histc and bin-width determination
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Hi
Given I have the following data:
A = [100 150 190 200 250 300 350 370 390 400 400]
And I want 3 bins: From 100 to 200, 200 to 300, and 300 to 400.
How do I do that with histc?
What I tried so far was:
edges = linspace(min(stim_durations),max(stim_durations),4);
counts = histc(stim_durations, edges);
But that results in 4 bins with "400" having its own bin..how can I solve that?
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Guillaume
2015년 5월 7일
편집: Guillaume
2015년 5월 7일
The simplest solution on a recent enough version of matlab (2014b or newer) is to use histcounts which behaves exactly as you want:
>>A = [100 150 190 200 250 300 350 370 390 400 400];
>>hiscounts(A, [100 200 300 400])
ans =
3 2 6
As per histc documentation, the last edge is its own bin and you can't do anything about that. If you want 400 to be included in the third bin and still use histc, you have to shift the edge of the 4th value slightly above 400 (and discard that last bin):
A = [100 150 190 200 250 300 350 370 390 400 400];
edges = linspace(min(A),max(A),4);
edges(end) = edges(end) + eps(edges(end)); %or any value greater than eps(400);
counts = histc(A, edges);
counts = counts(1:end-1)
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Image Analyst
2015년 5월 7일
histc IS available in older versions. It's now been deprecated in favor of histcounts, meaning that in new versions they recommend you use histcounts and not histc. I didn't mention it in my answer because it seems like hardly anyone except me has the latest version.
추가 답변 (1개)
Walter Roberson
2015년 5월 7일
edges = linspace(min(stim_durations),max(stim_durations),4);
edges(end) = [];
That is, construct 4 bins and discard the last of them. The final bin will be everything from the previous value upwards.
Please pay attention to whether you want your bins to be 100 <= x <= 200, 200 <= x <= 300, or if you want them to be 100 <= x < 200, 200 <= x < 300 or 100 < x <= 200, 200 < x <= 300 . Those are slightly different boundary conditions that can be pretty meaningful.
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