Cannot use 'histogram' to compute entropy

1 次查看（过去 30 天）

z8080 2021-9-9

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1449894-cannot-use-histogram-to-compute-entropy

评论： Walter Roberson 2021-9-10

I'd like to compute the entropy of various vectors. I was going to use something like:

X = randn(1,100);
h1 = histogram(X, 'Normalization', 'Probability');
probabilities = h1.Values;
entropy = -sum(probabilities .* log2(probabilities ))

The second command however gives the error:

Undefined function 'c:\Program Files\MATLAB\R2019b\toolbox\matlab\specgraph\histogram.m' for input arguments of type 'double'.

But surely that's exactly what the standard Matlab function 'histogram' expects?! Doing a

which histogram

indeed returns

C:\Program Files\MATLAB\R2019b\toolbox\matlab\specgraph\histogram.m

which is the newest file (by modified date) from several of that name that (sadly) exist in my Matlab folder. I believe this should be the standard Matlab function 'histogram'.

If on the other hand in the above example I use 'hist' instead of 'histogram', I get the scalar value for entropy that I expect. However, I know 'hist' is not recommended, not least because with it one cannot specify the normalization type.

So, my question is: is using 'hist' for computing probabilities ok, or should I try something else to be able to use 'histogram' instead?

13 个评论
显示 11更早的评论隐藏 11更早的评论

z8080 2021-9-9

编辑：z8080 2021-9-9

在 MATLAB Online 中打开

By "items with count 0" you mean items of the probabilities vector that are equal to 0?

Actually, for some reason all probabilities vectors that I get include at least one 0, however the corresponding entropy only is NaN for those where the 0 is mid-vector rather than when the 0s are at the end (due to padding when creating matrix rows of various length). It's not obvious which input signals lead to which situation.

For instance this code:

signal = [2.06623794644167,1.90137810157315,5.79024624899804,-10.0872171266770,3.72029696914678,1.88888495632052,-0.899652931772044,-1.01593746059118,6.02304846348934,-1.47602178714163,-1.26585426828863,-1.23193945906322,-1.58972503058801,-1.19316666394375,NaN,NaN,NaN,NaN,NaN,NaN,NaN]
% signal(isnan(signal)) = [];
h1 = histogram(signal, 'Normalization', 'Probability'); 
probabilities = h1.Values 
entropy = -sum(probabilities .* log2(probabilities )) % this returns a NaN
signal = [-6.30156202700424,2.31276368504634,2.53575767657180,3.85380766080873,-0.788693245197791,-3.14860106509044,5.94561552053542,-4.88433510447800,-0.536192076206886,4.93030202134006,-4.09979662991061,-0.597866984143998,-1.21151323711087,4.31812745523269,-1.74917050858314,-1.59171474633157,-0.845918488184140,NaN,NaN,NaN,NaN]
% signal(isnan(signal)) = [];
h1 = histogram(signal, 'Normalization', 'Probability'); 
probabilities = h1.Values 
entropy = -sum(probabilities .* log2(probabilities )) % this returns a non-NaN!

returns:

signal =
  Columns 1 through 17
    2.0662    1.9014    5.7902  -10.0872    3.7203    1.8889   -0.8997   -1.0159    6.0230   -1.4760   -1.2659   -1.2319   -1.5897   -1.1932       NaN       NaN       NaN
  Columns 18 through 21
       NaN       NaN       NaN       NaN
probabilities =
    0.0476         0    0.3333    0.1905    0.0952
entropy =
   NaN
signal =
  Columns 1 through 17
   -6.3016    2.3128    2.5358    3.8538   -0.7887   -3.1486    5.9456   -4.8843   -0.5362    4.9303   -4.0998   -0.5979   -1.2115    4.3181   -1.7492   -1.5917   -0.8459
  Columns 18 through 21
       NaN       NaN       NaN       NaN
probabilities =
    0.0476    0.4762    0.2381    0.0476
entropy =
    1.4210

Removing the NaNs from the signal vectors themselves makes no difference.

Thus, my questions:

1) I don't understand why a 0 in the middle leads to a NaN entropy, but a 0 at the end is acceptable.

2) Would it actually be correct to manually remove those mid-way 0s and compute entropy from the 'cleaned-up' probabilities vector?

z8080 2021-9-10

编辑：z8080 2021-9-10

Thanks a lot for this excellent answer and derivation. to answer my own question then, I guess that it is acceptable to manually remove all bins with a count of 0, to enable the computation of entropy based on the non-0 bins. This is in fact what you had answered me from the very beginning :)

Thanks again!

Walter Roberson 2021-9-10

Depending on your knowledge of the distribution, it might make sense to take ask for the counts, and take max(1,counts) to substitute a nominal hit for each bin, and then calculate probability from that, as adjusted_counts ./ sum(adjusted_counts) .

The fewer samples you have, the more that distorts the probabilities; the more samples you have, the less likely you are to need it.

But I do recommend figuring out the number of bits yourself somehow or else you are going to continue to be at the mercy of its undocumented method of selecting the number of bins.

请先登录，再进行评论。

请先登录，再回答此问题。

回答（0 个）

请先登录，再回答此问题。

类别

MATLAB Graphics 2-D and 3-D Plots Data Distribution Plots Histograms

在 Help Center 和 File Exchange 中查找有关 Histograms 的更多信息

产品

MATLAB

版本

R2019b

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by

Cannot use 'histogram' to compute entropy

13 个评论
显示 11更早的评论隐藏 11更早的评论

回答（0 个）

另请参阅

类别

标签

产品

版本

Community Treasure Hunt

Cannot use 'histogram' to compute entropy

13 个评论 显示 11更早的评论隐藏 11更早的评论

回答（0 个）

另请参阅

类别

标签

产品

版本

Community Treasure Hunt

13 个评论
显示 11更早的评论隐藏 11更早的评论