshgrid(xf0, yf0);
data = interp2(xf, yf, normFringe, xfi, yfi);
figure; surf(xfi, yfi, data,'edgecolor','none');
% tri = delaunay(xfi, yfi); trisurf(tri, xfi, yfi, data, 'edgecolor','none');
xlim([xboundl, xboundu])
ylim([yboundl, yboundu])
% colorbar
view(0,90)
colormap(hot)
axis equal
axis off
title('fringe thereo. ', ...
'fontsize', 18)
I didn't read your code because it is too long to do such a simple thing. I wrote mine and here is the result:
the code is
%spiral.m
function val = spiral(x,y)
r = sqrt( x*x + y*y);
a = atan2(y,x)*2+r;
x = r*cos(a);
y = r*sin(a);
val = exp(-x*x*y*y);
val = 1/(1+exp(-1000*(val)));
endfunction
%show.m
n=300;
l = 7;
A = zeros(n);
for i=1:n
for j=1:n
A(i,j) = spiral( 2*(i/n-0.5)*l,2*(j/n-0.5)*l);
end
end
imshow(A) %don't know if imshow is in matlab. I used octave.
the key for the sharpnes is line
val = 1/(1+exp(-1000*(val)));
It is logistic function. The number 1000 defines how sharp your image will be. So lower it for more blurry image or higher it for sharper.
I hope this answers your question ;)
Edit: It is real fun to play with. Here is another spiral:
function val = spiral(x,y)
s= 0.5;
r = sqrt( x*x + y*y);
a = atan2(y,x)*2+r*r*r;
x = r*cos(a);
y = r*sin(a);
val = 0;
if (abs(x)<s )
val = s-abs(x);
endif
if(abs(y)<s)
val =max(s-abs(y),val);
endif
%val = 1/(1+exp(-1*(val)));
endfunction
Edit2: Fun, fun, fun! Here the arms do not get thinner.
function val = spiral(x,y)
s= 0.1;
r = sqrt( x*x + y*y);
a = atan2(y,x)*2+r*r; % h
x = r*cos(a);
y = r*sin(a);
val = 0;
s = s*exp(r);
if (abs(x)<s )
val = s-abs(x);
endif
if(abs(y)<s)
val =max(s-abs(y),val);
endif
val = val/s;
val = 1/(1+exp(-10*(val)));
endfunction
Damn your question I really need to study for my exam, arghhh!
Edit3:
I vectorised the code and it runs much faster.
%spiral.m
function val = spiral(x,y)
s= 2;
r = sqrt( x.*x + y.*y);
a = atan2(y,x)*8+exp(r);
x = r.*cos(a);
y = r.*sin(a);
val = 0;
s = s.*exp(-0.1*r);
val = r;
val = (abs(x)<s ).*(s-abs(x));
val = val./s;
% val = 1./(1.+exp(-1*(val)));
endfunction
%show.m
n=1000;
l = 3;
A = zeros(n);
[X,Y] = meshgrid(-l:2*l/n:l);
A = spiral(X,Y);
imshow(A)
Sorry, can't post figures. But this might help. I wrote it for experiments with amplitude spatial modulators...
R=70; % radius of curvature of fresnel lens (in pixel units)
A=0; % oblique incidence by linear grating (1=oblique 0=collinear)
B=1; % expanding by fresnel lens (1=yes 0=no)
L=7; % topological charge
Lambda=30; % linear grating fringe spacing (in pixels)
aspect=1/2; % fraction of fringe period that is white/clear
xsize=1024; % resolution (xres x yres number data pts calculated)
ysize=768; %
% define the X and Y ranges (defined to skip zero)
xvec = linspace(-xsize/2, xsize/2, xsize); % list of x values
yvec = linspace(-ysize/2, ysize/2, ysize); % list of y values
% define the meshes - matrices linear in one dimension
[xmesh, ymesh] = meshgrid(xvec, yvec);
% calculate the individual phase components
vortexPh = atan2(ymesh,xmesh); % the vortex phase
linPh = -2*pi*ymesh; % a phase of linear grating
radialPh = (xmesh.^2+ymesh.^2); % a phase of defocus
% combine the phases with appropriate scales (phases are additive)
% the 'pi' at the end causes inversion of the pattern
Ph = L*vortexPh + A*linPh/Lambda + B*radialPh/R^2;
% transmittance function (the real part of exp(I*Ph))
T = cos(Ph);
% the binary version
binT = T > cos(pi*aspect);
% plot the pattern
% imagesc(binT)
imagesc(T)
colormap(gray)
I am working on rotating image manually in Matlab. Each time I run my code with a different image the previous images which are rotated are shown in the Figure. I couldn't figure it out. Any help would be appreciable.
The code is here:
[screenshot]
im1 = imread('gradient.jpg');
[h, w, p] = size(im1);
theta = pi/12;
hh = round( h*cos(theta) + w*abs(sin(theta))); %Round to nearest integer
ww = round( w*cos(theta) + h*abs(sin(theta))); %Round to nearest integer
R = [cos(theta) -sin(theta); sin(theta) cos(theta)];
T = [w/2; h/2];
RT = [inv(R) T; 0 0 1];
for z = 1:p
for x = 1:ww
for y = 1:hh
% Using matrix multiplication
i = zeros(3,1);
i = RT*[x-ww/2; y-hh/2; 1];
%% Nearest Neighbour
i = round(i);
if i(1)>0 && i(2)>0 && i(1)<=w && i(2)<=h
im2(y,x,z) = im1(i(2),i(1),z);
end
end
end
end
x=1:ww;
y=1:hh;
[X, Y] = meshgrid(x,y); % Generate X and Y arrays for 3-D plots
orig_pos = [X(:)' ; Y(:)' ; ones(1,numel(X))]; % Number of elements in array or subscripted array expression
orig_pos_2 = [X(:)'-(ww/2) ; Y(:)'-(hh/2) ; ones(1,numel(X))];
new_pos = round(RT*orig_pos_2); % Round to nearest neighbour
% Check if new positions fall from map:
valid_pos = new_pos(1,:)>=1 & new_pos(1,:)<=w & new_pos(2,:)>=1 & new_pos(2,:)<=h;
orig_pos = orig_pos(:,valid_pos);
new_pos = new_pos(:,valid_pos);
siz = size(im1);
siz2 = size(im2);
% Expand the 2D indices to include the third dimension.
ind_orig_pos = sub2ind(siz2,orig_pos(2*ones(p,1),:),orig_pos(ones(p,1),:), (1:p)'*ones(1,length(orig_pos)));
ind_new_pos = sub2ind(siz, new_pos(2*ones(p,1),:), new_pos(ones(p,1),:), (1:p)'*ones(1,length(new_pos)));
im2(ind_orig_pos) = im1(ind_new_pos);
imshow(im2);
There is a problem with the initialization of im2, or rather, the lack of it. im2 is created in the section shown below:
if i(1)>0 && i(2)>0 && i(1)<=w && i(2)<=h
im2(y,x,z) = im1(i(2),i(1),z);
end
If im2 exists before this code is run and its width or height is larger than the image you are generating the new image will only overwrite the top left corner of your existing im2. Try initializing im2 by adding adding
im2 = zeros(hh, ww, p);
before
for z = 1:p
for x = 1:ww
for y = 1:hh
...
As a bonus it might make your code a little faster since Matlab won't have to resize im2 as it grows in the loop.