Code Block of Operates in Image Processing
2 ビュー (過去 30 日間)
古いコメントを表示

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);
0 件のコメント
採用された回答
Image Analyst
2020 年 11 月 17 日
See the FAQ: How to process a sequence of files
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 件のコメント
Image Analyst
2020 年 11 月 19 日
Sorry, I don't know much about that topic and don't understand what or why you think something's wrong with the code.
その他の回答 (0 件)
参考
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!