Is NaN ok here?

%coin txt

5.00000000e-01 5.00000000e-01

%crime txt

1.66000000e-01 6.60000000e-02 1.40000000e-02 3.90000000e-02 1.40000000e-02 2.70000000e-02 7.20000000e-02 1.00000000e-03 9.00000000e-03 1.30000000e-02 5.50000000e-02 9.00000000e-03 2.80000000e-02 3.80000000e-02 2.60000000e-02 5.50000000e-02 9.60000000e-02 2.30000000e-02 3.50000000e-02 4.50000000e-02 5.40000000e-02 2.40000000e-02 1.00000000e-03 7.00000000e-03 2.00000000e-03 1.50000000e-02 7.00000000e-03 3.00000000e-03 0.00000000e+00 1.40000000e-02 1.80000000e-02 3.00000000e-03 5.00000000e-03 1.80000000e-02

%unfair.txt

9.99000000e-01 1.00000000e-03

%ventsel.txt

1.45000000e-01 6.40000000e-02 1.50000000e-02 3.90000000e-02 1.40000000e-02 2.60000000e-02 7.40000000e-02 8.00000000e-03 1.50000000e-02 6.40000000e-02 1.00000000e-02 2.90000000e-02 3.60000000e-02 2.60000000e-02 5.60000000e-02 9.50000000e-02 2.40000000e-02 4.10000000e-02 4.70000000e-02 5.60000000e-02 2.10000000e-02 2.00000000e-03 9.00000000e-03 4.00000000e-03 1.30000000e-02 6.00000000e-03 3.00000000e-03 1.50000000e-02 1.60000000e-02 3.00000000e-03 7.00000000e-03 1.90000000e-02

%ralph.txt

3.36184163e-02 1.43227594e-01 1.64970470e-01 1.34046838e-01 7.52792246e-02 6.54570283e-02 5.34771727e-03 9.39337045e-02 1.06252105e-01 1.77866902e-01

%ALPH_ENTROPY FUNCTION
function h = alph_entropy(P)
h = sum(-P .* log2(P));
end

%APLH_REDUNDANCY FUNCTION
function r = alph_redundancy(P)
r = sum(1 - (alph_entropy(P)./log2(P)));
end

%main.m

A = load("coin.txt",'-ascii') 
B = load("crime.txt",'-ascii') 
C = load("unfair.txt",'-ascii') 
D = load("ventsel.txt",'-ascii') 
E = load("ralph.txt",'-ascii') 

Z = [alph_entropy(A) alph_redundancy(A); alph_entropy(B) alph_redundancy(B) ; alph_entropy(C) alph_redundancy(C); alph_entropy(D) alph_redundancy(D); alph_entropy(E) alph_redundancy(E)];

save results.txt Z -ascii;

MATRIX RESULT

%results.txt

1.00000000e+00 4.00000000e+00

NaN NaN

1.14077577e-02 9.90444552e+00

4.41966505e+00 5.87982853e+01

3.06961940e+00 1.90521746e+01

Answer 1

[John D'Errico]

Sure. It does not bother me. Why does it bother you?

B = [1.66000000e-01 6.60000000e-02 1.40000000e-02 3.90000000e-02 1.40000000e-02 2.70000000e-02 7.20000000e-02 1.00000000e-03 9.00000000e-03 1.30000000e-02 5.50000000e-02 9.00000000e-03 2.80000000e-02 3.80000000e-02 2.60000000e-02 5.50000000e-02 9.60000000e-02 2.30000000e-02 3.50000000e-02 4.50000000e-02 5.40000000e-02 2.40000000e-02 1.00000000e-03 7.00000000e-03 2.00000000e-03 1.50000000e-02 7.00000000e-03 3.00000000e-03 0.00000000e+00 1.40000000e-02 1.80000000e-02 3.00000000e-03 5.00000000e-03 1.80000000e-02];
alph_entropy = @(P) sum(-P .* log2(P));

alph_entropy(B)
ans =
   NaN

Why is that?

B(29)
ans =
     0

In fact, B(29) was 0.00000000e+00.

Your entropy formula will generate NaN when any element is exactly zero. And since the redundancy code uses the entropy computation, it too results in NaN.

So what do you expect? Looks fine.

When you see a problem, LOOK AT YOUR DATA. THINK ABOUT WHAT YOUR CODE IS DOING.

You might decide if you can just drop any zero elements of the vector. Does that make sense?

[Vadim Potorocha]

it's homework task. I do not how to avoid this element. But i know it's possible

[John D'Errico]

If it is homework, then you might look at the formulas for entropy, and decide how zero impacts the result you would expect. Would deleting zero elements be a problem? You were given this as homework, so you are the one who is supposed to think.

[Vadim Potorocha]

yep deleting zero elements coulda be the answer. But i have another formula with limit in it, but in GNU Octave i have no idea how to code it lim(inf to 0) = -p/ln(2)

Answer 2

[KSSV]

You have alph_entropy output as 1. log2(1) will be zero and when it is divided i.e. when it is denominator; you will get NaN.

[Vadim Potorocha]

How can i avoid zero arguments in log2?

[Vadim Potorocha]

lim(inf)→0(−p/ln2) can help but i can't find how to code this function