Large amount of lists concatenation
As mentioned by @ScottBoston this is a graph problem, known as connected components, I suggest you used networkx as indicated by @ScottBoston, in case you cannot here is a version without networkx:
from itertools import combinations
def bfs(graph, start):
visited, queue = set(), [start]
while queue:
vertex = queue.pop(0)
if vertex not in visited:
visited.add(vertex)
queue.extend(graph[vertex] - visited)
return visited
def connected_components(G):
seen = set()
for v in G:
if v not in seen:
c = set(bfs(G, v))
yield c
seen.update(c)
def graph(edge_list):
result = {}
for source, target in edge_list:
result.setdefault(source, set()).add(target)
result.setdefault(target, set()).add(source)
return result
def concat(l):
edges = []
s = list(map(set, l))
for i, j in combinations(range(len(s)), r=2):
if s[i].intersection(s[j]):
edges.append((i, j))
G = graph(edges)
result = []
unassigned = list(range(len(s)))
for component in connected_components(G):
union = set().union(*(s[i] for i in component))
result.append(sorted(union))
unassigned = [i for i in unassigned if i not in component]
result.extend(map(sorted, (s[i] for i in unassigned)))
return result
print(concat([[1, 2], [3, 4, 5], [0, 4]]))
print(concat([[1], [1, 2], [0, 2]]))
print(concat([[1, 2], [2, 3], [3, 4]]))
Output
[[0, 3, 4, 5], [1, 2]]
[[0, 1, 2]]
[[1, 2, 3, 4]]
You can use networkx library because is a graph theory and connected components problem:
import networkx as nx
l = [[1,2],[3,4,5],[0,4]]
#l = [[1],[1,2],[0,2]]
#l = [[1, 2], [2, 3], [3, 4]]
G = nx.Graph()
#Add nodes to Graph
G.add_nodes_from(sum(l, []))
#Create edges from list of nodes
q = [[(s[i],s[i+1]) for i in range(len(s)-1)] for s in l]
for i in q:
#Add edges to Graph
G.add_edges_from(i)
#Find all connnected components in graph and list nodes for each component
[list(i) for i in nx.connected_components(G)]
Output:
[[1, 2], [0, 3, 4, 5]]
Output if uncomment line 2 and comment line 1:
[[0, 1, 2]]
Likewise for line 3:
[[1, 2, 3, 4]]