I have a dataset which consists of 2 types of items. One type (Type A) is easily matchable, and there are several groups within this set I would like to cluster. The second type (Type B) is randomly scattered about, and I would like to cluster all of these into a single bucket. I know how many clusters of Type A to expect.

The data items are all vectors of length 10ish (edit - it turns out I may have 100 elements in each vector! , so there are plenty of dimensions to play with. Even so, I expect there will be some Type B items which "fall into" one of the Type A buckets, and I can accept that.

A toy example:

import numpy as np
import random

observation_length = 10

# create some sample templates - 
data_templates = [
    np.random.random(size=(observation_length,)), # Type A1
    np.random.random(size=(observation_length,)), # Type A2
    None                                          # Type B - actual values created below

# generate some sample data
observations = []
for _ in range (100):
    observation = random.choice(data_templates)
    if observation is None:
        observation = np.random.random(size=(observation_length,))
    # in reality the observations will also have measurement noise

How would I cluster the observations into 3 buckets in such a system?


2 Answers 2


Looks like a case suitable for (H)DBSCAN. It assumes there exists a foreground (real/interesting data to be clustered) and background (noise). In your case group As would be foreground and B the background.


It might be useful to frame that problem as hierarchical clustering. In particular, use a "top-down" divisive approach. All observations start in one cluster, then clusters are split recursively down the hierarchy. The first level is "A" / "not A", then the second level is the split of "A" into "A1" and "A2". The divisive split decision will be based on a measure of dissimilarity between the observations.

  • $\begingroup$ Thanks - would that still work if I have say 20 "A groups" and one "not A" group? It feels unbalanced, but perhaps that is not a problem? $\endgroup$ Jul 18, 2023 at 10:57

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