Find duplicate items within a list of list of tuples Python

问题

I want to find the matching item from the below given list.My List may be super large.

The very first item in the tuple "N1_10" is duplicated and matched with another item in another array

tuple in 1st array in the ListA ('N1_10', 'N2_28')
tuple in 2nd array in the ListA ('N1_10', 'N3_98')

ListA  = [[('N1_10', 'N2_28'), ('N1_35', 'N2_44')],
          [('N1_22', 'N3_72'), ('N1_10', 'N3_98')],
          [('N2_33', 'N3_28'), ('N2_55', 'N3_62'), ('N2_61', 'N3_37')]]

what I want for the output is

output --> [('N1_10','N2_28','N3_98') , .... and the rest whatever match one of the key will get into same tuple]

If you guys think , changing the data structure of the ListA is better option , pls feel free to advise! Thanks for helping out!

SIMPLIFIED VERSION

List A = [[(a,x),(b,k),(c,l),(d,m)],[(e,d),(a,p),(g,s)],[...],[...]....]

wantedOutput --> [(a,x,p),(b,k),(c,l),(d,m,e),(g,s).....]

回答1:

Update: After rereading your question, it appears that you're trying to create equivalence classes, rather than collecting values for keys. If

[[(1, 2), (3, 4), (2, 3)]]

should become

[(1, 2, 3, 4)]

, then you're going to need to interpret your input as a graph and apply a connected components algorithm. You could turn your data structure into an adjacency list representation and traverse it with a breadth-first or depth-first search, or iterate over your list and build disjoint sets. In either case, your code is going to suddenly involve a lot of graph-related complexity, and it'll be hard to provide any output ordering guarantees based on the order of the input. Here's an algorithm based on a breadth-first search:

import collections

# build an adjacency list representation of your input
graph = collections.defaultdict(set)
for l in ListA:
    for first, second in l:
        graph[first].add(second)
        graph[second].add(first)

# breadth-first search the graph to produce the output
output = []
marked = set() # a set of all nodes whose connected component is known
for node in graph:
    if node not in marked:
        # this node is not in any previously seen connected component
        # run a breadth-first search to determine its connected component
        frontier = set([node])
        connected_component = []
        while frontier:
            marked |= frontier
            connected_component.extend(frontier)

            # find all unmarked nodes directly connected to frontier nodes
            # they will form the new frontier
            new_frontier = set()
            for node in frontier:
                new_frontier |= graph[node] - marked
            frontier = new_frontier
        output.append(tuple(connected_component))

Don't just copy this without understanding it, though; understand what it's doing, or write your own implementation. You'll probably need to be able to maintain this. (I would've used pseudocode, but Python is practically as simple as pseudocode already.)

In case my original interpretation of your question was correct, and your input is a collection of key-value pairs that you want to aggregate, here's my original answer:

Original answer

import collections

clusterer = collections.defaultdict(list)

for l in ListA:
    for k, v in l:
        clusterer[k].append(v)

output = clusterer.values()

defaultdict(list) is a dict that automatically creates a list as the value for any key that wasn't already present. The loop goes over all the tuples, collecting all values that match up to the same key, then creates a list of (key, value_list) pairs from the defaultdict.

(The output of this code is not quite in the form you specified, but I believe this form is more useful. If you want to change the form, that should be a simple matter.)

回答2:

Does output order matter? This is the simplest way I could think of:

ListA  = [[('N1_10', 'N2_28'), ('N1_35', 'N2_44')],[('N1_22', 'N3_72'), ('N1_10', 'N3_98')],
            [('N2_33', 'N3_28'), ('N2_55', 'N3_62'), ('N2_61', 'N3_37')]]

idx = dict()

for sublist in ListA:
    for pair in sublist:
        for item in pair:
            mapping = idx.get(item,set())
            mapping.update(pair)
            idx[item] = mapping 
            for subitem in mapping:
                submapping = idx.get(subitem,set())
                submapping.update(mapping)
                idx[subitem] = submapping


for x in set([frozenset(x) for x in idx.values()]):
    print list(x)

Output:

['N3_72', 'N1_22']
['N2_28', 'N3_98', 'N1_10']
['N2_61', 'N3_37']
['N2_33', 'N3_28']
['N2_55', 'N3_62']
['N2_44', 'N1_35']

回答3:

tupleList = [(1, 2), (3, 4), (1, 4), (3, 2), (1, 2), (7, 9), (9, 8), (5, 6)]

newSetSet = set ([frozenset (aTuple) for aTuple in tupleList])
setSet = set ()

while newSetSet != setSet:
    print '*'
    setSet = newSetSet
    newSetSet = set ()
    for set0 in setSet:
        merged = False
        for set1 in setSet:
            if set0 & set1 and set0 != set1:
                newSetSet.add (set0 | set1)
                merged = True
        if not merged:
            newSetSet.add (set0)

        print [tuple (element) for element in setSet]
        print [tuple (element) for element in newSetSet]
        print 

print [tuple (element) for element in newSetSet]

# Result:  [(1, 2, 3, 4), (5, 6), (8, 9, 7)]

来源：https://stackoverflow.com/questions/17482944/find-duplicate-items-within-a-list-of-list-of-tuples-python