Nrthugu

I tried to summurize everything that's been said and done about this topic in this question and in the duplicate one.

I tried to test and time every solution (all the code here).

Testing

This is the TestCase from the testing module:

class MergeTestCase(unittest.TestCase):



    def setUp(self):

        with open('./lists/test_list.txt') as f:

            self.lsts = json.loads(f.read())

        self.merged = self.merge_func(deepcopy(self.lsts))



    def test_disjoint(self):

        """Check disjoint-ness of merged results"""

        from itertools import combinations

        for a,b in combinations(self.merged, 2):

            self.assertTrue(a.isdisjoint(b))



    def test_coverage(self):    # Credit to katrielalex

        """Check coverage original data"""

        merged_flat = set()

        for s in self.merged:

            merged_flat |= s



        original_flat = set()

        for lst in self.lsts:

            original_flat |= set(lst)



        self.assertTrue(merged_flat == original_flat)



    def test_subset(self):      # Credit to WolframH

        """Check that every original data is a subset"""

        for lst in self.lsts:

            self.assertTrue(any(set(lst) <= e for e in self.merged))

This test is supposing a list of sets as result, so I couldn't test a couple of sulutions that worked with lists.

I couldn't test the following:

katrielalex

steabert

Among the ones I could test, two failed:

  -- Going to test: agf (optimized) --

Check disjoint-ness of merged results ... FAIL



  -- Going to test: robert king --

Check disjoint-ness of merged results ... FAIL

Timing

The performances are strongly related with the data test employed.

So far three answers tried to time theirs and others solution. Since they used different testing data they had different results.

1. 
   

   Niklas benchmark is very twakable. With his banchmark one could do different tests changing some parameters.

   
   
   

   I've used the same three sets of parameters he used in his own answer, and I put them in three different files:

   
   
   

   filename = './lists/timing_1.txt'
   
   class_count = 50,
   
   class_size = 1000,
   
   list_count_per_class = 100,
   
   large_list_sizes = (100, 1000),
   
   small_list_sizes = (0, 100),
   
   large_list_probability = 0.5,
   
   
   
   filename = './lists/timing_2.txt'
   
   class_count = 15,
   
   class_size = 1000,
   
   list_count_per_class = 300,
   
   large_list_sizes = (100, 1000),
   
   small_list_sizes = (0, 100),
   
   large_list_probability = 0.5,
   
   
   
   filename = './lists/timing_3.txt'
   
   class_count = 15,
   
   class_size = 1000,
   
   list_count_per_class = 300,
   
   large_list_sizes = (100, 1000),
   
   small_list_sizes = (0, 100),
   
   large_list_probability = 0.1,
   
   

   
   
   

   This are the results that I got:

   
   
   

   From file: timing_1.txt

   
   
   

   Timing with: >> Niklas << Benchmark
   
   Info: 5000 lists, average size 305, max size 999
   
   
   
   Timing Results:
   
   10.434  -- alexis
   
   11.476  -- agf
   
   11.555  -- Niklas B.
   
   13.622  -- Rik. Poggi
   
   14.016  -- agf (optimized)
   
   14.057  -- ChessMaster
   
   20.208  -- katrielalex
   
   21.697  -- steabert
   
   25.101  -- robert king
   
   76.870  -- Sven Marnach
   
   133.399  -- hochl
   
   

   
   
   

   From file: timing_2.txt

   
   
   

   Timing with: >> Niklas << Benchmark
   
   Info: 4500 lists, average size 305, max size 999
   
   
   
   Timing Results:
   
   8.247  -- Niklas B.
   
   8.286  -- agf
   
   8.637  -- Rik. Poggi
   
   8.967  -- alexis
   
   9.090  -- ChessMaster
   
   9.091  -- agf (optimized)
   
   18.186  -- katrielalex
   
   19.543  -- steabert
   
   22.852  -- robert king
   
   70.486  -- Sven Marnach
   
   104.405  -- hochl
   
   

   
   
   

   From file: timing_3.txt

   
   
   

   Timing with: >> Niklas << Benchmark
   
   Info: 4500 lists, average size 98, max size 999
   
   
   
   Timing Results:
   
   2.746  -- agf
   
   2.850  -- Niklas B.
   
