I did a gaussian filter and the image become index. I have to use imagesc to show the determine the color difference. How can I convert it to rgb so that I can do further process.
Edited Added some images, top is the 'original image', 'imshow(C)', 'imagesc(C)' respectively. Then I just want the 'C' variable to be like imagesc image. Is it possible??
Edited Here is my coding, see from gaussian onward
% Read Image
rgb = imread('barcode.jpg');
% Resize Image
rgb = imresize(rgb,0.33);
%figure(),imshow(rgb);
% Convert from RGB to Gray
Igray = rgb2gray(rgb);
BW2 = edge(Igray,'canny');
%figure(),imshow(BW2);
% Perform the Hough transform
[H, theta, rho] = hough(BW2);
% Find the peak pt in the Hough transform
peak = houghpeaks(H);
% Find the angle of the bars
barAngle = theta(peak(2));
J = imrotate(rgb,barAngle,'bilinear','crop');
%figure(),imshow(J);
Jgray = double(rgb2gray(J));
% Calculate the Gradients
[dIx, dIy] = gradient(Jgray);
%if min(dIx(:))<= -100 && max(dIx(:))>=100 || min(dIy(:))<=-100 && max(dIy(:))>=100
if barAngle <= 65 && barAngle >=-65 && min(dIx(:))<= -100
B = abs(dIx) - abs(dIy);
else
B = abs(dIy) - abs(dIx);
end
% Low-Pass Filtering
H = fspecial('gaussian', 20, 10);
C = imfilter(B, H);
C = imclearborder(C);
figure(),imshow(C);
figure(),imagesc(C);colorbar;
RGB = ind2rgb(X,map)
RGB is just eye-candy at this point, you can't magically add information that isn't there.
EDIT
In your code, C is a gray-scale image, because B is gray-scale which in terms is caused by the fact that it is composed from gradients dIx and dIy that originate from an image, you yourself make grayscale explicitly with the line Jgray = double(rgb2gray(J));
Related
I need to get a square region from input, and apply averaging filter to it.
Have tried this, but gives error:
I=imread('img1.jpg');
h=fspecial('average');
figure;
h_img = imshow(I);
sq=imrect();
mask = createMask(sq,h_img);
I2 = roifilt2(h,I,mask);
The last line gives error.
I think I know the problem.
You are probably trying to apply filter to RGB image.
Check the following code sample:
I = imread('peppers.png');
h = fspecial('average', 10);
figure;
h_img = imshow(I);
sq=imrect();
mask = createMask(sq,h_img);
if (ndims(I) == 3)
classI = class(I);
if (isequal(classI, 'uint8'))
I = double(I)/255; %Convert I to double before applying filter.
end
I2 = zeros(size(I));
I2(:,:,1) = roifilt2(h,I(:,:,1), mask); %Filter Red plain
I2(:,:,2) = roifilt2(h,I(:,:,2), mask); %Filter Green plain
I2(:,:,3) = roifilt2(h,I(:,:,3) ,mask); %Filter Blue plain
if (isequal(classI, 'uint8'))
I2 = uint8(round(double(I2)*255)); %Convert I2 to uint8 after applying filter.
end
end
figure;imshow(I2);
Selecting region:
Filter result:
I have a 128x128 grascale image that i wish to find the Hadamard transform of with Normal Hadamard, sequency, and dyadic ordering.
imdata = imread('origim.png'); %Load image
new = rgb2gray(imdata); %Convert to 2D Grayscale
N = 128;
H = hadamard(N); % Hadamard matrix
y = fwht(new,N,'sequency') %Perform Fast-walsh-hadamard-transform with order 128
imshow(y); %Display image transform
I may be doing it wrong, however y should be the transformed image if i understand the matlab walch transform correctly. When i try runing it i get an error with y = fwht(new,N,'sequency')
Before processing convert the image into double. Then put semicolon (;) at the end of y = fwht(new,N,'sequency'). Then you will get transformed image.
