Is NaN ok here?
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Vadim Potorocha
2020 年 11 月 20 日
コメント済み: Vadim Potorocha
2020 年 11 月 20 日
%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
0 件のコメント
採用された回答
John D'Errico
2020 年 11 月 20 日
編集済み: John D'Errico
2020 年 11 月 20 日
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?
3 件のコメント
John D'Errico
2020 年 11 月 20 日
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.
その他の回答 (1 件)
KSSV
2020 年 11 月 20 日
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.
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