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    發布日期: 2021-03-27
  • 語言: Matlab
  • 標簽: matlab??

資源簡介


這是效率很高的retinex方法 用MATLAB代碼實現 分享給大家

資源截圖

代碼片段和文件信息

I?=?imread(‘111.bmp‘);
subplot(121);
imshow(I);


I=double(I);
f=I(::1);
ff=I(::2);
fff=I(::3);
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
k1=4;
k2=5;
r=161;
alf=1458;
nn=floor((r+1)/2);
for?i=1:r
????for?j=1:r
????????b(ij)?=exp(-((i-nn)^2+(j-nn)^2)/(k1*alf))/(k2*pi*alf*10000);????????%高斯函數1
???end
end

k1=8;??????????????????????????????????????????????????????????????????%高斯函數2
k2=8;
r=161;
alf=1458;
nn=floor((r+1)/2);
for?i=1:r
????for?j=1:r
????????bb(ij)?=exp(-((i-nn)^2+(j-nn)^2)/(k1*alf))/(k2*pi*alf*10000);?????
???end
end

k1=0.5;
k2=0.5;
r=161;
alf=1458;
nn=floor((r+1)/2);
for?i=1:r
????for?j=1:r
????????bbb(ij)?=exp(-((i-nn)^2+(j-nn)^2)/(k1*alf))/(k2*pi*alf*10000);????????%高斯函數23
end
%%%%%%%%%%%對R分量的處理%%%%%%%%%%%%%
Img?=?double(f);
[mn]=size(f);

aa=125;
for?i=1:m
????for?j=1:n?
????????C(ij)=log(1+aa*(Img(ij)/I(ij)));
????end
end

K=imfilter(Imgb);
KK=imfilter(Imgbb);
KKK=imfilter(Imgbbb);

for?i

?屬性????????????大小?????日期????時間???名稱
-----------?---------??----------?-----??----

?????文件???????2775??2009-03-08?18:33??retinex_frankle_mccann.m

?????文件???????2612??2009-09-06?20:55??MSRCR.m

?????文件???????3183??2009-09-07?10:07??MSR_new.m

?????文件???????2693??2010-04-23?21:00??retinex.m

?????文件???????5392??2009-03-08?18:33??retinex_mccann99.m

-----------?---------??----------?-----??----

????????????????16655????????????????????5


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