Whats wrong with coding
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hi , I am trying to implement Image enhancement algorithm Dynamic Quadrant Histogram Equalization Plateau limit
I am trying to implement its first part please have a look on following code whats wrong in this code
clear;
clc;
p=imread('pout.tif');p=p(:,:,1);
h=imhist(p);
[m,n]=size(p);
Pic=zeros(m,n);
N=m*n;
max1=double(max(p(:)));
min1=double(min(p(:)));
aa=p;
ch=cumsum(h);
m0=min1;
m1=floor(0.25*(N));
m2=floor(0.5*(N));
m3=floor(0.75*(N));
m4=max1;
m=[m0 m1 m2 m3 m4];
L=256;
n0=0;
n1=m2*((m1-m0)/(m2-m0));
n2=m2;
n3=((L-1-m2)*((m3-m2)/(m4-m2)))+m2;
n4=L-1;
for j=0:4
a=sum(h(m(j):m(j+1)));
P(j)=a./(m(j+1)-m(j));
P(j)=h(h>P(j)); % clipped Histogram
M(j)=sum(P(m(j):m(j+1))); % total of Clipped histogram
Y(aa==j)=n(j+1)+(n(j+1)-n(j))*(P(j)./M(j)); %Transform function
end
imshow(Y)
following is the link to paper
댓글 수: 2
Walter Roberson
2013년 12월 16일
How does the output you get differ from your expectation? Are you receiving an error message? What does pout.tif look like?
답변 (2개)
David Sanchez
2013년 12월 17일
In your code
for j=0:4
a=sum(h(m(j):m(j+1)));
...
...
the index starts at j=0, which is not matlab's way of handling an array. It should start in 1.
m(j) for j=0, will return an error since m(0) (the 0th element of the array) does not exist.
댓글 수: 3
Image Analyst
2013년 12월 17일
You will find out after you look at this. Basically it's saying that m is negative, zero, or have some fractional part and that it's not strictly integers like it should be. I don't know why - I didn't run the code - but you will after you look at the link I gave and step through your code line by line and examine variables.
Image Analyst
2013년 12월 17일
편집: Image Analyst
2013년 12월 17일
OK, that's a different error than you originally had. If h is the histogram, then h has only 256 bins (elements). So why are you trying to set m=[25,15500,31000,46500,255]???? There is no 15500 bins so why are you trying to access it?
Try this:
m0=min1;
m1=floor(0.25*(max1-min1)+min1);
m2=floor(0.5*(max1-min1)+min1);
m3=floor(0.75*(max1-min1)+min1);
m4=max1;
댓글 수: 3
Image Analyst
2013년 12월 17일
Try this:
m0=min1;
m1=floor(0.25*(max1-min1)+min1);
m2=floor(0.5*(max1-min1)+min1);
m3=floor(0.75*(max1-min1)+min1);
m4=max1;
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