How to cluster data in a histogram ?

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Sim
Sim on 3 Nov 2022
Commented: Sim on 3 Nov 2022
How to cluster data in a histogram ?
Desired output
Input
A = duration({'00:01:01'
'00:00:53'
'00:00:55'
'00:00:54'
'00:00:54'
'00:00:53'
'02:45:08'
'00:01:33'
'00:00:57'
'00:00:58'
'00:00:51'
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'00:01:03'
'00:00:56'
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'00:26:52'
'00:01:12'
'00:00:41'
'00:00:56'
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'00:01:47'
'00:09:22'
'00:00:40'
'00:00:38'
'00:00:48'
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'00:00:42'
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'00:01:06'
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'00:00:47'
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'00:01:20'
'00:01:35'
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'00:29:36'
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'00:00:53'
'00:01:00'})
A = 98×1 duration array
00:01:01 00:00:53 00:00:55 00:00:54 00:00:54 00:00:53 02:45:08 00:01:33 00:00:57 00:00:58 00:00:51 00:00:45 00:01:03 00:00:56 00:00:45 00:26:52 00:01:12 00:00:41 00:00:56 00:01:16 00:01:47 00:09:22 00:00:40 00:00:38 00:00:48 00:00:38 00:00:42 00:00:42 00:01:06 00:01:00
binLimits = [min(A) max(A)];
binWidth = duration('00:00:5');
[counts,binEdges] = histcounts(A, 'BinLimits',binLimits,'BinWidth',binWidth);
histogram(A,'binEdges',binEdges,'FaceColor','k','EdgeColor','k');

Accepted Answer

Cris LaPierre
Cris LaPierre on 3 Nov 2022
You need to define the criteria for what makes a cluster and what makes an outlier. Once you know how you will define that, then you can use rmoutliers to apply your criteria and remove the outliers.
To me, it looks like you want to use quartiles. I find viewing this with a boxchart is easiest. Note that MATLAB does not accept the duration data type for outlier detection. Use the minutes function to covert your durations into numeric values, and vice versa.
A = duration({'00:01:01';'00:00:53';'00:00:55';'00:00:54';'00:00:54';'00:00:53';'02:45:08';'00:01:33';
'00:00:57';'00:00:58';'00:00:51';'00:00:45';'00:01:03';'00:00:56';'00:00:45';'00:26:52';'00:01:12';
'00:00:41';'00:00:56';'00:01:16';'00:01:47';'00:09:22';'00:00:40';'00:00:38';'00:00:48';'00:00:38';
'00:00:42';'00:00:42';'00:01:06';'00:01:00';'00:00:43';'00:00:47';'00:00:43';'00:00:50';'00:00:52';
'00:01:20';'00:01:35';'00:00:54';'00:01:05';'00:02:07';'00:00:43';'00:00:39';'00:29:36';'00:00:39';
'00:00:39';'00:01:01';'00:01:09';'00:01:12';'00:01:11';'00:01:12';'00:01:06';'00:01:06';'00:01:00';
'00:01:15';'00:01:08';'00:00:39';'00:00:59';'00:00:54';'00:01:25';'00:01:01';'00:01:03';'00:01:03';
'00:00:56';'00:01:19';'00:01:05';'00:01:00';'00:01:09';'00:01:12';'00:00:52';'00:00:40';'00:01:09';
'00:01:00';'00:01:04';'00:00:57';'00:02:07';'00:02:44';'00:00:51';'00:01:22';'00:01:10';'00:01:07';
'00:01:07';'00:00:48';'00:00:59';'00:01:02';'00:00:48';'00:00:49';'00:00:56';'00:01:03';'00:00:53';
'00:01:23';'00:00:40';'00:01:25';'00:01:15';'00:01:13';'00:02:14';'00:01:08';'00:00:53';'00:01:00'});
B = minutes(A)
B = 98×1
1.0167 0.8833 0.9167 0.9000 0.9000 0.8833 165.1333 1.5500 0.9500 0.9667
boxchart(B)
Here, outlers are indicated with the 'o' marker. Outliers are values that are more than 1.5 · IQR away from the top or bottom of the box. You indicate there should be 4, but we see the default definition classifies more, so you will need to use a custom definition.
I find using the Clean Outlier Data task in a live script is a quick and interactive way to find the desired threshold. Since tasks don't work, here, I'll just use rmoutliers with a threshold of 20 applied.
% Remove outliers
B= rmoutliers(B,"quartiles","ThresholdFactor",20);
% View results
boxchart(B)
% convert back to duration
A = minutes(B);
A.Format = 'hh:mm:ss'
A = 94×1 duration array
00:01:01 00:00:53 00:00:55 00:00:54 00:00:54 00:00:53 00:01:33 00:00:57 00:00:58 00:00:51 00:00:45 00:01:03 00:00:56 00:00:45 00:01:12 00:00:41 00:00:56 00:01:16 00:01:47 00:00:40 00:00:38 00:00:48 00:00:38 00:00:42 00:00:42 00:01:06 00:01:00 00:00:43 00:00:47 00:00:43
% Original code for creating a histogram
binLimits = [min(A) max(A)];
binWidth = duration('00:00:5');
[counts,binEdges] = histcounts(A, 'BinLimits',binLimits,'BinWidth',binWidth);
histogram(A,'binEdges',binEdges,'FaceColor','k','EdgeColor','k');

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