I am new to image processing, I have implemented a code for image warping and it works perfectly. I would like to improve the code by using linear interpolation to rotate the image WITHOUT using the built-in function (interp). Here is my code:
close all;
clear all;
img = 'woods.jpg';
input_image =double(imread(img))./255;
H=size(input_image,1);
W=size(input_image,2);
th=pi/4;
s0 = 2;
s1 = 2;
x0 = -W/2;
x1 = -H/2;
T=[1 0 x0 ; ...
0 1 x1 ; ...
0 0 1];
RST = [ (s0*cos(th)) (-s1*sin(th)) ((s0*x0*cos(th))-(s1*x1*sin(th))); ...
(s0*sin(th)) (s1*cos(th)) ((s0*x0*sin(th))+(s1*x1*cos(th))); ...
0 0 1];
M=inv(T)*R;
N = inv(M);
output_image=zeros(H,W,3);
for i=1:W
for j=1:H
x = [i ; j ; 1];
y = N * x;
a = y(1)/y(3);
b = y(2)/y(3);
a = round(a);
b = round(b);
if (a>0 && a<=W && b>0 && b<=H)
output_image(j,i,:)=input_image(b,a,:);
end
end
end
imgshow(output_image);
Check the following solution:
I verified implementation by comparing to Matalb build in function [imwarp][1].
close all;
clear all;
img = 'peppers.png';
input_image =double(imread(img))./255;
H=size(input_image,1); % height
W=size(input_image,2); % width
th=pi/4;
s0 = 2;
s1 = 2;
x0 = -W/2;
x1 = -H/2;
T=[1 0 x0 ; ...
0 1 x1 ; ...
0 0 1];
RST = [ (s0*cos(th)) (-s1*sin(th)) ((s0*x0*cos(th))-(s1*x1*sin(th))); ...
(s0*sin(th)) (s1*cos(th)) ((s0*x0*sin(th))+(s1*x1*cos(th))); ...
0 0 1];
M=inv(T)*RST;
N = inv(M);
output_image=zeros(H,W,3);
for i=1:W
for j=1:H
x = [i ; j ; 1];
y = N * x;
a = y(1)/y(3);
b = y(2)/y(3);
%Nearest neighbor
%a = round(a);
%b = round(b);
x1 = floor(a);
y1 = floor(b);
x2 = x1 + 1;
y2 = y1 + 1;
%Bi-linear interpolation ilsutration:
%Image coordinates style (horizontal index first)
%
%(x1,y1) | (x2,y1)
% | 1-dy
% 1-dx | dx
% ------(a,b)------------
% |
% |
% |
% | dy
% |
% |
%(x1,y2) | (x2,y2)
if ((x1 >= 1) && (y1 >= 1) && (x2 <= W) && (y2 <= H))
%Load 2x2 pixels
i11 = input_image(y1, x1, :); %Top left pixel
i21 = input_image(y2, x1, :); %Bottom left pixel
i12 = input_image(y1, x2, :); %Top right pixel
i22 = input_image(y2, x2, :); %Bottom right pixel
%Interpolation wieghts
dx = x2 - a;
dy = y2 - b;
%Bi-lienar interpolation
output_image(j, i, :) = i11*dx*dy + i21*dx*(1-dy) + i12*(1-dx)*dy + i22*(1-dx)*(1-dy);
end
end
end
imshow(output_image);
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%Verify implementation by comparing with Matalb build in function imwarp:
tform = affine2d(M');
ref_image = imwarp(input_image, tform, 'OutputView', imref2d(size(input_image)), 'Interp', 'linear');
figure;imshow(ref_image)
figure;imshow(output_image - ref_image)
max_diff = max(abs(output_image(:) - ref_image(:)));
disp(['Maximum difference from imwarp = ', num2str(max_diff)]);
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
Result:
Remark:
I missed the following execelnt post that resizes an image using bilinear interpolation (bilinear interpolation is explained better):
Resize an image with bilinear interpolation without imresize
Related
I am getting an error when I run this code for disc waves. The code is attached.
The Error is in line 137 and 292. Please help in resolving issue.
function waves
% WAVES Wave equation in one and two space dimensions.
% The two-dimensional domains include a pi-by-pi square, a unit disc,
% a three-quarter circular sector and the L-shaped union of three squares.
% The eigenfunctions of the square are sin(m*x)*sin(n*y). With polar
% coordinates, the eigenfunctions of the disc and the sector involve Bessel
% functions. The eigenfunctions of the L-shaped domain also involve
% Bessel functions and are computed by the MATLAB function membranetx.m.
