Get confused (twice) with entropy of grayscale image

First, the quote from help says: "entropy converts any class other than logical to uint8 for the histogram count calculation"

I did following 2 tests:

1) A = 0:255; % here A is double and

entropy(A) gives 0.0369

2) A = uint8(0:255);

entropy(A) gives 8

It looks like it doesn't convert to uint8.

Second, I tried formula -sum(p.*log2(p))

[p,x] = imhist(A);

And then

-sum(p.*log2(p)) gives 0, which is obvious because histogram of the given A makes 256 bins with 1 in every bin, and log2(1) = 0.

What was wrong in my tests?

Thanks, Pavel

[pavel]

No more confusion! The answers are in the "entropy" code!

Answer 1

[Jeff E]

This boils down to the way IMHIST calculates the bins for an image of type double. For some (all?) functions, Matlab assumes pixel values of this type fall within a range of zero to one. IMTOOL, for example, scales its display to this range as a default. IMHIST does the same, as can be seen if you call it directly on your vector A:

A_counts = imhist(A);

you will see that all the pixels fall in the highest bin.

[pavel]

It means that the note in the help on "entropy" is wrong, doesn't it?

[Jeff E]

I wouldn't say it's wrong, but the conversion they are performing isn't the one you are expecting. For example, look at the result of :

im2double(uint8(0:255));

Again, Matlab assumes an image of type double will fall in the range of zero to one. In this case, given a uint8 image, it essentially performs the conversion:

B_conv = double(B) ./ 255 ;

This remaps the uint8 values of 0-255, into the double range of 0-1.

>> entropy(0:255)

ans =

      0.0369

>> entropy(uint8(0:255))

ans =

       8

>> entropy(im2double(uint8(0:255)))

ans =

       8