Suppose I have a multitude of sets with (unordered) combinations of elements and I want to determine which elements tend to appear together.

For example

Given the following sets:


Element-pairs where both elements tend to occur in the same sets will have lower distance:

# Low-distance pairs:
{a,e}, {b,f}, {d,c}

# Medium-distance pairs
{a,g}, {b,j}, {d,m}, ...

# High-distance pairs:
{g,h}, {j,n}, {b,f}, ...


I'm implementing DBSCAN with a custom distance metric. I use the following distance metric between two elements:

d(a,b) = 1 - numsets(a, b) / (numsets(a,!b) + numsets(b,!a))

Where d(a,b) denote the distance between elements a and b. While numsets denotes how many sets fulfill some conditions:

  • numsets(a, b) - the number of sets that contain a and b
  • numsets(a,!b) - the number of sets that contain a but not b

This solution should achieve the goal however it's not a pretty solution and I couldn't find this problem on SE. In terms of solving the problem, is there a more sensible distance metric? In terms of implementation, is there a nicer way to do this?


That is a data mining problem, specifically affinity analysis.

One common method to solve it is the Apriori algorithm.

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Notice that your metric might suffer division by zero, or have negative values. It's pretty close to Jaccard distance, so maybe consider that. See also http://curtis.ml.cmu.edu/w/courses/index.php/Co-occurrence_metrics

You might also have success with a graph clustering approach, see for a start

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Levenshtein distance (and its cousing Jaro, Hemming etc...)

Levenshtein distance for measuring the difference between two sequences between two words is the minimum number of single-character edits (insertions, deletions or substitutions) required to change one word (your case one set of characters) into the other.

There are a couple implementations, for example here, its the "edit_distance" function.

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    $\begingroup$ Given that the sets of combinations are unordered, Levenstein is not a good metric. $\endgroup$ – DannyDannyDanny Dec 30 '19 at 10:37
  • $\begingroup$ True, but you can sort them (if thats possible given the task-in the example yes) $\endgroup$ – Noah Weber Dec 30 '19 at 10:39
  • 1
    $\begingroup$ I've tried to mull it over. Levenstein measures distance between sequences. I need the distance between elements. $\endgroup$ – DannyDannyDanny Dec 30 '19 at 11:01

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