   2.887  -- Rik. Poggi
   
   2.972  -- alexis
   
   3.077  -- ChessMaster
   
   3.174  -- agf (optimized)
   
   5.811  -- katrielalex
   
   7.208  -- robert king
   
   9.193  -- steabert
   
   23.536  -- Sven Marnach
   
   37.436  -- hochl
   
   

   
   


2. 
   

   With Sven's testing data I got the following results:

   
   
   

   Timing with: >> Sven << Benchmark
   
   Info: 200 lists, average size 10, max size 10
   
   
   
   Timing Results:
   
   2.053  -- alexis
   
   2.199  -- ChessMaster
   
   2.410  -- agf (optimized)
   
   3.394  -- agf
   
   3.398  -- Rik. Poggi
   
   3.640  -- robert king
   
   3.719  -- steabert
   
   3.776  -- Niklas B.
   
   3.888  -- hochl
   
   4.610  -- Sven Marnach
   
   5.018  -- katrielalex
   
   

   
   


3. 
   

   And finally with Agf's benchmark I got:

   
   
   

   Timing with: >> Agf << Benchmark
   
   Info: 2000 lists, average size 246, max size 500
   
   
   
   Timing Results:
   
   3.446  -- Rik. Poggi
   
   3.500  -- ChessMaster
   
   3.520  -- agf (optimized)
   
   3.527  -- Niklas B.
   
   3.527  -- agf
   
   3.902  -- hochl
   
   5.080  -- alexis
   
   15.997  -- steabert
   
   16.422  -- katrielalex
   
   18.317  -- robert king
   
   1257.152  -- Sven Marnach
   
   

   
   

As I said at the beginning all the code is available at this git repository. All the merging functions are in a file called core.py, every function there with its name ending with _merge will be auto loaded during the tests, so it shouldn't be hard to add/test/improve your own solution.

Let me also know if there's something wrong, it's been a lot of coding and I could use a couple of fresh eyes :)

Using Matrix Manipulations

Let me preface this answer with the following comment:

THIS IS THE WRONG WAY TO DO THIS. IT IS PRONE TO NUMERICAL INSTABILITY AND IS MUCH SLOWER THAN THE OTHER METHODS PRESENTED, USE AT YOUR OWN RISK.

That being said, I couldn't resist solving the problem from a dynamical point of view (and I hope you'll get a fresh perspective on the problem). In theory this should work all the time, but eigenvalue calculations can often fail. The idea is to think of your list as a flow from rows to columns. If two rows share a common value there is a connecting flow between them. If we were to think of these flows as water, we would see that the flows cluster into little pools when they there is a connecting path between them. For simplicity, I'm going to use a smaller set, though it works with your data set as well:

from numpy import where, newaxis

from scipy import linalg, array, zeros



X = [[0,1,3],[2],[3,1]]

We need to convert the data into a flow graph. If row i flows into value j we put it in the matrix. Here we have 3 rows and 4 unique values:

A = zeros((4,len(X)), dtype=float)

for i,row in enumerate(X):

    for val in row: A[val,i] = 1

In general, you'll need to change the 4 to capture the number of unique values you have. If the set is a list of integers starting from 0 as we have, you can simply make this the largest number. We now perform an eigenvalue decomposition. A SVD to be exact, since our matrix is not square.

S  = linalg.svd(A)

We want to keep only the 3x3 portion of this answer, since it will represent the flow of the pools. In fact we only want the absolute values of this matrix; we only care if there is a flow in this cluster space.

M  = abs(S[2])

We can think of this matrix M as a Markov matrix and make it explicit by row normalizing. Once we have this we compute the (left) eigenvalue decomp. of this matrix.

M /=  M.sum(axis=1)[:,newaxis]

U,V = linalg.eig(M,left=True, right=False)

V = abs(V)

Now a disconnected (non-ergodic) Markov matrix has the nice property that, for each non-connected cluster, there is a eigenvalue of unity. The eigenvectors associated with these unity values are the ones we want:

idx = where(U > .999)[0]

C = V.T[idx] > 0

I have to use .999 due to the aforementioned numerical instability. At this point, we are done! Each independent cluster can now pull the corresponding rows out:

for cluster in C:

    print where(A[:,cluster].sum(axis=1))[0]

Which gives, as intended:

[0 1 3]

[2]

Change X to your lst and you'll get: [ 0 1 3 4 5 10 11 16] [2 8].

