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\n<\/p>\n
To group the sublists in a general way you can:<\/p>\n
def linking_sublists(lists):\n index = {}\n sets = []\n for l in lists:\n found = None\n for i in l:\n if i in index:\n # this list has an element we have already seen\n if found:\n # combine two sets\n to_remove = index[i]\n if found != to_remove:\n for j in index[i]:\n found.add(j)\n index[j] = found\n to_remove.clear()\n else:\n found = index[i]\n\n if found is not None:\n s = found\n for i in l:\n s.add(i)\n else:\n s = set(l)\n sets.append(s)\n for i in l:\n index[i] = s\n\n return [list(sorted(s)) for s in sets if s]\n<\/code><\/pre>\nHow:<\/h3>\n
This function uses sets and an index dict<\/code><\/a> to group any list with matching elements into sets and track which elements are already in a set<\/code><\/a>.<\/p>\nTest Code:<\/h3>\nlist_2 = [[2, 3], [4, 3], [6, 5], [7, 6], [7, 8], [13, 14], [30, 32]]\nprint(linking_sublists(list_2))\n\nlist_3 = [[2, 3], [4, 3], [6, 5], [7, 6], [7, 8], [30, 32], [4, 5], [3, 4]]\nprint(linking_sublists(list_3))\n<\/code><\/pre>\nResults:<\/h3>\n[[2, 3, 4], [5, 6, 7, 8], [13, 14], [30, 32]]\n[[2, 3, 4, 5, 6, 7, 8], [30, 32]]\n<\/code><\/pre>\n<\/p><\/div>\n