% 2-D eigenvalues and eigenfunctions
m = 11; % Determines number of grid points
speed = 1;
bvals = [1; 0; 0; 0; 0];
t = 0;
while bvals(5) == 0
% Initialize figure
shg
clf reset
set(gcf,'doublebuffer','on','menubar','none','tag','', ...
'numbertitle','off','name','Waves','colormap',hot(64));
for k= 1:5
b(k) = uicontrol('style','toggle','value',bvals(k), ...
'units','normal','position',[.15*k .01 .14 .05]);
end
set(b(1),'style','pop','string', ...
{'1-d','square','disc','sector'})
set(b(2),'string','modes/wave')
set(b(3),'string','slower')
set(b(4),'string','faster')
set(b(5),'string','close')
if bvals(3)==1
speed = speed/sqrt(2);
set(b(3),'value',0);
end
if bvals(4)==1
speed = speed*sqrt(2);
set(b(4),'value',0);
end
bvals = cell2mat(get(b,'value'));
region = bvals(1);
modes = bvals(2)==0;
if region == 1
% 1-D
x = (0:4*m)/(4*m)*pi;
orange = [1 1/3 0];
gray = get(gcf,'color');
if modes
% 1-D modes
for k = 1:4
subplot(2,2,k)
h(k) = plot(x,zeros(size(x)));
axis([0 pi -3/2 3/2])
set(h(k),'color',orange,'linewidth',3)
set(gca,'color',gray','xtick',[],'ytick',[])
end
delta = 0.005*speed;
bvs = bvals;
while all(bvs == bvals)
t = t + delta;
for k = 1:4
u = sin(k*t)*sin(k*x);
set(h(k),'ydata',u)
end
drawnow
bvs = cell2mat(get(b,'value'));
end
else
% 1-D wave
h = plot(x,zeros(size(x)));
axis([0 pi -9/4 9/4])
set(h,'color',orange,'linewidth',3)
set(gca,'color',gray','xtick',[],'ytick',[])
delta = 0.005*speed;
a = 1./(1:4);
bvs = bvals;
while all(bvs == bvals)
t = t + delta;
u = zeros(size(x));
for k = 1:4
u = u + a(k)*sin(k*t)*sin(k*x);
end
set(h,'ydata',u)
drawnow
bvs = cell2mat(get(b,'value'));
end
end
elseif region <= 5
switch region
case 2
% Square
x = (0:2*m)/(2*m)*pi;
y = x';
lambda = zeros(4,1);
V = cell(4,1);
k = 0;
for i = 1:2
for j = 1:2
k = k+1;
lambda(k) = i^2 + j^2;
V{k} = sin(i*y)*sin(j*x);
end
end
ax = [0 pi 0 pi -1.75 1.75];
case 3
% Disc, mu = zeros of J_0(r) and J_1(r)
mu = [bjzeros(0,2) bjzeros(1,2)];
[r,theta] = meshgrid((0:m)/m,(-m:m)/m*pi);
x = r.*cos(theta);
y = r.*sin(theta);
V = cell(4,1);
k = 0;
for j = 0:1
for i = 1:2
k = k+1;
if j == 0
V{k} = besselj(0,mu(k)*r);
else
V{k} = besselj(j,mu(k)*r).*sin(j*theta);
end
V{k} = V{k}/max(max(abs(V{k})));
end
end
lambda = mu.^2;
ax = [-1 1 -1 1 -1.75 1.75];
case 4
% Circular sector , mu = zeros of J_(2/3)(r) and J_(4/3)(r)
mu = [bjzeros(2/3,2) bjzeros(4/3,2)];
[r,theta] = meshgrid((0:m)/m,(3/4)*(0:2*m)/m*pi);
x = r.*cos(theta+pi);
y = r.*sin(theta+pi);
V = cell(4,1);
k = 0;
for j = 1:2
for i = 1:2
k = k+1;
alpha = 2*j/3;
V{k} = besselj(alpha,mu(k)*r).*sin(alpha*theta);
V{k} = V{k}/max(max(abs(V{k})));
end
end
lambda = mu.^2;
ax = [-1 1 -1 1 -1.75 1.75];
case 5\
% L-membrane
x = (-m:m)/m;
y = x';
lambda = zeros(4,1);
V = cell(4,1);
for k = 1:4
[L lambda(k)] = membranetx(k,m,9,9);
L(m+2:2*m+1,m+2:2*m+1) = NaN;
V{k} = rot90(L,-1);
end
ax = [-1 1 -1 1 -1.75 1.75];
end
if modes
% 2-D modes
p = [.02 .52 .02 .52];
q = [.52 .52 .02 .02];
for k = 1:4
axes('position',[p(k) q(k) .46 .46]);
h(k) = surf(x,y,zeros(size(V{k})));
axis(ax)
axis off
view(225,30);
caxis([-1.5 1]);
end
delta = .08*speed;
mu = sqrt(lambda(:));
bvs = bvals;
while all(bvs == bvals)
t = t + delta;
for k = 1:4
U = 1.5*sin(mu(k)*t)*V{k};
set(h(k),'zdata',U)
set(h(k),'cdata',U)
end
drawnow