Addendum

Why might this be useful? I don't know where your underlying data comes from, but what happens when the connections are not absolute? Say row 1 has entry 3 80% of the time - how would you generalize the problem? The flow method above would work just fine, and would be completely parametrized by that .999 value, the further away from unity it is, the looser the association.

Visual Representation

Since a picture is worth 1K words, here are the plots of the matrices A and V for my example and your lst respectively. Notice how in V splits into two clusters (it is a block-diagonal matrix with two blocks after permutation), since for each example there were only two unique lists!

Faster Implementation

In hindsight, I realized that you can skip the SVD step and compute only a single decomp:

M = dot(A.T,A)

M /=  M.sum(axis=1)[:,newaxis]

U,V = linalg.eig(M,left=True, right=False)

The advantage with this method (besides speed) is that M is now symmetric, hence the computation can be faster and more accurate (no imaginary values to worry about).

EDIT: OK, the other questions has been closed, posting here.

Nice question! It's much simpler if you think of it as a connected-components problem in a graph. The following code uses the excellent networkx graph library and the pairs function from this question.

def pairs(lst):

    i = iter(lst)

    first = prev = item = i.next()

    for item in i:

        yield prev, item

        prev = item

    yield item, first



lists = [[1,2,3],[3,5,6],[8,9,10],[11,12,13]]



import networkx

g = networkx.Graph()

for sub_list in lists:

    for edge in pairs(sub_list):

            g.add_edge(*edge)



networkx.connected_components(g)

[[1, 2, 3, 5, 6], [8, 9, 10], [11, 12, 13]]

Explanation

We create a new (empty) graph g. For each sub-list in lists, consider its elements as nodes of the graph and add an edge between them. (Since we only care about connectedness, we don't need to add all the edges -- only adjacent ones!) Note that add_edge takes two objects, treats them as nodes (and adds them if they aren't already there), and adds an edge between them.

Then, we just find the connected components of the graph -- a solved problem! -- and output them as our intersecting sets.

Here's my answer. I haven't checked it against today's batch of answers.

The intersection-based algorithms are O(N^2) since they check each new set against all the existing ones, so I used an approach that indexes each number and runs on close to O(N) (if we accept that dictionary lookups are O(1)). Then I ran the benchmarks and felt like a complete idiot because it ran slower, but on closer inspection it turned out that the test data ends up with only a handful of distinct result sets, so the quadratic algorithms don't have a lot work to do. Test it with more than 10-15 distinct bins and my algorithm is much faster. Try test data with more than 50 distinct bins, and it is enormously faster.

(Edit: There was also a problem with the way the benchmark is run, but I was wrong in my diagnosis. I altered my code to work with the way the repeated tests are run).

def mergelists5(data):

    """Check each number in our arrays only once, merging when we find

    a number we have seen before.

    """



    bins = range(len(data))  # Initialize each bin[n] == n

    nums = dict()



    data = [set(m) for m in data ]  # Convert to sets    

    for r, row in enumerate(data):

        for num in row:

            if num not in nums:

                # New number: tag it with a pointer to this row's bin

                nums[num] = r

                continue

            else:

                dest = locatebin(bins, nums[num])

                if dest == r:

                    continue # already in the same bin



                if dest > r:

                    dest, r = r, dest   # always merge into the smallest bin



                data[dest].update(data[r]) 

                data[r] = None

                # Update our indices to reflect the move

                bins[r] = dest

                r = dest 



    # Filter out the empty bins

    have = [ m for m in data if m ]

    print len(have), "groups in result"

    return have





def locatebin(bins, n):

    """

    Find the bin where list n has ended up: Follow bin references until

    we find a bin that has not moved.

