CODEWITHSUNDEEP
algorithmlistcluster-analysis
Erlich
Erlich

Reputation: 345

How to cluster lists with shared elements

This is a problem I can't currently find on Leetcode or StackOverflow. Lets say you have a set of lists of numbers or references as so:

[[1,2,3],[3,4,5],[6,7,8],[8,9],[10],[7]]

What is the fastest algorithm to merge these lists such that the output would be:

[[1,2,3,4,5],[6,7,8,9],[10]]

Many thanks.

Upvotes: 3

Views: 295

Answers (3)

Has QUIT--Anony-Mousse
Has QUIT--Anony-Mousse

Reputation: 77474

This is actually not a clustering problem, but a set union problem.

By the names "union find" or "disjoint-set" you can find some well discussed approaches to make these things fast.

Upvotes: 0

Paddy3118
Paddy3118

Reputation: 4772

RosettaCode has a task set consolidation that has a Python example that can be modified to work with lists:

def consolidate(lists):
    setlist = [set(lst) for lst in lists if lst]
    for i, s1 in enumerate(setlist):
        if s1:
            for s2 in setlist[i+1:]:
                intersection = s1.intersection(s2)
                if intersection:
                    s2.update(s1)
                    s1.clear()
                    s1 = s2
    return sorted(sorted(s) for s in setlist if s)

print(consolidate([[1,2,3],[3,4,5],[6,7,8],[8,9],[10],[7]]))

Output:

[[1, 2, 3, 4, 5], [6, 7, 8, 9], [10]]

Upvotes: 0

David Eisenstat
David Eisenstat

Reputation: 65498

Prepare a graph from the lists as follows and find the connected components using depth-first search.

Each list gives rise to undirected edges that connected the first element to the others, e.g.,

[1,2,3] -> [(1,2), (1,3)]
[3,4,5] -> [(3,4), (3,5)]
[6,7,8] -> [(6,7), (6,8)]
[8,9]   -> [(8,9)]
[10]    -> []
[7]     -> []

Then run depth-first search to find connected components. In Python, everything goes something like this.

import collections
def merge(lsts):
    neighbors = collections.defaultdict(set)
    for lst in lsts:
        if not lst:
            continue
        for x in lst:
            neighbors[x].add(lst[0])
            neighbors[lst[0]].add(x)
    visited = set()
    for v in neighbors:
        if v in visited:
            continue
        stack = [v]
        component = []
        while stack:
            w = stack.pop()
            if w in visited:
                continue
            visited.add(w)
            component.append(w)
            stack.extend(neighbors[w])
        yield component

Upvotes: 3

Related Questions