Code Block of Operates in Image Processing

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Rooter Boy
Rooter Boy el 17 de Nov. de 2020
Comentada: Image Analyst el 19 de Nov. de 2020
I tried this:
%Take negative
%positiveImage = imread('cameraman.tif');
%negativeImage = 255 - positiveImage;
%imshow(negativeImage)
%or
1) a=imread('cameraman.tif');
d(:, :, 3) =255 - a(: ,:, 3);
d(: ,:, 2) = 255 - a(:, :, 2);
d(:, :, 1) =255 -a(:, :, 1);
imshow([a,d])
%log transformation with c=10
clc; clear all; close all;
f=imread('cameraman.tif')
g=rgb2gray(f);
c=input('Enter the constant value, c= 10');
[M,N]= size(g);
for x=1:M
for y=1:N
m= double(g(x,y));
z(x,y)= c. *log10(1+m);
end
end
imshow(f), figure, imshow(z);
%power-law with c=10, gamma=0.5
clc; clear all; close all;
RGB=imread(' cameraman.tif');
I=rgb2gray(RGB);
I=im2double(I);
[m,n]= size(I);
c=10;
g=[0.5];
for r= 1:length(g)
for p=1:m
for q=1:n
I3=(p,q) =c *I(p,q). ^ g(r);
end
end
figure, imshow(I3); title('Power law transformation'); xlabel('Gamma='), ylabel(g(r));
end
%contrast stretching
I=imread('cameraman.tif');
[m,n]=size(I);
figure, imshow(I);
minp=min(min(I));
maxp=max(max(I));
maxp=double(maxp);
minp=double(minp);
c=10;
d=255;
for i=1:m
for j=1:n
sonuc=(((d-c)/(maxp-minp))*(double(I(i,j))-minp))+c;
B(i,j)=round(sonuc);
end
end
B=uint8(B);
figure, imshow(B);

Respuesta aceptada

Image Analyst
Image Analyst el 17 de Nov. de 2020
In the middle of the loop, call a function that processes one image only and returns an output image. Then call sprintf() to create an output name, then call imwrite() to save the output image to that output file name.
  9 comentarios
Rooter Boy
Rooter Boy el 18 de Nov. de 2020
Editada: Rooter Boy el 19 de Nov. de 2020
Sir, could you help me when you are available? you are professional at matrix operates in matlab.
Image Analyst
Image Analyst el 19 de Nov. de 2020
Sorry, I don't know much about that topic and don't understand what or why you think something's wrong with the code.

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