    """

    while bins[n] != n:

        n = bins[n]

    return n

This new function only does the minimum necessary number of disjointness tests, something the other similar solutions fail to do. It also uses a deque to avoid as many linear time operations as possible, like list slicing and deletion from early in the list.

from collections import deque



def merge(lists):

    sets = deque(set(lst) for lst in lists if lst)

    results = 

    disjoint = 0

    current = sets.pop()

    while True:

        merged = False

        newsets = deque()

        for _ in xrange(disjoint, len(sets)):

            this = sets.pop()

            if not current.isdisjoint(this):

                current.update(this)

                merged = True

                disjoint = 0

            else:

                newsets.append(this)

                disjoint += 1

        if sets:

            newsets.extendleft(sets)

        if not merged:

            results.append(current)

            try:

                current = newsets.pop()

            except IndexError:

                break

            disjoint = 0

        sets = newsets

    return results

The less overlap between the sets in a given set of data, the better this will do compared to the other functions.

Here is an example case. If you have 4 sets, you need to compare:

1, 2

1, 3

1, 4

2, 3

2, 4

3, 4

If 1 overlaps with 3, then 2 needs to be re-tested to see if it now overlaps with 1, in order to safely skip testing 2 against 3.

There are two ways to deal with this. The first is to restart the testing of set 1 against the other sets after each overlap and merge. The second is to continue with the testing by comparing 1 with 4, then going back and re-testing. The latter results in fewer disjointness tests, as more merges happen in a single pass, so on the re-test pass, there are fewer sets left to test against.

The problem is to track which sets have to be re-tested. In the above example, 1 needs to be re-tested against 2 but not against 4, since 1 was already in its current state before 4 was tested the first time.

The disjoint counter allows this to be tracked.

My answer doesn't help with the main problem of finding an improved algorithm for recoding into FORTRAN; it is just what appears to me to be the simplest and most elegant way to implement the algorithm in Python.

According to my testing (or the test in the accepted answer), it's slightly (up to 10%) faster than the next fastest solution.

def merge0(lists):

    newsets, sets = [set(lst) for lst in lists if lst], 

    while len(sets) != len(newsets):

        sets, newsets = newsets, 

        for aset in sets:

            for eachset in newsets:

                if not aset.isdisjoint(eachset):

                    eachset.update(aset)

                    break

            else:

                newsets.append(aset)

    return newsets

No need for the un-Pythonic counters (i, range) or complicated mutation (del, pop, insert) used in the other implementations. It uses only simple iteration, merges overlapping sets in the simplest manner, and builds a single new list on each pass through the data.

My (faster and simpler) version of the test code:

import random



tenk = range(10000)

lsts = [random.sample(tenk, random.randint(0, 500)) for _ in range(2000)]



setup = """

def merge0(lists):

  newsets, sets = [set(lst) for lst in lists if lst], 

  while len(sets) != len(newsets):

    sets, newsets = newsets, 

    for aset in sets:

      for eachset in newsets:

        if not aset.isdisjoint(eachset):

          eachset.update(aset)

          break

      else:

        newsets.append(aset)

  return newsets



def merge1(lsts):

  sets = [set(lst) for lst in lsts if lst]

  merged = 1

  while merged:

    merged = 0

    results = 

    while sets:

      common, rest = sets[0], sets[1:]

      sets = 

      for x in rest:

        if x.isdisjoint(common):

          sets.append(x)

        else:

          merged = 1

          common |= x

      results.append(common)

    sets = results

  return sets



lsts = """ + repr(lsts)



import timeit

print timeit.timeit("merge0(lsts)", setup=setup, number=10)

print timeit.timeit("merge1(lsts)", setup=setup, number=10)

This would be my updated approach:

def merge(data):

    sets = (set(e) for e in data if e)

    results = [next(sets)]

    for e_set in sets:

        to_update = 

        for i,res in enumerate(results):

            if not e_set.isdisjoint(res):

                to_update.insert(0,i)



        if not to_update:

            results.append(e_set)

        else:

            last = results[to_update.pop(-1)]

            for i in to_update:

                last |= results[i]

                del results[i]

            last |= e_set



    return results

Note: During the merging empty lists will be removed.

Update: Reliability.

