MATLAB: Finding maximum abs value of four images

finding maximum abs value of four images

I have four 16×16 images.I want to form another image by taking the maximum abs value of all four values of a particular pixel e.g. by comparing I1(3,3),I2(3,3),I3(3,3) and I4(3,3),I will be getting I5(3,3).

my piece of code is:

for k:4
.......
figure,imshow(B);
    s=size(B)
   eval(sprintf('im_%d=B;',k));
end

Best Answer

No loops necessary:

CombinedImages = cat(3, I1, I2, I3, I4)
I5 = max(abs(CombinedImages),[],3)

Related Solutions

MATLAB: Hello… the problem is i have 4 images…i need to subtract an image 2 with image 1, image 3 with image 1 and image 4 with image 1…so how can i do this process in one loop

image_name = {'first' 'second' 'third' 'fourth'};  % give names with extension
I1 = imread(image_name{1}) ;  % read first image 
for i = 2:4   % loop for rest of images 
    I2 = imread(image_name{i}) ; % read i'th image 
    I12 = I2-I1 ;               % substract
    % do what you want 
end

MATLAB: Unable to find explicit solution to differential eq

Hi Ditlev,

I believe this does have an analytic solution that is not overly complicated, of the form

x = (a+be^(-ct)) / (f+ge^(-ct))

First of all the scaling of this problem is not too good so I went to the variables (I used y in place of your x2)

u = km*t         x = (kp/km) x_old        y = (kp/km) y_old

in which case the equations are

x' = 2y - 2x^2
y' = -y + x^2

where differentiation is d/du. The starting value for the new x is

(kp/km)*1e-9 = 740.7

The most complicated the algebra gets is the roots of a quadratic eqn. The code below plots both a numerical and analytic solution.

Eventually the solution gets to the fixed point

x = r1 = 18.9966      y = -r1/2 + C = 360.8721

at which point x^2 = y and both derivatives are zero.

kp = 2*10^7;
km = 0.27*10^-4;
x0 = (kp/km)*1e-9   % starting point for x
y0 = 0;
% numerical version
[unum xynum] = ode45(@(u,xy) fun(u,xy), [0 .01], [x0 y0]);         % xy = [x; y]
%analytic version
C = x0/2;
r = roots([1,1/2,-C]);
r1 = max(r);
r2 = min(r);
alpha = 2*(r1-r2);
D = (x0-r1)/(x0-r2);
u = 0:1e-4:.01;
x = (r1-D*r2*exp(-alpha*u))./(1-D*exp(-alpha*u));   % solution
y = -x/2 + C;
figure(1)
plot(unum,xynum,u,x,'o',u,y,'o')
grid on
function dxy = fun(u,xy)
  dxy = zeros(2,1);
  dxy(1) = 2*xy(2)-2*xy(1)^2;
  dxy(2) = -xy(2) + xy(1)^2;
end

Best Answer

Related Solutions

MATLAB: Hello… the problem is i have 4 images…i need to subtract an image 2 with image 1, image 3 with image 1 and image 4 with image 1…so how can i do this process in one loop

MATLAB: Unable to find explicit solution to differential eq

Related Question