Already checked This SO Question , but my question is how to make changes appear to image after the calculation , and code will explain better , i have marked the line where is am confused:
% code written by Zulqurnain Jutt
img1 = imread('rotme.jpg');
img = img1;
[r,m] = size(img);
promt = 'Enter Angle to Rotate an image:=:';
x = input(promt);
% making antirotate matrix
coss = cos(x);
sinn = sin(x);
antirotate = floor( [coss -sinn 0;sinn coss 0; 0 0 1] );
for i=1:r
for j=1:m
%forming point matrix
pointMat = [i-1;j-1;1];
%multiplying both matrix
mul = floor(antirotate*pointMat);
% swapping points
if mul(1)<0
mul(1) = abs(mul(1));
end
if mul(2)<0
mul(2) = abs(mul(2));
end
img( (mul(1)+1) , (mul(2)+1) ) = img(i,j); % Here Lies my problem
end
end
imshow(img);
Problem 1: "Why the output in imshow after assigning the new pixel values giving me three different images as output red , green , blue. is there any way to fix it and get one image as output?"
Problem 2: "i am converting negative coordinates to positive is this okay?"
Note: it is homework question, and i am not hidding it. but is need help
Related
I have 2 greyscale images that i am trying to align using scalar scaling 1 , rotation matrix [2,2] and translation vector [2,1]. I can calculate image1's transformed coordinates as
y = s*R*x + t;
Below the resulting images are shown.
The first image is image1 before transformation,
the second image is image1 (red) with attempted interpolation using interp2 shown on top of image2 (green)
The third image is when i manually insert the pixel values from image1 into an empty array (that has the same size as image2) using the transformed coordinates.
From this we can see that the coordinate transformation must have been successful, as the images are aligned although not perfectly (which is to be expected since only 2 coordinates were used in calculating s, R and t) .
How come interp2 is not producing a result more similar to when i manually insert pixel values?
Below the code for doing this is included:
Interpolation code
function [transformed_image] = interpolate_image(im_r,im_t,s,R,t)
[m,n] = size(im_t);
% doesn't help if i use get_grid that the other function is using here
[~, grid_xr, grid_yr] = get_ipgrid(im_r);
[x_t, grid_xt, grid_yt] = get_ipgrid(im_t);
y = s*R*x_t + t;
yx = reshape(y(1,:), m,n);
yy = reshape(y(2,:), m,n);
transformed_image = interp2(grid_xr, grid_yr, im_r, yx, yy, 'nearest');
end
function [x, grid_x, grid_y] = get_ipgrid(image)
[m,n] = size(image);
[grid_x,grid_y] = meshgrid(1:n,1:m);
x = [reshape(grid_x, 1, []); reshape(grid_y, 1, [])]; % X is [2xM*N] coordinate pairs
end
The manual code
function [transformed_image] = transform_image(im_r,im_t,s,R,t)
[m,n] = size(im_t);
[x_t, grid_xt, grid_yt] = get_grid(im_t);
y = s*R*x_t + t;
ymat = reshape(y',m,n,2);
yx = ymat(:,:,1);
yy = ymat(:,:,2);
transformed_image = zeros(m,n);
for i = 1:m
for j = 1:n
% make sure coordinates are inside
if (yx(i,j) < m & yy(i,j) < n & yx(i,j) > 0.5 & yy(i,j) > 0.5)
transformed_image(round(yx(i,j)),round(yy(i,j))) = im_r(i,j);
end
end
end
end
function [x, grid_x, grid_y] = get_grid(image)
[m,n] = size(image);
[grid_y,grid_x] = meshgrid(1:n,1:m);
x = [grid_x(:) grid_y(:)]'; % X is [2xM*N] coordinate pairs
end
Can anyone see what i'm doing wrong with interp2? I feel like i have tried everything
Turns out i got interpolation all wrong.
In my question i calculate the coordinates of im1 in im2.
However the way interpolation works is that i need to calculate the coordinates of im2 in im1 such that i can map the image as shown below.
This means that i also calculated the wrong s,R and t since they were used to transform im1 -> im2, where as i needed im2 -> im1. (this is also called the inverse transform). Below is the manual code, that is basically the same as interp2 with nearest neighbour interpolation
function [transformed_image] = transform_image(im_r,im_t,s,R,t)
[m,n] = size(im_t);
[x_t, grid_xt, grid_yt] = get_grid(im_t);
y = s*R*x_t + t;
ymat = reshape(y',m,n,2);
yx = ymat(:,:,1);
yy = ymat(:,:,2);
transformed_image = zeros(m,n);
for i = 1:m
for j = 1:n
% make sure coordinates are inside
if (yx(i,j) < m & yy(i,j) < n & yx(i,j) > 0.5 & yy(i,j) > 0.5)
transformed_image(i,j) = im_r(round(yx(i,j)),round(yy(i,j)));
end
end
end
end
I am trying to do a translation, euclidian, similarity, affine and projective transformation on an image pixel by pixel. the input for my program is the image name and the transformation matrix.