bvs = cell2mat(get(b,'value'));
end
else
% 2-D wave
h = surf(x,y,zeros(size(V{1})));
axis(ax);
axis off
view(225,30);
caxis([-1.5 1]);
delta = .02*speed;
mu = sqrt(lambda(:));
a = 1.25./(1:4);
bvs = bvals;
while all(bvs == bvals)
t = t + delta;
U = zeros(size(V{1}));
for k = 1:4
U = U + a(k)*sin(mu(k)*t)*V{k};
end
set(h,'zdata',U)
set(h,'cdata',U)
drawnow
bvs = cell2mat(get(b,'value'));
end
end
elseif region == 6
figure
bizcard
set(b(1),'value',1)
end
% Retain uicontrol values
bvals = cell2mat(get(b,'value'));
end
close
% -------------------------------
function z = bjzeros(n,k)
% BJZEROS Zeros of the Bessel function.
% z = bjzeros(n,k) is the first k zeros of besselj(n,x)
% delta must be chosen so that the linear search can take
% steps as large as possible without skipping any zeros.
% delta is approx bjzero(0,2)-bjzero(0,1)
delta = .99*pi;
Jsubn = inline('besselj(n,x)''x','n');
a = n+1;
fa = besselj(n,a);
z = zeros(1,k);
j = 0;
while j < k
b = a + delta;
fb = besselj(n,b);
if sign(fb) ~= sign(fa)
j = j+1;
z(j) = fzerotx(Jsubn,[a b],n);
end
a = b;
fa = fb;
end
I have a data cube a of radius w and for every element of that cube, I would like to add the element and all surrounding values within a cube of radius r, where r < w. The result should be returned in an array of the same shape, b.
As a simple example, suppose:
a = numpy.ones(shape=(2*w,2*w,2*w),dtype='float32')
kernel = numpy.ones(shape=(2*r,2*r,2*r),dtype='float32')
b = convolve(a,kernel,mode='constant',cval=0)
then b would have the value (2r)(2r)(2r) for all the indices not on the edge.
Currently I am using a loop to do this and it is very slow, especially for larger w and r. I tried scipy convolution but got little speedup over the loop. I am now looking at numba's parallel computation feature but cannot figure out how to rewrite the code to work with numba. I have a Nvidia RTX card so CUDA GPU calculations are also possible.
Suggestions are welcome.
Here is my current code:
for x in range(0,w*2):
print(x)
for y in range(0,w*2):
for z in range(0,w*2):
if x >= r:
x1 = x - r
else:
x1 = 0
if x < w*2-r:
x2 = x + r
else:
x2 = w*2 - 1
if y >= r:
y1 = y - r
else:
y1 = 0
if y < w*2-r:
y2 = y + r
else:
y2 = w*2 - 1
if z >= r:
z1 = z - r
else:
z1 = 0
if z < w*2-r:
z2 = z + r
else:
z2 = w*2 - 1
b[x][y][z] = numpy.sum(a[x1:x2,y1:y2,z1:z2])
return b
Here is a very simple version of your code so that it works with numba. I was finding speed-ups of a factor of 10 relative to the pure numpy code. However, you should be able to get even greater speed-ups using a FFT convolution algorithm (e.g. scipy's fftconvolve). Can you share your attempt at getting convolution to work?
from numba import njit
#njit
def sum_cubes(a,b,w,r):
for x in range(0,w*2):
#print(x)
for y in range(0,w*2):
for z in range(0,w*2):
if x >= r:
x1 = x - r
else:
x1 = 0
if x < w*2-r:
x2 = x + r
else:
x2 = w*2 - 1
if y >= r:
y1 = y - r
else:
y1 = 0
if y < w*2-r:
y2 = y + r
else:
y2 = w*2 - 1
if z >= r:
z1 = z - r
else:
z1 = 0
if z < w*2-r:
z2 = z + r
else:
z2 = w*2 - 1
b[x,y,z] = np.sum(a[x1:x2,y1:y2,z1:z2])
return b
EDIT: Your original code has a small bug in it. The way numpy indexing works, the final line should be
b[x,y,z] = np.sum(a[x1:x2+1,y1:y2+1,z1:z2+1])
unless you want the cube to be off-centre.
Assuming you do want the cube to be centred, then a much faster way to do this calculation is using scipy's uniform filter:
from scipy.ndimage import uniform_filter
def sum_cubes_quickly(a,b,w,r):
b = uniform_filter(a,mode='constant',cval=0,size=2*r+1)*(2*r+1)**3
return b
A few quick runtime comparisons for randomly generated data with w = 50, r = 10:
Original raw numpy code - 15.1 sec
Numba'd numpy code - 8.1 sec
uniform_filter - 13.1 ms
clear
I = imread('256.jpg');
%imshow(I);
center = 128;
[x, y] = size(I); % declare image size array
Original = [x, y];
Rotated_I = zeros(x,y); %declare size of array to store pixel
theta = 90;
for row = 1:y
for column = 1:x
x_original = (column - 128) * cos(theta) - (row - 128)*sin(theta);
y_original = (column - 128) * sin(theta) + (row - 128)*cos(theta); % reverse rotate
p = floor(x_original);
q = floor(y_original);
a = y_original - p;
b = x_original - q; %
Rotated_I(column, row) = (1-a)*((1-b)*Original(p,q)+b*Original(p,q+1))+a*((q-b)*Original(p+1,q)+b*Original(p+1,q+1)); % Find pixel using bilinear interpolation
end
end
imshow(Rotated_I);
I tried to rotate image using reverse rotate and bilinear interpolation, but only i see is error message. It says "the first index exceeds array". Is there anything wrong in my code?
Here is a working version with a number of changes. The main difference is that it checks whether a coordinate exists in the original image before adding that to the rotate image. This allows for arbitrary rotations, like 45 degrees. Also, images in MATLAB have y as the first dimension and x as the second, so are accessed as I(y, x) or I(row, column).
clear
I = imread('256.jpg');
% imshow(I);
center = 128;
[y, x] = size(I); % in MATLAB, images are y-by-x in size (ie. y is dimension 1)
Original = I; % Original needs to be the image I
Rotated_I = zeros(y, x);
theta = 90;
for row = 1:y
for column = 1:x
x_original = (column - center) * cosd(theta) - (row - center)*sind(theta) + center; % theta is in degrees so use cosd and sind
y_original = (column - center) * sind(theta) + (row - center)*cosd(theta) + center; % also add center back on
p = floor(y_original); % x_original and y_original were swapped here
q = floor(x_original); % x_original and y_original were swapped here
a = y_original - p;
b = x_original - q;
% check if the coordinate is in the original image to prevent errors
if p > 0 && p <= y && q > 0 && q <= x
Rotated_I(row, column) = Rotated_I(row, column) + (1-a)*(1-b)*Original(p,q);
end
if p > 0 && p <= y && q+1 > 0 && q+1 <= x
Rotated_I(row, column) = Rotated_I(row, column) + (1-a)*b*Original(p,q+1);
end
if p+1 > 0 && p+1 <= y && q > 0 && q <= x
Rotated_I(row, column) = Rotated_I(row, column) + a*(1-b)*Original(p+1,q);
end
if p+1 > 0 && p+1 <= y && q+1 > 0 && q+1 <= x
Rotated_I(row, column) = Rotated_I(row, column) + a*b*Original(p+1,q+1);
end
end
end
% convert to uint image so it displays properly (double expects values from 0 to 1)
imshow(uint8(Rotated_I));
I do not know if you necessarily want to have your own implementation or not. But if not, you could always use imrotate:
Rotated_I = imrotate(I, 90, 'bilinear', 'crop');
90 => Degrees of rotation
'bilinear' => Bilinear interpolation (alternatives: nearest, bicubic)
'crop' => Maintain the pixel size of the rotated image the same as the input image
imrotate is part of the Image Processing Toolbox.
Actually i have two intersecting circles as specified in the figure
i want to find the area of each part separately using Monte carlo method in Matlab .
