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How to get entropy values of different images to a graph?

3 vues (au cours des 30 derniers jours)
Ishara Nipuni
Ishara Nipuni le 25 Nov 2018
Commenté : KALYAN ACHARJYA le 25 Nov 2018
Hi,
I have calculated the entropy values of different images using the code I have shown below. Now I need to get the entropy values o several images to a graph. Can you please help me on that?
%read te image
I = imread('152.bmp');
subplot(2,2,1),imshow(I,map1)
I2 = imread('006.bmp');
subplot(2,2,3),imshow(I2,map2)
x = entropy(I);
x2 = entropy(I2);
[Height,Width] = size(I);
[m,Binsx] = imhist(I);
%normalize the counts
m = m/(Height*Width);
%get the sum of the histogram values
sprintf('the sum of the histogram value is = %g',sum(m));
subplot(2,2,2),plot(Binsx,m,'k')
xlabel('pixel value'),ylabel('relative count')
H = sum(-m.*log2(m));
sprintf('the entropy of the image is = %g',H)
%calculate height & width of the image
[Height,Width] = size(I2);
%get the histogram of the image in 256 bins
[n,Binsx] = imhist(I2);
%normalize the counts
n = n/(Height*Width);
%get the sum of the histogram values
sprintf('the sum of the histogram value is = %g',sum(n));
%
subplot(2,2,4),plot(Binsx,n,'k')
xlabel('pixel value'),ylabel('relative count')
H2 = sum(-n.*log2(n));
sprintf('the entropy of the image is = %g',H2)
%get the graph of the entropy values
figure(),plot(H,H2);
xlabel('image number'),ylabel('entropy value')
  1 commentaire
Image Analyst
Image Analyst le 25 Nov 2018
Do you want both entropy values on one graph? Why not use bar() instead of plot()?

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KALYAN ACHARJYA
KALYAN ACHARJYA le 25 Nov 2018
I=imread('test.png');
I2=imread('test.jpg');
x=entropy(I);
x2=entropy(I2);
bar([x x2]);
set(gca,'xticklabel');
title('Entropy of Two Images');
rrr.png
Another Way:
I=imread('test.png');
I2=imread('test.jpg');
x=entropy(I);
x2=entropy(I2);
bar(1,x);
hold on;
bar(2, x2,'r');
bar_name={'Image 1' 'Image 2'};
legend(bar_name,'location','northwest');
title('Entropy of Two Images');
yy.png
  2 commentaires
Ishara Nipuni
Ishara Nipuni le 25 Nov 2018
Thank you so much for the help. It will be very helpfull for me. :-) :-)
KALYAN ACHARJYA
KALYAN ACHARJYA le 25 Nov 2018
Welcome @Ishara
I'm glad, it helped.
Good Wishes
Regards,
Kalyan Acharjya

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