You need two tests for a 100% reliabilty of success:

• 
  

  Check that all the resulting sets are mutually disjointed:

  
  
  

  merged = [{0, 1, 3, 4, 5, 10, 11, 16}, {8, 2}, {8}]
  
  
  
  from itertools import combinations
  
  for a,b in combinations(merged,2):
  
      if not a.isdisjoint(b):
  
          raise Exception(a,b)    # just an example
  
  

  
  



• Check that the merged set cover the original data. (as suggested by katrielalex)

I think this will take some time, but maybe it'll be worth it if you want to be 100% sure.

Here's a function (Python 3.1) to check if the result of a merge function is OK. It checks:

• Are the result sets disjoint? (number of elements of union == sum of numbers of elements)


• Are the elements of the result sets the same as of the input lists?


• Is every input list a subset of a result set?


• Is every result set minimal, i.e. is it impossible to split it into two smaller sets?


• It does not check if there are empty result sets - I don't know if you want them or not...

.

from itertools import chain



def check(lsts, result):

    lsts = [set(s) for s in lsts]

    all_items = set(chain(*lsts))

    all_result_items = set(chain(*result))

    num_result_items = sum(len(s) for s in result)

    if num_result_items != len(all_result_items):

        print("Error: result sets overlap!")

        print(num_result_items, len(all_result_items))

        print(sorted(map(len, result)), sorted(map(len, lsts)))

    if all_items != all_result_items:

        print("Error: result doesn't match input lists!")

    if not all(any(set(s).issubset(t) for t in result) for s in lst):

        print("Error: not all input lists are contained in a result set!")



    seen = set()

    todo = list(filter(bool, lsts))

    done = False

    while not done:

        deletes = 

        for i, s in enumerate(todo): # intersection with seen, or with unseen result set, is OK

            if not s.isdisjoint(seen) or any(t.isdisjoint(seen) for t in result if not s.isdisjoint(t)):

                seen.update(s)

                deletes.append(i)

        for i in reversed(deletes):

            del todo[i]

        done = not deletes

    if todo:

        print("Error: A result set should be split into two or more parts!")

        print(todo)

lists = [[1,2,3],[3,5,6],[8,9,10],[11,12,13]]



import networkx as nx

g = nx.Graph()



for sub_list in lists:

    for i in range(1,len(sub_list)):

        g.add_edge(sub_list[0],sub_list[i])



print nx.connected_components(g)

#[[1, 2, 3, 5, 6], [8, 9, 10], [11, 12, 13]]

Performance:

5000 lists, 5 classes, average size 74, max size 1000

15.2264976415

Performance of merge1:

print timeit.timeit("merge1(lsts)", setup=setup, number=10)

5000 lists, 5 classes, average size 74, max size 1000

1.26998780571

So it is 11x slower than the fastest.. but the code is much more simple and readable!

This is slower than the solution offered by Niklas (I got 3.9s on the test.txt instead of 0.5s for his solution), but yields the same result and might be easier to implement in e.g. Fortran, since it doesn't use sets, only sorting of the total amount of elements and then a single run through all of them.

It returns a list with the ids of the merged lists, so also keeps track of empty lists, they stay unmerged.

def merge(lsts):

        # this is an index list that stores the joined id for each list

        joined = range(len(lsts))

        # create an ordered list with indices

        indexed_list = sorted((el,index) for index,lst in enumerate(lsts) for el in lst)

        # loop throught the ordered list, and if two elements are the same and

        # the lists are not yet joined, alter the list with joined id

        el_0,idx_0 = None,None

        for el,idx in indexed_list:

                if el == el_0 and joined[idx] != joined[idx_0]:

                        old = joined[idx]

                        rep = joined[idx_0]

                        joined = [rep if id == old else id for id in joined]

                el_0, idx_0 = el, idx

        return joined

Firstly I'm not exactly sure if the benchmarks are fair:

Adding the following code to the start of my function:

c = Counter(chain(*lists))

    print c[1]

"88"

This means that out of all the values in all the lists, there are only 88 distinct values. Usually in the real world duplicates are rare, and you would expect a lot more distinct values. (of course i don't know where your data from so can't make assumptions).

Because Duplicates are more common, it means sets are less likely to be disjoint. This means the set.isdisjoint() method will be much faster because only after a few tests it will find that the sets aren't disjoint.

