Dividing part of a histogram

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davut
davut el 14 de Jul. de 2023
Respondida: Image Analyst el 14 de Jul. de 2023
I want to plot a histogram but with divided parts. For example; it has four colums of age: 0-4, 4-8, 8-12 and 12-16. I want to divide 0-4 column into 5 parts, 4-8 into 10 parts. Is it possible?
  5 comentarios
davut
davut el 14 de Jul. de 2023
like this ( I did it this manually)
Dyuman Joshi
Dyuman Joshi el 14 de Jul. de 2023
It's not clear to me what your data is or what you are plotting.
Is your data the numbers in circles?

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Respuestas (4)

Florian Bidaud
Florian Bidaud el 14 de Jul. de 2023
Editada: Florian Bidaud el 14 de Jul. de 2023
Let's say your data is stored like :
y = [2 8 5 10]
Where y(1) is your 0-4 range, y(2) your 5-8 range etc.
Then you could do something like
y_new = [];
for i = 1:2
y_new = [y_new repmat(y(i),1,5*i)];
end
y_new = [y_new y_new(3) y_new(4)]
and then use y_new to plot your histogram
However, from what I understood from your graph, your new histogram wouldn't represent the same thing as before, as each colomn would be equal to the total of the range.

Image Analyst
Image Analyst el 14 de Jul. de 2023
Specify the "edges" you want for the bins as an input to histogram or histcounts
  1 comentario
Steven Lord
Steven Lord el 14 de Jul. de 2023
Or if you've already created a histogram, set its BinEdges property to the more refined bin edge vector or call the morebins or fewerbins functions with the histogram handle as input.
x = randn(1, 1e5);
h = histogram(x, -3:0.1:3); % bins with width of 0.1
figure % Making a second copy so you can compare original and modified
h = histogram(x, -3:0.1:3);
h.BinEdges = -3:0.25:3; % bins with width of 0.25
figure % third copy
h = histogram(x, -3:0.1:3);
fprintf("Before morebins call, histogram has %d bins.\n", h.NumBins)
Before morebins call, histogram has 60 bins.
morebins(h);
fprintf("After morebins call, histogram has %d bins.\n", h.NumBins)
After morebins call, histogram has 66 bins.
h.BinWidth % no longer width of 0.1
ans = 0.0909

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davut
davut el 14 de Jul. de 2023
no any useful answers unluckily
  1 comentario
Dyuman Joshi
Dyuman Joshi el 14 de Jul. de 2023
You have not answered my queries nor have you provided enough information for us to give a particular solution for your question.
@Image Analyst and @Steven Lord have given a general solution to the problem you have posted. If you want a "useful" solution, you will have specify "useful" details.

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Image Analyst
Image Analyst el 14 de Jul. de 2023
I think you may have overlooked my suggestion to use linspace to compute the edges, or you just coudn't figure it out. So here it is:
% For example; it has four colums of age: 0-4, 4-8, 8-12 and 12-16.
% I want to divide 0-4 column into 5 parts, and
% 4-8 into 10 parts. Is it possible?
edges1 = linspace(0, 4, 5+1);
edges2 = linspace(4, 8, 10+1);
edges = unique([edges1, edges2, 8, 12, 16])
edges = 1×18
0 0.8000 1.6000 2.4000 3.2000 4.0000 4.4000 4.8000 5.2000 5.6000 6.0000 6.4000 6.8000 7.2000 7.6000 8.0000 12.0000 16.0000
% Create sample data.
data = 5 + 2 * randn(1, 1000);
% Compute and plot histogram.
histogram(data, edges);
grid on;
ylabel('Count');
xlabel('Data Value');
Note that there are 5 bins between 0 and 4, 10 bins between 4 and 8, and then one bin from 8-12, and one bin from 12 to 16, just like you said you wanted.

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