Question

我正在使用nltk的wordnet API。当我将一个synset与另一个synset进行比较时，得到了None，但是当我比较它们时，我得到一个浮点值。

它们不应该给出相同的价值吗？有没有解释或者这是wordnet的错误？

示例：

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

Answer 1

从技术上讲，没有虚拟根，car和automobile同义词集之间没有链接：

>>> 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

现在，让我们仔细看看虚拟根问题。首先，NLTK中有一个简洁的函数，说明一个synset是否需要一个虚拟根：

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

接下来，当您查看path_similarity代码（http://nltk.googlecode.com/svn-/trunk/doc/api/nltk.corpus.reader.wordnet-pysrc.html#Synset.path_similarity）时，您会看到：

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)

因此对于automobile synset，当您尝试simulate_root=simulate_root and self._needs_root()时，此参数True始终为y.path_similarity(x)，当您尝试x.path_similarity(y)时，它始终为False 1}}因为x._needs_root()是False：

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

现在当path_similarity()传递给shortest_path_distance()（https://nltk.googlecode.com/svn/trunk/doc/api/nltk.corpus.reader.wordnet-pysrc.html#Synset.shortest_path_distance）然后传递给hypernym_distances()时，它会尝试调用上位词列表来检查它们的距离，而不会simulate_root = True automobile同义词不会连接到car，反之亦然：

>>> 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)])

理论上，右path_similarity为0 / None，但由于simulate_root=simulate_root and self._needs_root()参数，

NLTK API中的

nltk.corpus.wordnet.path_similarity()不可交换。

但是代码也没有错误/错误，因为通过遍历根的任何synset距离的比较将会很远，因为虚拟*ROOT*的位置永远不会改变，所以最好的做法是这样做来计算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))

Answer 2

我认为这不是wordnet本身的错误。在您的情况下，汽车被指定为动词和汽车作为名词，因此您需要查看同义词集以查看图表的样子并确定网络是否正确标记。

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

wordnet路径相似性是否可交换？

2 个答案: