How can I get data from histogram ?

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Tania Islam
Tania Islam el 13 de Dic. de 2019
Comentada: Tania Islam el 16 de Dic. de 2019
Hi,
I want a CDF curve from my dataset. For that, at first I am plotting the dataset into histogram using histogram function. My dataset contain error of times of 1000 samples. Among 1000, we may have only distinct 200 values (others are repetition, so I want the numbe of occurrance of each values). Thats why I plot them to histogram to get the values and the propbabaility of the values.
Priviously I was using %[counts_5_10, centers_5_10] = hist(data);. This code was fine for me. How can I get this information from the new histogram function?
In the previous code I cannot fixed the binwidth.
Following is my old and new code.
I cannot extract the X-axes dataset (unique values among the 1000)
old code:
%[counts_5_10, centers_5_10] = hist(data);
probability=counts_L_7(1,:)/1000;
CDF=cumsum(probability/sum(probability));
New code:
h = histogram(data,'BinWidth',0.005);
N= h.Values% repetition of number of error values
Edges = h.BinEdges% this is the error values
probability=N/1000;
CDF=cumsum(probability./sum(probability));

Respuesta aceptada

Ridwan Alam
Ridwan Alam el 13 de Dic. de 2019
Editada: Ridwan Alam el 13 de Dic. de 2019
I believe you are looking for histcounts() instead of histogram().
[counts, centers] = histcounts(data,'BinWidth',0.005);
probability=counts/1000; % =counts/sum(counts);
CDF=cumsum(probability/sum(probability));
Btw, if you don't need probability, the CDF can be directly calculated as follows:
[CDF, centers] = histcounts(data,'BinWidth',0.005,'normalization','cdf');
  9 comentarios
Ridwan Alam
Ridwan Alam el 14 de Dic. de 2019
Editada: Ridwan Alam el 15 de Dic. de 2019
Adding Adam's suggestion to the answer:
BinWidth = 0.005;
[CDF, edges] = histcounts(data,'BinWidth',BinWidth,'normalization','cdf'); % feel free to use the other solution using 'counts'
centers = edges(2:end) - BinWidth/2;
% with this 'centers' will have the same size as CDF.
Tania Islam
Tania Islam el 16 de Dic. de 2019
Thank you so much for your kind suggestions and time.

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