Having said all that, I do believe the methods presented that use disjoint are the fastest anyway, but i'm just saying, instead of being 20x faster maybe they should only be 10x faster than the other methods with different benchmark testing.

Anyway, i Thought I would try a slightly different technique to solve this, however the merge sorting was too slow, this method is about 20x slower than the two fastest methods using the benchmarking:

I thought I would order everything

import heapq

from itertools import chain

def merge6(lists):

    for l in lists:

        l.sort()

    one_list = heapq.merge(*[zip(l,[i]*len(l)) for i,l in enumerate(lists)]) #iterating through one_list takes 25 seconds!!

    previous = one_list.next()



    d = {i:i for i in range(len(lists))}

    for current in one_list:

        if current[0]==previous[0]:

            d[current[1]] = d[previous[1]]

        previous=current



    groups=[ for i in range(len(lists))]

    for k in d:

        groups[d[k]].append(lists[k]) #add a each list to its group



    return [set(chain(*g)) for g in groups if g] #since each subroup in each g is sorted, it would be faster to merge these subgroups removing duplicates along the way.





lists = [[1,2,3],[3,5,6],[8,9,10],[11,12,13]]

print merge6(lists)

"[set([1, 2, 3, 5, 6]), set([8, 9, 10]), set([11, 12, 13])]""







import timeit

print timeit.timeit("merge1(lsts)", setup=setup, number=10)

print timeit.timeit("merge4(lsts)", setup=setup, number=10)

print timeit.timeit("merge6(lsts)", setup=setup, number=10)

5000 lists, 5 classes, average size 74, max size 1000

1.26732238315

5000 lists, 5 classes, average size 74, max size 1000

1.16062907437

5000 lists, 5 classes, average size 74, max size 1000

30.7257182826

Just for fun...

def merge(mylists):

    results, sets = , [set(lst) for lst in mylists if lst]

    upd, isd, pop = set.update, set.isdisjoint, sets.pop

    while sets:

        if not [upd(sets[0],pop(i)) for i in xrange(len(sets)-1,0,-1) if not isd(sets[0],sets[i])]:

            results.append(pop(0))

    return results

and my rewrite of the best answer

def merge(lsts):

  sets = map(set,lsts)

  results = 

  while sets:

    first, rest = sets[0], sets[1:]

    merged = False

    sets = 

    for s in rest:

      if s and s.isdisjoint(first):

        sets.append(s)

      else:

        first |= s

        merged = True

    if merged: sets.append(first)

    else: results.append(first)

  return results

Here's an implementation using a disjoint-set data structure (specifically a disjoint forest), thanks to comingstorm's hint at merging sets which have even one element in common. I'm using path compression for a slight (~5%) speed improvement; it's not entirely necessary (and it prevents find being tail recursive, which could slow things down). Note that I'm using a dict to represent the disjoint forest; given that the data are ints, an array would also work although it might not be much faster.

def merge(data):

    parents = {}

    def find(i):

        j = parents.get(i, i)

        if j == i:

            return i

        k = find(j)

        if k != j:

            parents[i] = k

        return k

    for l in filter(None, data):

        parents.update(dict.fromkeys(map(find, l), find(l[0])))

    merged = {}

    for k, v in parents.items():

        merged.setdefault(find(v), ).append(k)

    return merged.values()

This approach is comparable to the other best algorithms on Rik's benchmarks.

My solution, works well on small lists and is quite readable without dependencies.

def merge_list(starting_list):

    final_list = 

    for i,v in enumerate(starting_list[:-1]):

        if set(v)&set(starting_list[i+1]):

            starting_list[i+1].extend(list(set(v) - set(starting_list[i+1])))

        else:

            final_list.append(v)

    final_list.append(starting_list[-1])

    return final_list

Benchmarking it:

lists = [[1,2,3],[3,5,6],[8,9,10],[11,12,13]]

%timeit merge_list(lists)

100000 loops, best of 3: 4.9 µs per loop

This can be solved in O(n) by using the union-find algorithm. Given the first two rows of your data, edges to use in the union-find are the following pairs:
(0,1),(1,3),(1,0),(0,3),(3,4),(4,5),(5,10)