Code Block of Operates in Image Processing

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Rooter Boy
Rooter Boy am 17 Nov. 2020
Kommentiert: Image Analyst am 19 Nov. 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);

Akzeptierte Antwort

Image Analyst
Image Analyst am 17 Nov. 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 Kommentare
Rooter Boy
Rooter Boy am 18 Nov. 2020
Bearbeitet: Rooter Boy am 19 Nov. 2020
Sir, could you help me when you are available? you are professional at matrix operates in matlab.
Image Analyst
Image Analyst am 19 Nov. 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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