How can i calculate entropy of an image by using the entropy's type?

Hello ,

I have a grayscale image 256x256 and i want to calculate its entropy . Basically i want to calculate entropy using this type E= .But i dont really know which value corresponds to which variable . For example i assumed that p=1/256 because this is the propability of a random number between 0...255 .But i am not sure.

Can someone help?

Answer 1

[Image Analyst]

There is an entropy function you know:

e = entropy(grayImage);

[Gn Gnk]

Thank you , i tried the entropy function(and it perfectly works) but i was looking for answer that follows the entropy type i wrote above

[Image Analyst]

The help defines it:

Entropy is defined as -sum(p.*log2(p)), where p contains the normalized histogram counts returned from imhist.

so tell me why that formula does not exactly match the formula you gave? Because I'm not seeing it. They look identical to me.

[Gn Gnk]

Ok i see . Thank you a lot.

Answer 2

[KALYAN ACHARJYA]

"But i dont really know which value corresponds to which variable"

E=

Let's suppose any gray scale Image of size 10x10

image =
   254    20    12   101   227   185   188    52   200    74
    85   198   170    15   204    28   248    22    25   154
    76   231   154   199   187    30   221   197    75   246
    15   136   134    86    13   164    22    52    60   110
    76    27   186   155    18    84    93    99   135   177
    11   211   181   189    22   167    94   141    23   194
   129    86   200    26   204   191   175    58   103   110
   194    75    73    32   241   149   153   164    26   167
   161   191   177   140   175   189   202   124    28    28
    23     2   142   124    33    60    94    38   200   239

1. The First Step to find the probaililities of all probable pixles (From 0 to 255),

Example: The Probaibility of 0 is (P0) = Total Numbers of Occurance of 0 pixel value in the image/size of the image

As there is no "0" pixel within the image, hence P0=0/10x10=0

Simmilarly, the Probaibility of 1 is (P1) = Total Numbers of Occurance of 1 pixel value in the image/size of the image

As there is no "1" pixel within the image, hence P1=0/10x10=0

Same way, the Probaibility of 2 is (P12) = Total Numbers of Occurance of 2 pixel value in the image/size of the image

As there is one "2" pixel within the image, hence P2=1/10x10= 0.01

.............

so on..same way,calculate the probabilities of all pixels values (P3,P4.....P255)

At the end, you may get an array as follows

p=[P0,P1,P2.........P255]
p=[0,0,0.01..........]

2. Second Step: Find the log2 of the p

logP=[log2(P0),log2(P1),.........log(P255)]

Please note here I have considering logP as a variable, In MATLAB, you have to use log(p), where p is 1D array.

In MATLAB, you can directly use log2(p), more about log2 MATLAB fucction here

3. Third Step: Multiple the Probailities with log2(p), its like that, considering log base 2

mul_result=[P0*log2(P0),P1*log2(P1),P2*log2(P2).........P255*log2(P255)]

In MATLAB

mul_result=p.*log2(p);

4. Fourth Step, Add all the result values of the Multiplication result in step 3, afterwards the entropy result as per formula is

Entropy_val=-sum(mul_result); 

So in the mentioned formula, k represents all pixel values from 0 to 288 (in the case of uint8 image), p represents the probabilities of individual pixels, you have to get the log (p) of same individual pixels, sum all the data at the end, and put the minus sign, which is a result.

You can implement this in MATLAB on your own, please practice. Or directly you can use MATLAB inbuilt function Entropy here

Good Luck

Kalyan :)

[Gn Gnk]

Thank you a lot for your detailed answer !!Isn't it too difficult and time consuming to do that for 255 pixels ? I mean it would take me ages to do this .

[KALYAN ACHARJYA]

"Isn't it too difficult and time consuming to do that for 255 pixels ? I mean it would take me ages to do this"

Of course not, you wouldn't do it manually, well, I've shown such a way to get understanding. As you have asked for ""But i dont really know which value corresponds to which variable"

Using MATLAB or any other tool, its hardly take fractions of second

p=imhist(image)/r*c;

The complete Entropy Function is (Own)

grayImage=rgb2gray(imread('Baboon.jpeg'));
[r,c]=size(grayImage);
prob=imhist(grayImage)/(r*c);
prob=prob(prob>0);
Entropy_own=-sum(prob.*log2(prob))

MATLAB Inbuilt function

Entropy_matlab=entropy(grayImage)

Results:

Entropy_own =
    7.2373
Entropy_matlab =
    7.2373

:)

[RENJI]

What does this negative value signify? Output obtained shows a positive number.

[Image Analyst]

@RENJI what negative value? The entropies are positive. Do you mean the negative sign before the sum? Well the probabilities are less than 1 and the log of a number less than 1 is negative so the negative sign makes the sum positive. Is that what you mean?

[RENJI]

Yes. I was thinking of the negative sign before the sum only. Thankyou very much.