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Background

I'm implementing the DBScan algorithm. I have trained it to cluster a small dataset of random clusters, and want to be able to get a decimal for its accuracy of clustering the groups.

Motivation

This is for some simple unittesting that checks it can cluster basic, separate classes (i.e. checking in my CI/CD that the accuracy is appropriately high).

The Problem

The problem is that it's outputs are its own cluster-indexes (the classes it's discovered/chosen). These do not align with the original integer labels I generated earlier.

An Example to Clear Things Up

Say, I have some original labels:

y_true = np.array([0, 1, 1, 0, 2, 1, 2])

and I have some cluster-predicted outputs

y_predicted = np.array([1, 0, 0, 1, 2, 0, 2])

You can see it has clustered values correctly, however they don't align with the original y_true array values. Therefore we can't use the normal accuracy function of np.mean(y_true == y_predicted).

The Challenge

Whilst preserving the outlier class of -1, how can I check that the model is significantly accurate, despite the fact that its generated classes do not align with original inputs. I understand that clustering algorithms, in practice are not used in this way but this is for testing that my implementation is proper, in my CI/CD automated testing.

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  • $\begingroup$ After sorting y_true and y_pred and take the average of absolute difference between y_true[i], y_pred[i], assuming the cluster's have similar sizes i.e., len(y_true) == len(y_pred)? $\endgroup$ Commented Aug 7, 2023 at 19:13
  • $\begingroup$ Thank you, I'll try that now! $\endgroup$ Commented Aug 7, 2023 at 19:15
  • $\begingroup$ However, given that the clusters often have similar sizes, could this be a problem?@shaikmoeed The numpy code I've used is np.mean(np.sort(y_true) == np.sort(y_prediction)) to give a value from 0 to 1 $\endgroup$ Commented Aug 7, 2023 at 19:15
  • $\begingroup$ I have found that this doesn't quite work. y1 = [1, 0, 2, 0, 1]; y2 = [2, 1, 0, 1, 2] The value ends up as 0.6 rather than 1 as it should be here. $\endgroup$ Commented Aug 7, 2023 at 19:24
  • $\begingroup$ Yeah, that's true. Your requirement seems to be close with adjusted_rand_score. $\endgroup$ Commented Aug 7, 2023 at 19:35

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