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copied!<p>This relies on the list of lists being sorted in ascending length (and the outputs needs sorting), but works with all provided input as of time of posting.</p> <pre><code>def removeFromList(elementsToRemove): def closure(list): for element in elementsToRemove: if list[0] != element: return else: list.pop(0) return closure def func(listOfLists): result = [] for i, thisList in enumerate(listOfLists): result.append(thisList) map(removeFromList(thisList), listOfLists[i+1:]) return result </code></pre> <p>You can remove the closure using more for loops, but it looked too ugly, and too deeply nested.</p> <p>Edit: based on updates, this is the naive solution to the actual problem:</p> <pre><code>from itertools import combinations def isSubsetOrDisjointTo(listA, listB): return all(item in listB for item in listA) or all(item not in listB for item in listA) def func(nodesToJoin): #flatten list, extracting duplicates allNodes = sorted(set(itertools.chain(*nodesToJoin))) result = [] seen = set() for length in xrange(max(map(len, nodesToJoin)), 0, -1): #start with longest possible, work to shortest for sublist in combinations(allNodes, length): if any(item in seen for item in sublist): #skip possible sublists with options we've already seen in the result continue if all(isSubsetOrDisjointTo(sublist, node) for node in nodesToJoin): result.append(sublist) seen.update(sublist) return result </code></pre> <p>There are probably a number of ways this can be optimised or improved.</p>

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