Just try the below code.
imdata = imread('peppers.png'); %Load image
new = rgb2gray(imdata); %Convert to 2D Grayscale
neww = im2double(new);
N = 128;
H = hadamard(N); % Hadamard matrix
y = fwht(neww,N,'sequency'); %Perform Fast-walsh-hadamard-transform with order 128
imshow(y); %Display image transform
I have this code converting a fisheye image into rectangular form but the code is only able to perform this operation on a grayscale image. Can anybody help converting the code to perform the operation on a RGB image. The code is as follows:
edit: I have updated the code to contain a functionality which performs interpolation in each color channel. But this seem to disform the output image. See pictures below
function imP = FISHCOLOR (imR)
rMin=0.1;
rMax=1;
[Mr, Nr, Dr] = size(imR); % size of rectangular image
xRc = (Mr+1)/2; % co-ordinates of the center of the image
yRc = (Nr+1)/2;
sx = (Mr-1)/2; % scale factors
sy = (Nr-1)/2;
M=size(imR,1);N=size(imR,2);
dr = (rMax - rMin)/(M-1);
dth = 2*pi/N;
r=rMin:dr:rMin+(M-1)*dr;
th=(0:dth:(N-1)*dth)';
[r,th]=meshgrid(r,th);
x=r.*cos(th);
y=r.*sin(th);
xR = x*sx + xRc;
yR = y*sy + yRc;
imP =zeros(M, N); % initialize the final matrix
for k=1:3 % colors
T = imR(:,:,k);
Ichannel = interp2(T,xR,yR);
imP(:,:,k)= Ichannel; % add k channel
end
SOLVED
Input image <- Image link
Grayscale output, what i would like in color <- Image link
Try changing these three lines:
[Mr Nr] = size(imR); % size of rectangular image
...
imP = zeros(M, N);
...
imP = interp2(imR, xR, yR); %interpolate (imR, xR, yR);
...to these:
[Mr Nr Pr] = size(imR); % size of rectangular image
...
imP = zeros(M, N, Pr);
...
for dim = 1:Pr
imP(:,:,dim) = interp2(imR(:,:,dim), xR, yR); %interpolate (imR, xR, yR);
end
I am trying to rotate an image with Matlab without using imrotate function. I actually made it by using transformation matrix.But it is not good enough.The problem is, the rotated image is "sliding".Let me tell you with pictures.
This is my image which I want to rotate:
But when I rotate it ,for example 45 degrees, it becomes this:
I am asking why this is happening.Here is my code,is there any mathematical or programming mistakes about it?
image=torso;
%image padding
[Rows, Cols] = size(image);
Diagonal = sqrt(Rows^2 + Cols^2);
RowPad = ceil(Diagonal - Rows) + 2;
ColPad = ceil(Diagonal - Cols) + 2;
imagepad = zeros(Rows+RowPad, Cols+ColPad);
imagepad(ceil(RowPad/2):(ceil(RowPad/2)+Rows-1),ceil(ColPad/2):(ceil(ColPad/2)+Cols-1)) = image;
degree=45;
%midpoints
midx=ceil((size(imagepad,1)+1)/2);
midy=ceil((size(imagepad,2)+1)/2);
imagerot=zeros(size(imagepad));
%rotation
for i=1:size(imagepad,1)
for j=1:size(imagepad,2)
x=(i-midx)*cos(degree)-(j-midy)*sin(degree);
y=(i-midx)*sin(degree)+(j-midy)*cos(degree);
x=round(x)+midx;
y=round(y)+midy;
if (x>=1 && y>=1)
imagerot(x,y)=imagepad(i,j); % k degrees rotated image
end
end
end
figure,imagesc(imagerot);
colormap(gray(256));
The reason you have holes in your image is because you are computing the location in imagerot of each pixel in imagepad. You need to do the computation the other way around. That is, for each pixel in imagerot interpolate in imagepad. To do this, you just need to apply the inverse transform, which in the case of a rotation matrix is just the transpose of the matrix (just change the sign on each sin and translate the other way).
