Is wordnet path similarity commutative?

Question

I am using the wordnet API from nltk. When I compare one synset with another I got None but when I compare them the other way around I get a float value.

Shouldn't they give the same value? Is there an explanation or is this a bug of wordnet?

Example:

wn.synset('car.n.01').path_similarity(wn.synset('automobile.v.01')) # None
wn.synset('automobile.v.01').path_similarity(wn.synset('car.n.01')) # 0.06666666666666667

Buddhi

Answer 1

Technically without the dummy root, both car and automobile synsets would have no link to each other:

>>> from nltk.corpus import wordnet as wn
>>> x = wn.synset('car.n.01')
>>> y = wn.synset('automobile.v.01')
>>> print x.shortest_path_distance(y)
None
>>> print y.shortest_path_distance(x)
None

Now, let's look at the dummy root issue closely. Firstly, there is a neat function in NLTK that says whether a synset needs a dummy root:

>>> x._needs_root()
False
>>> y._needs_root()
True

Next, when you look at the path_similarity code (http://nltk.googlecode.com/svn-/trunk/doc/api/nltk.corpus.reader.wordnet-pysrc.html#Synset.path_similarity), you can see:

def path_similarity(self, other, verbose=False, simulate_root=True):
  distance = self.shortest_path_distance(other, \
               simulate_root=simulate_root and self._needs_root())

  if distance is None or distance < 0:
    return None
  return 1.0 / (distance + 1)

So for automobile synset, this parameter simulate_root=simulate_root and self._needs_root() will always be True when you try y.path_similarity(x) and when you try x.path_similarity(y) it will always be False since x._needs_root() is False:

>>> True and y._needs_root()
True
>>> True and x._needs_root()
False

Now when path_similarity() pass down to shortest_path_distance() (https://nltk.googlecode.com/svn/trunk/doc/api/nltk.corpus.reader.wordnet-pysrc.html#Synset.shortest_path_distance) and then to hypernym_distances(), it will try to call for a list of hypernyms to check their distances, without simulate_root = True, the automobile synset will not connect to the car and vice versa:

>>> y.hypernym_distances(simulate_root=True)
set([(Synset('automobile.v.01'), 0), (Synset('*ROOT*'), 2), (Synset('travel.v.01'), 1)])
>>> y.hypernym_distances()
set([(Synset('automobile.v.01'), 0), (Synset('travel.v.01'), 1)])
>>> x.hypernym_distances()
set([(Synset('object.n.01'), 8), (Synset('self-propelled_vehicle.n.01'), 2), (Synset('whole.n.02'), 8), (Synset('artifact.n.01'), 7), (Synset('physical_entity.n.01'), 10), (Synset('entity.n.01'), 11), (Synset('object.n.01'), 9), (Synset('instrumentality.n.03'), 5), (Synset('motor_vehicle.n.01'), 1), (Synset('vehicle.n.01'), 4), (Synset('entity.n.01'), 10), (Synset('physical_entity.n.01'), 9), (Synset('whole.n.02'), 7), (Synset('conveyance.n.03'), 5), (Synset('wheeled_vehicle.n.01'), 3), (Synset('artifact.n.01'), 6), (Synset('car.n.01'), 0), (Synset('container.n.01'), 4), (Synset('instrumentality.n.03'), 6)])

So theoretically, the right path_similarity is 0 / None , but because of the simulate_root=simulate_root and self._needs_root() parameter,

nltk.corpus.wordnet.path_similarity() in NLTK's API is not commutative.

BUT the code is also not wrong/bugged, since comparison of any synset distance by going through the root will be constantly far since the position of the dummy *ROOT* will never change, so the best of practice is to do this to calculate path_similarity:

>>> from nltk.corpus import wordnet as wn
>>> x = wn.synset('car.n.01')
>>> y = wn.synset('automobile.v.01')

# When you NEVER want a non-zero value, since going to 
# the *ROOT* will always get you some sort of distance 
# from synset x to synset y
>>> max(wn.path_similarity(x,y), wn.path_similarity(y,x))

# when you can allow None in synset similarity comparison
>>> min(wn.path_similarity(x,y), wn.path_similarity(y,x))

Sestina

Answer 2

I don't think it is a bug in wordnet per se. In your case, automobile is specified as a verb and car as noun, so you will need to look through the synset to see what the graph looks like and decide if the nets are labeled correctly.

A = 'car.n.01'
B = 'automobile.v.01'
C = 'automobile.n.01'


wn.synset(A).path_similarity(wn.synset(B)) 
wn.synset(B).path_similarity(wn.synset(A)) 


wn.synset(A).path_similarity(wn.synset(C)) # is 1
wn.synset(C).path_similarity(wn.synset(A)) # is also 1

Tombolo

In path_similarity there is a default variable simulate_root=True, that creates a dummy root in the tree of the taxonomies, thus verbs and nouns can connect with each other, thus the function should always return a value. If we dont create a dummy root then always verbs and nouns do not connect with each other e.g. wn.synset(A).path_similarity(wn.synset(B), simulate_root=False)) # None wn.synset(B).path_similarity(wn.synset(A), simulate_root=False)) # None – Buddhi

Also the way the path_similarity is defined doesn't take into account direction of edges, it is: 1/(1+length_of_shortest_path(A,B)) – Buddhi

Thank you for those comments. I'll update my answer accordingly – Tombolo

Actually, wn.synset('car.n.01') and wn.synset('automobile.n.01') refers to the same synset. That's why their path similarity is 1. Try: >>> wn.synsets('automobile') and you will see it. – Sestina