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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
2 Commenti
Walter Roberson
il 16 Dic 2013
How does the output you get differ from your expectation? Are you receiving an error message? What does pout.tif look like?
Risposte (2)
David Sanchez
il 17 Dic 2013
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.
3 Commenti
Image Analyst
il 17 Dic 2013
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.
Muhammad Ali Qadar
il 17 Dic 2013
Modificato: Muhammad Ali Qadar
il 17 Dic 2013
Image Analyst
il 17 Dic 2013
Modificato: Image Analyst
il 17 Dic 2013
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;
3 Commenti
Image Analyst
il 17 Dic 2013
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;
Muhammad Ali Qadar
il 18 Dic 2013
Modificato: Muhammad Ali Qadar
il 18 Dic 2013
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