The code doesn't draw the rectangle or the circles correctly so
i guess what is wrong is my calculation for the x and y and i am not much aware about the geometry equations for solving it so i need help about the equations.
this is my code so far :
n=1000;
%supposing that a rectangle will contain both circles so :
% the mid point of the distance between 2 circles will be (0,6)
% then by adding the radius of the left and right circles the total distance
% will be 27 , 11 from the left and 16 from the right
% width of rectangle = 24
x=27.*rand(n-1)-11;
y=24.*rand(n-1)+2;
count=0;
for i=1:n
if((x(i))^2+(y(i))^2<=25 && (x(i))^2+(y(i)-12)^2<=100)
count=count+1;
figure(2);
plot(x(i),y(i),'b+')
hold on
elseif(~(x(i))^2+(y(i))^2<=25 &&(x(i))^2+(y(i)-12)^2<=100)
figure(2);
plot(x(i),y(i),'y+')
hold on
else
figure(2);
plot(x(i),y(i),'r+')
end
end
Here are the errors I found:
x = 27*rand(n,1)-5
y = 24*rand(n,1)-12
The rectangle extents were incorrect, and if you use rand(n-1) will give you a (n-1) by (n-1) matrix.
and
first If:
(x(i))^2+(y(i))^2<=25 && (x(i)-12)^2+(y(i))^2<=100
the center of the large circle is at x=12 not y=12
Second If:
~(x(i))^2+(y(i))^2<=25 &&(x(i)-12)^2+(y(i))^2<=100
This code can be improved by using logical indexing.
For example, using R, you could do (Matlab code is left as an excercise):
n = 10000
x = 27*runif(n)-5
y = 24*runif(n)-12
plot(x,y)
r = (x^2 + y^2)<=25 & ((x-12)^2 + y^2)<=100
g = (x^2 + y^2)<=25
b = ((x-12)^2 + y^2)<=100
points(x[g],y[g],col="green")
points(x[b],y[b],col="blue")
points(x[r],y[r],col="red")
which gives:
Here is my generic solution for any two circles (without any hardcoded value):
function [ P ] = circles_intersection_area( k1, k2, N )
%CIRCLES_INTERSECTION_AREA Summary...
% Adnan A.
x1 = k1(1);
y1 = k1(2);
r1 = k1(3);
x2 = k2(1);
y2 = k2(2);
r2 = k2(3);
if sqrt((x1-x2)*(x1-x2) + (y1-y2)*(y1-y2)) >= (r1 + r2)
% no intersection
P = 0;
return
end
% Wrapper rectangle config
a_min = x1 - r1 - 2*r2;
a_max = x1 + r1 + 2*r2;
b_min = y1 - r1 - 2*r2;
b_max = y1 + r1 + 2*r2;
% Monte Carlo algorithm
n = 0;
for i = 1:N
rand_x = unifrnd(a_min, a_max);
rand_y = unifrnd(b_min, b_max);
if sqrt((rand_x - x1)^2 + (rand_y - y1)^2) < r1 && sqrt((rand_x - x2)^2 + (rand_y - y2)^2) < r2
% is a point in the both of circles
n = n + 1;
plot(rand_x,rand_y, 'go-');
hold on;
else
plot(rand_x,rand_y, 'ko-');
hold on;
end
end
P = (a_max - a_min) * (b_max - b_min) * n / N;
end
Call it like: circles_intersection_area([-0.4,0,1], [0.4,0,1], 10000) where the first param is the first circle (x,y,r) and the second param is the second circle.
Without using For loop.
n = 100000;
data = rand(2,n);
data = data*2*30 - 30;
x = data(1,:);
y = data(2,:);
plot(x,y,'ro');
inside5 = find(x.^2 + y.^2 <=25);
hold on
plot (x(inside5),y(inside5),'bo');
hold on
inside12 = find(x.^2 + (y-12).^2<=144);
plot (x(inside12),y(inside12),'g');
hold on
insidefinal1 = find(x.^2 + y.^2 <=25 & x.^2 + (y-12).^2>=144);
insidefinal2 = find(x.^2 + y.^2 >=25 & x.^2 + (y-12).^2<=144);
% plot(x(insidefinal1),y(insidefinal1),'bo');
hold on
% plot(x(insidefinal2),y(insidefinal2),'ro');
insidefinal3 = find(x.^2 + y.^2 <=25 & x.^2 + (y-12).^2<=144);
% plot(x(insidefinal3),y(insidefinal3),'ro');
area1=(60^2)*(length(insidefinal1)/n);
area3=(60^2)*(length(insidefinal2)/n);
area2= (60^2)*(length(insidefinal3)/n);
I'm trying to find the maximum width of the blob by counting the number of white pixels of each line in the blob I wrote the code, however, it never stops. how it can be fixed?
For y = 0 To bmp.ScaleHeight - 1
sum = 0
For x = 0 To bmp.ScaleWidth - 1
pixel1 = bmp.Point(x, y)
If pixel1 = vbWhite Then
sum = sum + 1
If bmp.Point(x + 1, y) = vbBlack Then
If sum > max Then
Lmax = sum
y1 = y
x2 = x
x1 = x2 - sum
End If
End If
End If
Next x
Next y