Loop over pixels in imagerot:
imagerot=zeros(size(imagepad)); % midx and midy same for both
for i=1:size(imagerot,1)
for j=1:size(imagerot,2)
x= (i-midx)*cos(rads)+(j-midy)*sin(rads);
y=-(i-midx)*sin(rads)+(j-midy)*cos(rads);
x=round(x)+midx;
y=round(y)+midy;
if (x>=1 && y>=1 && x<=size(imagepad,2) && y<=size(imagepad,1))
imagerot(i,j)=imagepad(x,y); % k degrees rotated image
end
end
end
Also note that your midx and midy need to be calculated with size(imagepad,2) and size(imagepad,1) respectively, since the first dimension refers to the number of rows (height) and the second to width.
NOTE: The same approach applies when you decide to adopt an interpolation scheme other than nearest neighbor, as in Rody's example with linear interpolation.
EDIT: I'm assuming you are using a loop for demonstrative purposes, but in practice there is no need for loops. Here's an example of nearest neighbor interpolation (what you are using), keeping the same size image, but you can modify this to produce a larger image that includes the whole source image:
imagepad = imread('peppers.png');
[nrows ncols nslices] = size(imagepad);
midx=ceil((ncols+1)/2);
midy=ceil((nrows+1)/2);
Mr = [cos(pi/4) sin(pi/4); -sin(pi/4) cos(pi/4)]; % e.g. 45 degree rotation
% rotate about center
[X Y] = meshgrid(1:ncols,1:nrows);
XYt = [X(:)-midx Y(:)-midy]*Mr;
XYt = bsxfun(#plus,XYt,[midx midy]);
xout = round(XYt(:,1)); yout = round(XYt(:,2)); % nearest neighbor!
outbound = yout<1 | yout>nrows | xout<1 | xout>ncols;
zout=repmat(cat(3,1,2,3),nrows,ncols,1); zout=zout(:);
xout(xout<1) = 1; xout(xout>ncols) = ncols;
yout(yout<1) = 1; yout(yout>nrows) = nrows;
xout = repmat(xout,[3 1]); yout = repmat(yout,[3 1]);
imagerot = imagepad(sub2ind(size(imagepad),yout,xout,zout(:))); % lookup
imagerot = reshape(imagerot,size(imagepad));
imagerot(repmat(outbound,[1 1 3])) = 0; % set background value to [0 0 0] (black)
To modify the above to linear interpolation, compute the 4 neighboring pixels to each coordinate in XYt and perform a weighted sum using the fractional components product as the weights. I'll leave that as an exercise, since it would only serve to bloat my answer further beyond the scope of your question. :)
The method you are using (rotate by sampling) is the fastest and simplest, but also the least accurate.
Rotation by area mapping, as given below (this is a good reference), is much better at preserving color.
But: note that this will only work on greyscale/RGB images, but NOT on colormapped images like the one you seem to be using.
image = imread('peppers.png');
figure(1), clf, hold on
subplot(1,2,1)
imshow(image);
degree = 45;
switch mod(degree, 360)
% Special cases
case 0
imagerot = image;
case 90
imagerot = rot90(image);
case 180
imagerot = image(end:-1:1, end:-1:1);
case 270
imagerot = rot90(image(end:-1:1, end:-1:1));
% General rotations
otherwise
% Convert to radians and create transformation matrix
a = degree*pi/180;
R = [+cos(a) +sin(a); -sin(a) +cos(a)];
% Figure out the size of the transformed image
[m,n,p] = size(image);
dest = round( [1 1; 1 n; m 1; m n]*R );
dest = bsxfun(#minus, dest, min(dest)) + 1;
imagerot = zeros([max(dest) p],class(image));
% Map all pixels of the transformed image to the original image
for ii = 1:size(imagerot,1)
for jj = 1:size(imagerot,2)
source = ([ii jj]-dest(1,:))*R.';
if all(source >= 1) && all(source <= [m n])
% Get all 4 surrounding pixels
C = ceil(source);
F = floor(source);
% Compute the relative areas
A = [...
