binning data in equally spaced intervals
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I would like to bin the data in 8 equally spaced bins. I have a the data accessible in vectors [X= conc, Y= alt] this what I have so far to create the bins
binedge = linspace(min(alt),max(alt),6)
then I want to take the average alt and conc of each each bin and plot it. Thanks
4 comentarios
per isakson
el 11 de Mzo. de 2015
The description and examples of the documentation histc, Histogram bin counts (not recommended; use histcounts) are better than mine.
histcounts was introduced in R2014b. I've never used it.
Respuestas (2)
Josh Meyer
el 27 de Mzo. de 2015
As others noted, histcounts was introduced in R2014b and provides a great deal more flexibility for problems like this.
To use 8 bins, just do:
% Assume the first column is X, second column is Y
data = [rand(100,1), rand(100,1)];
% Find the bin placement based on the X values
[N,edges,bins] = histcounts(data(:,1),8);
The third output, bins, describes the bin placement of each element. So this makes finding the average X and Y value in each bin simple.
for n = 1:8
bin_means(:,n) = mean(data(bins==n,:))';
end
bin_means =
0.0751 0.1979 0.3342 0.4658 0.5691 0.7113 0.8691 0.9676
0.5100 0.6264 0.4949 0.5172 0.5323 0.5556 0.4381 0.6514
3 comentarios
Andrew Hurford
el 6 de Ag. de 2020
This worked well for me - but for 1 result in a bin, mean defaults to selecting data across the data, so set the direction flag to 1
Rahul Kesarkar
el 7 de Abr. de 2021
How would it work if i have to bin data for a particular date range?
For example range : - weekday 12/06/2017 20:00 to 13/06/2017 08:00 and repeat this for every weekday. Weekend 9/06/2017 20:00 to 12/06/2017 08:00. Sum up all the values in the bin.
Stephen23
el 11 de Mzo. de 2015
Editada: Stephen23
el 11 de Mzo. de 2015
You can use histc to get bin the data, and then accumarray to get the mean of all of the values in each bin:
>> X = 0.25:0.5:5
X =
0.25 0.75 1.25 1.75 2.25 2.75 3.25 3.75 4.25 4.75
>> [B,idx] = histc(X,0:5);
>> V = accumarray(idx(:),X,[],@mean)
V =
0.5000
1.5000
2.5000
3.5000
4.5000
[B,~,idx] = histcounts(...);
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