hi ..i am writing a code for quadrant dynamic histogram equalization...but i am getting an error while mapping back to the original image as 'Attempted to access map(0); index must be a positive integer or logical.'..i have attached the code

clc;
close all;
clear all;
WAAD_img= imread('C:\Users\Public\Pictures\Sample Pictures\image14.jpg');
WAAD_img= rgb2gray(WAAD_img);
% WAAD_hist=Fn_get_hist(WAAD_img,R,C);
WAAD_hist=imhist(uint8(WAAD_img))';
m = zeros(1,5);
[R C] = size(WAAD_img);

for i = 1 : 256
    if WAAD_hist(i)>R*C/100000;     m(1) = i; break;
    end
end
for i = 256 :-1: 1
    if WAAD_hist(i)>R*C/100000;     m(5) = i; break;
    end
end

cdf_data = cumsum(WAAD_hist)/(R*C);
[val idx] = find(cdf_data >= 0.25 &cdf_data < 0.50);
if size(idx) ~= 0
    m(2) = idx(1);
    m(3) = idx(end);
else
    [val idx] = find(cdf_data > 0);
    m(2) = m(1);
    m(3) = m(1);
end
[val idx] = find(cdf_data >= 0.75);
m(4) = idx(1)-1;

% cliping 
avgFreqOfimage = floor(R*C/256);
chist_data = zeros(1,256);
for i = 1 :256
    if WAAD_hist(i) >= avgFreqOfimage
        chist_data(i) = avgFreqOfimage; 
    else 
    chist_data(i) = WAAD_hist(i);
    end
end

span= zeros(1,4);
for i = 1 : 4
    span(i) = m(i+1)-m(i);
end

range = round(256*span/sum(span));
sepPosition = cumsum([0 range]);
new_cdf= zeros(1,256);
map = zeros(1,256);

for i = 1 : 4
     if i==1      
         temp = cumsum(chist_data(m(i):m(i+1)));
        new_cdf(m(i):m(i+1)) = temp/temp(end);
        map(m(i):m(i+1)) = round(range(i)*new_cdf(m(i):m(i+1))); 
     else
         temp = cumsum(chist_data(m(i):m(i+1)));
        new_cdf(m(i):m(i+1)) = temp/temp(end);
        map(m(i):m(i+1)) = round(range(i)*new_cdf(m(i):m(i+1))+sepPosition(i));        
    end   
end

QDHE_img = zeros(R,C);
for i = 1: R
    for j = 1: C
        QDHE_img(i,j) = map(WAAD_img(i,j));
    end
end
QDHE_img = uint8(QDHE_img);

        figure, imshow(QDHE_img)

[Lalitha Susmitha Kothamasu]

I am not able to understand the approach you used in the code can u please help me

Answer 1

[Walter Roberson]

QDHE_img(i,j) = map(1 + int16(WAAD_img(i,j)));

[vishwas h s]

thanks Mr.walter for the answer.....

[DHIVYA BHARKAVI]

Sir, please help to project in this QDHE version of matlab 2013a

Answer 2

[DHIVYA BHARKAVI]

Sir, Please help to my project your Quandrant Dynamic histogram Equalization in Version Matlab 2013a

[DGM]

The attached code (and the identical code in the other redundant questions-as-answers that have been deleted) is basically the same as OP's code with Walter's fix included. It runs on R2009b and R2015b just fine so long as the input image depth is appropriate, and I see no reason it shouldn't be the same in R2013a.

Was the code throwing an error because the input to rgb2gray() was already single channel? Probably. Certainly, cameraman.tif would be. The call to rgb2gray() should only be done if the image is RGB.

% any practical use of rgb2gray() requires extra code
if size(WAAD_img,3) == 3
    WAAD_img = rgb2gray(WAAD_img);
elseif size(WAAD_img,3) ~= 1
    error('this image is neither RGB nor I')
end

Was the code throwing an error because IPT wasn't installed? Maybe. rgb2gray() and imshow() would still be in IPT at the time.

Was it something else? This is why you need to actually describe the problem. "Please help" isn't effective at troubleshooting in a timely manner.