CODEWITHSUNDEEP
python
Abhishek Tumuluru
Abhishek Tumuluru

Reputation: 155

Removing duplicate edges from graph in Python list

My program returns a list of tuples, which represent the edges of a graph, in the form of:

[(i, (e, 130)), (e, (i, 130)), (g, (a, 65)), (g, (d, 15)), (a, (g, 65))]

So, (i, (e, 130)) means that 'i' is connected to 'e' and is 130 units away.

Similarly, (e, (i, 130)) means that 'e' is connected to 'i' and is 130 units away. So essentially, both these tuples represent the same thing.

How would I remove any one of them from this list? Desired output:

[(i, (e, 130)), (g, (a, 65)), (g, (d, 15))]

I tried writing an equals function. Would this be of any help?

def edge_equal(edge_tuple1, edge_tuple2):
    return edge_tuple1[0] == edge_tuple2[1][0] and edge_tuple2[0] == edge_tuple1[1][0]

Upvotes: 10

Views: 5259

Answers (4)

Michael Hoff
Michael Hoff

Reputation: 6358

If a tuple (n1, (n2, distance)) represents a bidirectional connection, I would introduce a normalization property which constraints the ordering of the two nodes in the tuple. This way, each possible edge has exactly one unique representation.

Consequently, a normalization function would map a given, possibly un-normalized, edge to the normalized variant. This function can then be used to normalize all given edges. Duplicates can now be eliminated in several ways. For instance, convert the list to a set.

def normalize(edge):
    n1, (n2, dist) = edge
    if n1 > n2: # use a custom compare function if desired
        n1, n2 = n2, n1
    return (n1, (n2, dist))

edges = [('i', ('e', 130)), ('e', ('i', 130)), ('g', ('a', 65)), ('g', ('d', 15)), ('a', ('g', 65))]

unique_edges = set(map(normalize, edges))

# set([('e', ('i', 130)), ('d', ('g', 15)), ('a', ('g', 65))])

Upvotes: 8

Moses Koledoye
Moses Koledoye

Reputation: 78564

Reconstruct each edge to take its alternate form and check if the alternate form is already in a new set. If it is not then add to the set:

lst = [('i', ('e', 130)), ('e', ('i', 130)), ('g', ('a', 65)), ('g', ('d', 15)), ('a', ('g', 65))]

r = set()
for e, v in lst:
    if (v[0], (e, v[1])) in r:
        continue
    r.add((e, v))

print(list(r))
# [('i', ('e', 130)), ('g', ('a', 65)), ('g', ('d', 15))]

Upvotes: 3

be_good_do_good
be_good_do_good

Reputation: 4441

edges = [(i, (e, 130)), (e, (i, 130)), (g, (a, 65)), (g, (d, 15)), (a, (g, 65))]

for each in edges:
    try:
        edges.remove((each[1][0], (each[0], each[1][1])))
    except ValueError:
        pass

reverse the vectors and remove them as you traverse

Upvotes: 2

James
James

Reputation: 2741

The simplest solution to write would be simply to iterator and check equality of all of them:

def edge_equal(edge_tuple1, edge_tuple2):
        return edge_tuple1[0] == edge_tuple2[1][0] and edge_tuple2[0] == edge_t\
uple1[1][0]

new = []

for i in range(len(graph)):
    found_equal = False
    for e in range(i,len(graph)):
        if edge_equal(graph[i],graph[e]):
            found_equal = True
            break
    if not found_equal:
        new.append(graph[i])

print new

Upvotes: 2

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