This is my code
function imagetrans(name, m)
Image = imread(name);
[rows, cols] = size(Image);
newImage(1:rows,1:cols) = 1;
for row = 1 : rows
for col = 1 : cols
if(Image(row,col) == 0)
point = [row;col;1];
answer = m * point;
answer = int8(answer);
newx = answer(1,1);
newy = answer(2,1);
newImage(newx,newy) = 0;
end
end
end
imshow(newImage);
end
This is the image
Right now I am testing just a translation matrix.
matrix =
1 0 7
0 1 2
0 0 1
when I pass the image and matrix through the function my result is just a little black line
What am I doing wrong?
Using the matlab debugger, I noticed that that you are casting to int8, which is too small too represent all indices. So, you should use int32/int64 or uint32/uint64 instead, i.e.
answer = uint8(answer);
should be
answer = uint32(answer);
Please try to use the Matlab debugger before asking the question: why does it not work?
Based on Shai and Biguri's codes and comments, I have finished a color picture like this:
A problem arises, how to remove the white edge and make it smooth? One solution may be to build 3x3 matrix or bigger and average. But the calculations should be large for every white-edge points. Or there may be some useful functions in Matlab to deal well with this problems?
If you have a license for the image processing toolbox, you can try using for example medfilt2 to apply a median filter on the image. A 11 by 11 median filter should do the trick. It is not very difficult to reimplement the filter yourself if you don't have the toolbox.
This is just one of the possibilities, you can use many different filters that will have different impacts on sharpness ang edge removal.
Edit:
Here is a quick median filter implementation (it may contain errors and could be optimized):
function ret = imageMedianFilter(im, np)
if(size(np,2) == 1)
npx = np;
npy = np;
else
npx = np(1);
npy = np(2);
end
ret = zeros(size(im,1),size(im,2));
for xpos = 1:size(im,1)
for ypos = 1:size(im,2)
curval = double(0);
if(xpos + npx - 1) > size(im,1)
npixels_x = size(im,1) - xpos + 1;
else
npixels_x = npx;
end
if(ypos + npy - 1) > size(im,2)
npixels_y = size(im,2) - ypos + 1;
else
npixels_y = npy;
end
a = im(xpos:xpos+npixels_x-1 , ypos:ypos+npixels_y-1);
a = reshape(a,1,size(a,1)*size(a,2));
curval = median(a);
ret(xpos , ypos) = curval;
end
end
ret = uint8(ret);
end
You can use it on R,G and B components as shown by Rotem below:
RGB = cat(3, imageMedianFilter(RGB(:,:,1), [11,11]), imageMedianFilter(RGB(:,:,2), [11,11]), imageMedianFilter(RGB(:,:,3), [11,11]));
(assuming your image is named RGB).
Here is my solution. I take n*n patch to average the near RGB. But there is a problem arising. Why the right down side of processed picture showing black lines?
clc;clf;close all;clear all;
img = imread('sample2color_t1.bmp'); %// read image
bw = img(:,:,1) > 128; %// convert to binary mask
[lb,lab] = bwlabel(bw,4); %// extract distinct regions
[a,b,c]=size(img);
R=ones(a,b);
G=ones(a,b);
B=ones(a,b);
%I have omitted other colors process codes. Below it is the white edges code.
r=[];c=[];
[r,c] = find(lb ==0);
for i=1:length(r)
R(r(i),c(i))=1;
G(r(i),c(i))=1;
B(r(i),c(i))=1;
end
scale=5;%步长1,8连通
for i=1:length(r)
sumR=0;sumG=0;sumB=0;
for j=0:2*scale
for k=0:2*scale
sumR=sumR+R(r(i)-scale+j,c(i)-scale+k);
sumG=sumG+G(r(i)-scale+j,c(i)-scale+k);
sumB=sumB+B(r(i)-scale+j,c(i)-scale+k);
end
end
R(r(i),c(i))=sumR/(2*scale+1)^2;
G(r(i),c(i))=sumG/(2*scale+1)^2;
B(r(i),c(i))=sumB/(2*scale+1)^2;
end
imPaint=cat(3,R,G,B);
figure;
imshow(imPaint);
Hello so I have an original image and I cropped a part of that image(template) and wrote a code with SAD algorithim to detect on the original image the part i cropped (template) where it exists and draw a rectangle on it.
