I am having a problem determining the affine transformation matrix of below image.
Original Image:
Affine Transformed Image:
I determined 2 points on the images to solve affine transformation matrix, but the results I get does not convert original to desired.
The code is below, it first solves the matrix and use it to transform original to transformed version:
% Pixel values
p1_aff = [164; 470];
p1_nor = [1; 512];
p2_aff = [470; 164];
p2_nor = [512; 1];
%p3_aff = [131;68];
%p3_nor = [166;61];
%p4_aff = [328;90];
%p4_nor = [456;59];
%Transformation matrix
syms a11 a12 a21 a22;
A = [a11 a12; a21 a22];
%Solving matrix
[Sx, Sy, Sz, Sk] = solve(A*p1_nor==p1_aff, A*p2_nor==p2_aff);
a11_d = double(Sx);
a12_d = double(Sy);
a21_d = double(Sz);
a22_d = double(Sk);
lena = imread('lena512.png');
new_aff = uint8(zeros(size(lena)));
for i = 1:size(lena,1)
for j = 1:size(lena,2)
% Applying affine transformation
new_coord = [a11_d a12_d 0; a21_d a22_d 0; 0 0 1]*[i; j; 1];
% Nearest-Neighbor interpolation for placing new pixels
if(round(new_coord(1)) > 0 && round(new_coord(2)) > 0)
new_aff(round(new_coord(1)),round(new_coord(2))) = lena(i,j);
end
end
end
imwrite(new_aff, 'lenaAffine_new.png');
At the end of above code, I get this image:
Does anyone understand what is wrong here? I am going crazy.
Two corresponding points are not enough to determine affine transformation. You need at least 6 matches (If I'm not mistaken).
Use cpselect gui to select corresponding points:
>> [input_points, base_points] = cpselect(oim2, oim1, 'Wait', true);
Then you can transform using:
>> T = cp2tform( input_points, base_points, 'affine' );
>> aim2 = tformarray( oim2, T, makeresampler('cubic','fill'), [2 1], [2 1], size(oim1(:,:,1)'), [], 0 );
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'm trying to perform an image rotation without embedded Matlab's function.
But I'm still getting this error:
Error using .'
Transpose on ND array is not defined. Use PERMUTE instead.
Error in interp2 (line 130)
V = V.';
But I don't know why there is such a mistake and likewise I don't know how to customize the function interp2 or PERMUTE to make it functional (I have used help in Matlab).
Could you please help to customize the code?
Thanks in advance!
clc; clear all; close all;
input_image = imread('mri.png');
Z = double(input_image);
Size = size(Z);
[X,Y] = meshgrid(1:Size(2), 1:Size(1));
%Center of an image
c = Size(end:-1:1)/2;
%Angle of rotation
angle = 45;
t = angle*pi/180;
%Making the rotation
ct = cos(t);
st = sin(t);
Xi = c(1) + ct*(X - c(1)) - st*(Y - c(2));
Yi = c(2) + st*(X - c(1)) + ct*(Y - c(2));
%Interpolation
Zi = interp2(X, Y, Z, Xi, Yi);
figure()
subplot(121); imshow(I); title('Original image');
subplot(122); imshow(uint8(Zi)); title('Rotated image without embedded
function');
Z is a 3D matrix and interp2 only works for 2D matrices. So you have to do the interpolation for each colour separately, and recombine them:
%Interpolation
Zir = interp2(X, Y, Z(:,:,1), Xi, Yi);
Zig = interp2(X, Y, Z(:,:,2), Xi, Yi);
Zib = interp2(X, Y, Z(:,:,3), Xi, Yi);
Zi = cat(3, Zir, Zig, Zib);
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?
Code is below. I'm looping through an input image 1 pixel at a time and determining its RGB value. Afterwards i'm trying to find the average RGB value for the image overall. For some reason the averaging portion of my code isnt working though.
im = imread(filename);
[width, height, depth] = size(im);
count = 0;
r=0;
g=0;
b=0;
for x = 1 : width
for y = 1: height
r = r + im(x,y,1);
g = g + im(x,y,2);
b = b + im(x,y,3);
count = count + 1;
end
end
%find averages of each RGB value.
r2 = r/count;
g2 = g/count;
b2 = b/count;
Why not vectorizing and using mean?
mean( reshape( im, [], 3 ), 1 )
The following code would work as well;
pep = imread('peppers.png');
mean(mean(pep))
This will return a 1x1x3 vector which will be the mean values of R, G, and B respectively.
using Hough Transform, how can I detect and get coordinates of (x0,y0) and "a" and "b" of an ellipse in 2D space?
