Merge tuples in list with similar elements - python-3.x

Question

I have to merge all the tuples containing atleast one element of each other.
tups=[(1,2),(2,3),(8,9),(4,5),(15,12),(9,6),(7,8),(3,11),(1,15)]
first tuple (1,2) should be merged with (2,3),(3,11),(1,15),(15,12) since each of these tuples contains similar items of the preceding tuple. so the final ouput should be
lst1 = [1,2,3,11,12,15]
lst2=[6,7,8,9] since (8,9),(9,6) and (7,8) have matching elements
My code so far:
finlst=[]
for items in range(len(tups)):
for resid in range(len(tups)):
if(tups[items] != tups[resid] ):
if(tups[items][0]==tups[resid][0] or tups[items][0]==tups[resid][1]):
finlst.append(list(set(tups[items]+tups[resid])))

Answer 1

You could do it like this, using sets that are expanded with matching tuples:
tups = [(1, 2), (2, 3), (8, 9), (4, 5), (15, 12), (9, 6), (7, 8), (3, 11), (1, 15)]
groups = []
for t in tups:
for group in groups:
# find a group that has at least one element in common with the tuple
if any(x in group for x in t):
# extend the group with the items from the tuple
group.update(t)
# break from the group-loop as we don’t need to search any further
break
else:
# otherwise (if the group-loop ended without being cancelled with `break`)
# create a new group from the tuple
groups.append(set(t))
# output
for group in groups:
print(group)
{1, 2, 3, 11, 15}
{8, 9, 6, 7}
{4, 5}
{12, 15}
Since this solution iterates the original tuple list once and in order, this will not work for inputs where the connections are not directly visible. For that, we could use the following solution instead which uses fixed-point iteration to combine the groups for as long as that still works:
tups = [(1, 2), (3, 4), (1, 4)]
import itertools
groups = [set(t) for t in tups]
while True:
for a, b in itertools.combinations(groups, 2):
# if the groups can be merged
if len(a & b):
# construct new groups list
groups = [g for g in groups if g != a and g != b]
groups.append(a | b)
# break the for loop and restart
break
else:
# the for loop ended naturally, so no overlapping groups were found
break

Answer 2

I found the solution, it's more of graph theory problem related to connectivity, Connectivity-Graph Theory
We can use NetworkX for this, it's pretty much guaranteed to be correct:
def uniqueGroup(groups):
# grp=[]
# for group in groups:
# grp.append(list(group))
# l=groups
import networkx
from networkx.algorithms.components.connected import connected_components
def to_graph(groups):
G = networkx.Graph()
for part in groups:
# each sublist is a bunch of nodes
G.add_nodes_from(part)
# it also imlies a number of edges:
G.add_edges_from(to_edges(part))
return G
def to_edges(groups):
"""
treat `l` as a Graph and returns it's edges
to_edges(['a','b','c','d']) -> [(a,b), (b,c),(c,d)]
"""
it = iter(groups)
last = next(it)
for current in it:
yield last, current
last = current
G = to_graph(groups)
return connected_components(G)
Output:
tups = [(1, 2),(3,4),(1,4)]
uniqueGroup(tups)
{1, 2, 3, 4}