((C(2)-source(2))*(C(1)-source(1))),...
((source(2)-F(2))*(source(1)-F(1)));
((C(2)-source(2))*(source(1)-F(1))),...
((source(2)-F(2))*(C(1)-source(1)))];
% Extract colors and re-scale them relative to area
cols = bsxfun(#times, A, double(image(F(1):C(1),F(2):C(2),:)));
% Assign
imagerot(ii,jj,:) = sum(sum(cols),2);
end
end
end
end
subplot(1,2,2)
imshow(imagerot);
Output:
Rotates colored image according to angle given by user without any cropping of image in matlab.
Output of this program is similar to output of inbuilt command "imrotate" .This program dynamically creates background according to angle input given by user.By using rotation matrix and origin shifting, we get relation between coordinates of initial and final image.Using relation between coordinates of initial and final image, we now map the intensity values for each pixel.
img=imread('img.jpg');
[rowsi,colsi,z]= size(img);
angle=45;
rads=2*pi*angle/360;
%calculating array dimesions such that rotated image gets fit in it exactly.
% we are using absolute so that we get positve value in any case ie.,any quadrant.
rowsf=ceil(rowsi*abs(cos(rads))+colsi*abs(sin(rads)));
colsf=ceil(rowsi*abs(sin(rads))+colsi*abs(cos(rads)));
% define an array withcalculated dimensionsand fill the array with zeros ie.,black
C=uint8(zeros([rowsf colsf 3 ]));
%calculating center of original and final image
xo=ceil(rowsi/2);
yo=ceil(colsi/2);
midx=ceil((size(C,1))/2);
midy=ceil((size(C,2))/2);
% in this loop we calculate corresponding coordinates of pixel of A
% for each pixel of C, and its intensity will be assigned after checking
% weather it lie in the bound of A (original image)
for i=1:size(C,1)
for j=1:size(C,2)
x= (i-midx)*cos(rads)+(j-midy)*sin(rads);
y= -(i-midx)*sin(rads)+(j-midy)*cos(rads);
x=round(x)+xo;
y=round(y)+yo;
if (x>=1 && y>=1 && x<=size(img,1) && y<=size(img,2) )
C(i,j,:)=img(x,y,:);
end
end
end
imshow(C);
Check this out.
this is fastest way that you can do.
img = imread('Koala.jpg');
theta = pi/10;
rmat = [
cos(theta) sin(theta) 0
-sin(theta) cos(theta) 0
0 0 1];
mx = size(img,2);
my = size(img,1);
corners = [
0 0 1
mx 0 1
0 my 1
mx my 1];
new_c = corners*rmat;
T = maketform('affine', rmat); %# represents translation
img2 = imtransform(img, T, ...
'XData',[min(new_c(:,1)) max(new_c(:,1))],...
'YData',[min(new_c(:,2)) max(new_c(:,2))]);
subplot(121), imshow(img);
subplot(122), imshow(img2);
I am trying to rotate an image with Matlab without using imrotate function. I actually made it by using transformation matrix.But it is not good enough.The problem is, the rotated image is "sliding".Let me tell you with pictures.
This is my image which I want to rotate:
But when I rotate it ,for example 45 degrees, it becomes this:
I am asking why this is happening.Here is my code,is there any mathematical or programming mistakes about it?