This is ellipse01.bmp:
I = imread('ellipse01.bmp');
[m n] = size(I);
c=0;
for i=1:m
for j=1:n
if I(i,j)==1
c=c+1;
p(c,1)=i;
p(c,2)=j;
end
end
end
Edges=transpose(p);
Size_Ellipse = size(Edges);
B = 1:ceil(Size_Ellipse(1)/2);
Acc = zeros(length(B),1);
a1=0;a2=0;b1=0;b2=0;
Ellipse_Minor=[];Ellipse_Major=[];Ellipse_X0 = [];Ellipse_Y0 = [];
Global_Threshold = ceil(Size_Ellipse(2)/6);%Used for Major Axis Comparison
Local_Threshold = ceil(Size_Ellipse(1)/25);%Used for Minor Axis Comparison
[Y,X]=find(Edges);
Limit=numel(Y);
Thresh = 150;
Para=[];
for Count_01 =1:(Limit-1)
for Count_02 =(Count_01+1):Limit
if ((Count_02>Limit) || (Count_01>Limit))
continue
end
a1=Y(Count_01);b1=X(Count_01);
a2=Y(Count_02);b2=X(Count_02);
Dist_01 = (sqrt((a1-a2)^2+(b1-b2)^2));
if (Dist_01 >Global_Threshold)
Center_X0 = (b1+b2)/2;Center_Y0 = (a1+a2)/2;
Major = Dist_01/2.0;Alpha = atan((a2-a1)/(b2-b1));
if(Alpha == 0)
for Count_03 = 1:Limit
if( (Count_03 ~= Count_01) || (Count_03 ~= Count_02))
a3=Y(Count_03);b3=X(Count_03);
Dist_02 = (sqrt((a3 - Center_Y0)^2+(b3 - Center_X0)^2));
if(Dist_02 > Local_Threshold)
Cos_Tau = ((Major)^2 + (Dist_02)^2 - (a3-a2)^2 - (b3-b2)^2)/(2*Major*Dist_02);
Sin_Tau = 1 - (Cos_Tau)^2;
Minor_Temp = ((Major*Dist_02*Sin_Tau)^2)/(Major^2 - ((Dist_02*Cos_Tau)^2));
if((Minor_Temp>1) && (Minor_Temp<B(end)))
Acc(round(Minor_Temp)) = Acc(round(Minor_Temp))+1;
end
end
end
end
end
Minor = find(Acc == max(Acc(:)));
if(Acc(Minor)>Thresh)
Ellipse_Minor(end+1)=Minor(1);Ellipse_Major(end+1)=Major;
Ellipse_X0(end+1) = Center_X0;Ellipse_Y0(end+1) = Center_Y0;
for Count = 1:numel(X)
Para_X = ((X(Count)-Ellipse_X0(end))^2)/(Ellipse_Major(end)^2);
Para_Y = ((Y(Count)-Ellipse_Y0(end))^2)/(Ellipse_Minor(end)^2);
if (((Para_X + Para_Y)>=-2)&&((Para_X + Para_Y)<=2))
Edges(X(Count),Y(Count))=0;
end
end
end
Acc = zeros(size(Acc));
end
end
end
Although this is an old question, perhaps what I found can help someone.
The main problem of using the normal Hough Transform to detect ellipses is the dimension of the accumulator, since we would need to vote for 5 variables (the equation is explained here):
There is a very nice algorithm where the accumulator can be a simple 1D array, for example, and that runs in . If you wanna see code, you can look at here (the image used to test was that posted above).
If you use circle for rough transform is given as rho = xcos(theta) + ysin(theta)
For ellipse since it is
You could transform the equation as
rho = axcos(theta) + bysin(theta)
Although I am not sure if you use standard Hough Transform, for ellipse-like transforms, you could manipulate the first given function.
If your ellipse is as provided, being a true ellipse and not a noisy sample of points;
the search for the two furthest points gives the ends of the major axis,
the search for the two nearest points gives the ends of the minor axis,
the intersection of these lines (you can check it's a right angle) occurs at the centre.
If you know the 'a' and 'b' of an ellipse then you can rescale the image by factor of a/b in one direction and look for circle. I am still thinking about what to do when a and b are unknown.
If you know that it is circle then use Hough transform for circles. Here is a sample code:
int accomulatorResolution = 1; // for each pixel
int minDistBetweenCircles = 10; // In pixels
int cannyThresh = 20;
int accomulatorThresh = 5*_accT+1;
int minCircleRadius = 0;
int maxCircleRadius = _maxR*10;
cvClearMemStorage(storage);
circles = cvHoughCircles( gryImage, storage,
CV_HOUGH_GRADIENT, accomulatorResolution,
minDistBetweenCircles,
cannyThresh , accomulatorThresh,
minCircleRadius,maxCircleRadius );
// Draw circles
for (int i = 0; i < circles->total; i++){
float* p = (float*)cvGetSeqElem(circles,i);
// Draw center
cvCircle(dstImage, cvPoint(cvRound(p[0]),cvRound(p[1])),
1, CV_RGB(0,255,0), -1, 8, 0 );
// Draw circle
cvCircle(dstImage, cvPoint(cvRound(p[0]),cvRound(p[1])),
cvRound(p[2]),CV_RGB(255,0,0), 1, 8, 0 );
}