The code doesnt have any errors , but the rectangle draw doesnt match the template , so I guess the 'output ' variable is the problem can you please help me
I2=imread('img.PNG');
I2=rgb2gray(I2);
[r,c]= size(I2);
%padding the image
%padding
B = padarray(I2,[24 24],'replicate' ,'both');
%imshow(B);
%creating template
temp=imread('crop_img.PNG');
temp= rgb2gray(temp);
%imshow(temp)
size(temp)
output = zeros (size(I2));
K = size(temp)
x=1;
y=1;
for i = 25 : r-24
for j = 25: c-24
%dif = temp -I2(i:i+47,j:j+47) ;
K = imabsdiff(temp,B(i-24:i+24,j-24:j+24));
output(i-24, j-24) = sum (K(:));
end
end
%gettting min value in output
min_value = output(1,1)
for i=1 : r
for j=1 :c
if(output(i,j)<min_value)
min_value=output(i,j);
row=i;
col=j;
end
end
end
row
col
output(1,465)
output(6,200)
%draw rectangle on matching area
%Create the shape inserter object.
shapeInserter = vision.ShapeInserter;
%Define the rectangle dimensions as [x y width height].
rectangle = int32([row col 24 24]);
%Draw the rectangle and display the result.
J = step(shapeInserter, I2, rectangle);
imshow(J);
There is an issue with image coordinates vs. matrix coordinates - you need to flip your row and column variables. Also, with shapeInserter, the coordinates are for the corner of the rectangle, so to make something centred on your output, you'd need something along the lines of:
rectangle = uint8([col-12 row-12 24 24]);
In MATLAB, it is usually not required to loop over every pixel of an image - it's much more efficient to work on the whole image at once. For example, you don't need a loop here:
min_value = output(1,1)
for i=1 : r
for j=1 :c
if(output(i,j)<min_value)
min_value=output(i,j);
row=i;
col=j;
end
end
end
This can be replaced by:
min_value = min(output(:));
[row,col] = find(output==min_value,1,'first');
I'm using normxcorr2 to find the area that exactly match with my pattern and i also want to find the other area(in the red rectangle) that is look like the pattern. I think it will be works if i can find the next maximum and so on and that value must not in the first maximum area or the first one that it has been detected but i can't do it. Or if you have any idea that using normxcorr2 to find the others area please advise me, I don't have any idea at all.
Here's my code. I modified from this one http://www.mathworks.com/products/demos/image/cross_correlation/imreg.html
onion = imread('pattern103.jpg'); %pattern image
peppers = imread('rsz_1jib-159.jpg'); %Original image
onion = rgb2gray(onion);
peppers = rgb2gray(peppers);
%imshow(onion)
%figure, imshow(peppers)
c = normxcorr2(onion,peppers);
figure, surf(c), shading flat
% offset found by correlation
[max_c, imax] = max(abs(c(:)));
[ypeak, xpeak] = ind2sub(size(c),imax(1));
corr_offset = [(xpeak-size(onion,2))
(size(onion,1)-ypeak)]; %size of window show of max value
offset = corr_offset;
xoffset = offset(1);
yoffset = offset(2);
xbegin = round(xoffset+1); fprintf(['xbegin = ',num2str(xbegin)]);fprintf('\n');
xend = round(xoffset+ size(onion,2));fprintf(['xend = ',num2str(xbegin)]);fprintf('\n');
ybegin = round(yoffset+1);fprintf(['ybegin = ',num2str(ybegin)]);fprintf('\n');
yend = round(yoffset+size(onion,1));fprintf(['yend = ',num2str(yend)]);fprintf('\n');
% extract region from peppers and compare to onion
extracted_onion = peppers(ybegin:yend,xbegin:xend,:);
if isequal(onion,extracted_onion)
disp('pattern103.jpg was extracted from rsz_org103.jpg')
end
recovered_onion = uint8(zeros(size(peppers)));
recovered_onion(ybegin:yend,xbegin:xend,:) = onion;
figure, imshow(recovered_onion)
[m,n,p] = size(peppers);
mask = ones(m,n);
i = find(recovered_onion(:,:,1)==0);
mask(i) = .2; % try experimenting with different levels of
% transparency
% overlay images with transparency
figure, imshow(peppers(:,:,1)) % show only red plane of peppers
hold on
h = imshow(recovered_onion); % overlay recovered_onion
set(h,'AlphaData',mask)