image=torso;
%image padding
[Rows, Cols] = size(image);
Diagonal = sqrt(Rows^2 + Cols^2);
RowPad = ceil(Diagonal - Rows) + 2;
ColPad = ceil(Diagonal - Cols) + 2;
imagepad = zeros(Rows+RowPad, Cols+ColPad);
imagepad(ceil(RowPad/2):(ceil(RowPad/2)+Rows-1),ceil(ColPad/2):(ceil(ColPad/2)+Cols-1)) = image;
degree=45;
%midpoints
midx=ceil((size(imagepad,1)+1)/2);
midy=ceil((size(imagepad,2)+1)/2);
imagerot=zeros(size(imagepad));
%rotation
for i=1:size(imagepad,1)
for j=1:size(imagepad,2)
x=(i-midx)*cos(degree)-(j-midy)*sin(degree);
y=(i-midx)*sin(degree)+(j-midy)*cos(degree);
x=round(x)+midx;
y=round(y)+midy;
if (x>=1 && y>=1)
imagerot(x,y)=imagepad(i,j); % k degrees rotated image
end
end
end
figure,imagesc(imagerot);
colormap(gray(256));
The reason you have holes in your image is because you are computing the location in imagerot of each pixel in imagepad. You need to do the computation the other way around. That is, for each pixel in imagerot interpolate in imagepad. To do this, you just need to apply the inverse transform, which in the case of a rotation matrix is just the transpose of the matrix (just change the sign on each sin and translate the other way).
Loop over pixels in imagerot:
imagerot=zeros(size(imagepad)); % midx and midy same for both
for i=1:size(imagerot,1)
for j=1:size(imagerot,2)
x= (i-midx)*cos(rads)+(j-midy)*sin(rads);
y=-(i-midx)*sin(rads)+(j-midy)*cos(rads);
x=round(x)+midx;
y=round(y)+midy;
if (x>=1 && y>=1 && x<=size(imagepad,2) && y<=size(imagepad,1))
imagerot(i,j)=imagepad(x,y); % k degrees rotated image
end
end
end
Also note that your midx and midy need to be calculated with size(imagepad,2) and size(imagepad,1) respectively, since the first dimension refers to the number of rows (height) and the second to width.
NOTE: The same approach applies when you decide to adopt an interpolation scheme other than nearest neighbor, as in Rody's example with linear interpolation.
EDIT: I'm assuming you are using a loop for demonstrative purposes, but in practice there is no need for loops. Here's an example of nearest neighbor interpolation (what you are using), keeping the same size image, but you can modify this to produce a larger image that includes the whole source image:
imagepad = imread('peppers.png');
[nrows ncols nslices] = size(imagepad);
midx=ceil((ncols+1)/2);
midy=ceil((nrows+1)/2);
Mr = [cos(pi/4) sin(pi/4); -sin(pi/4) cos(pi/4)]; % e.g. 45 degree rotation
% rotate about center
[X Y] = meshgrid(1:ncols,1:nrows);
XYt = [X(:)-midx Y(:)-midy]*Mr;
XYt = bsxfun(#plus,XYt,[midx midy]);
xout = round(XYt(:,1)); yout = round(XYt(:,2)); % nearest neighbor!
outbound = yout<1 | yout>nrows | xout<1 | xout>ncols;
zout=repmat(cat(3,1,2,3),nrows,ncols,1); zout=zout(:);
xout(xout<1) = 1; xout(xout>ncols) = ncols;
yout(yout<1) = 1; yout(yout>nrows) = nrows;
xout = repmat(xout,[3 1]); yout = repmat(yout,[3 1]);
imagerot = imagepad(sub2ind(size(imagepad),yout,xout,zout(:))); % lookup
imagerot = reshape(imagerot,size(imagepad));
imagerot(repmat(outbound,[1 1 3])) = 0; % set background value to [0 0 0] (black)
To modify the above to linear interpolation, compute the 4 neighboring pixels to each coordinate in XYt and perform a weighted sum using the fractional components product as the weights. I'll leave that as an exercise, since it would only serve to bloat my answer further beyond the scope of your question. :)
The method you are using (rotate by sampling) is the fastest and simplest, but also the least accurate.
Rotation by area mapping, as given below (this is a good reference), is much better at preserving color.
But: note that this will only work on greyscale/RGB images, but NOT on colormapped images like the one you seem to be using.
image = imread('peppers.png');
figure(1), clf, hold on
subplot(1,2,1)
imshow(image);
degree = 45;
switch mod(degree, 360)
% Special cases
case 0
imagerot = image;
case 90
imagerot = rot90(image);
case 180
imagerot = image(end:-1:1, end:-1:1);
case 270
imagerot = rot90(image(end:-1:1, end:-1:1));
% General rotations
otherwise
% Convert to radians and create transformation matrix
a = degree*pi/180;
R = [+cos(a) +sin(a); -sin(a) +cos(a)];
% Figure out the size of the transformed image
[m,n,p] = size(image);
dest = round( [1 1; 1 n; m 1; m n]*R );
dest = bsxfun(#minus, dest, min(dest)) + 1;
imagerot = zeros([max(dest) p],class(image));
% Map all pixels of the transformed image to the original image
for ii = 1:size(imagerot,1)
for jj = 1:size(imagerot,2)
source = ([ii jj]-dest(1,:))*R.';
if all(source >= 1) && all(source <= [m n])
% Get all 4 surrounding pixels
C = ceil(source);
F = floor(source);
% Compute the relative areas
A = [...
((C(2)-source(2))*(C(1)-source(1))),...
((source(2)-F(2))*(source(1)-F(1)));
((C(2)-source(2))*(source(1)-F(1))),...
((source(2)-F(2))*(C(1)-source(1)))];
% Extract colors and re-scale them relative to area
cols = bsxfun(#times, A, double(image(F(1):C(1),F(2):C(2),:)));
% Assign
imagerot(ii,jj,:) = sum(sum(cols),2);
end
end
end
end
subplot(1,2,2)
imshow(imagerot);
Output:
Rotates colored image according to angle given by user without any cropping of image in matlab.
Output of this program is similar to output of inbuilt command "imrotate" .This program dynamically creates background according to angle input given by user.By using rotation matrix and origin shifting, we get relation between coordinates of initial and final image.Using relation between coordinates of initial and final image, we now map the intensity values for each pixel.
img=imread('img.jpg');
[rowsi,colsi,z]= size(img);
angle=45;
rads=2*pi*angle/360;
%calculating array dimesions such that rotated image gets fit in it exactly.
% we are using absolute so that we get positve value in any case ie.,any quadrant.
rowsf=ceil(rowsi*abs(cos(rads))+colsi*abs(sin(rads)));
colsf=ceil(rowsi*abs(sin(rads))+colsi*abs(cos(rads)));
% define an array withcalculated dimensionsand fill the array with zeros ie.,black
C=uint8(zeros([rowsf colsf 3 ]));
%calculating center of original and final image
xo=ceil(rowsi/2);
yo=ceil(colsi/2);
midx=ceil((size(C,1))/2);
midy=ceil((size(C,2))/2);
% in this loop we calculate corresponding coordinates of pixel of A
% for each pixel of C, and its intensity will be assigned after checking
% weather it lie in the bound of A (original image)
for i=1:size(C,1)
for j=1:size(C,2)
x= (i-midx)*cos(rads)+(j-midy)*sin(rads);
y= -(i-midx)*sin(rads)+(j-midy)*cos(rads);
x=round(x)+xo;
y=round(y)+yo;
if (x>=1 && y>=1 && x<=size(img,1) && y<=size(img,2) )
C(i,j,:)=img(x,y,:);
end
end
end
imshow(C);
Check this out.
this is fastest way that you can do.
img = imread('Koala.jpg');
theta = pi/10;
rmat = [
cos(theta) sin(theta) 0
-sin(theta) cos(theta) 0
0 0 1];
mx = size(img,2);
my = size(img,1);
corners = [
0 0 1
mx 0 1
0 my 1
mx my 1];
new_c = corners*rmat;
T = maketform('affine', rmat); %# represents translation
img2 = imtransform(img, T, ...
'XData',[min(new_c(:,1)) max(new_c(:,1))],...
'YData',[min(new_c(:,2)) max(new_c(:,2))]);
subplot(121), imshow(img);
subplot(122), imshow(img2);