WordNet层次结构的索引级别

如何解决WordNet层次结构的索引级别

我想在WordNet名词数据集中找到每个节点的“级别”。根据{{​​3}}，“ entity.n.01”是根节点，因此我们只需要找到每个节点到此根的距离即可。但是，当我运行此代码时，会得到一些意外的答案。例如，

使用他们的入门代码，我编写了以下内容：

#import data
import re
import pandas
from nltk.corpus import wordnet as wn
from tqdm import tqdm
try:
    wn.all_synsets
except LookupError as e:
    import nltk
    nltk.download('wordnet')

# make sure each edge is included only once
edges = set()
for synset in tqdm(wn.all_synsets(pos='n')):
    # write the transitive closure of all hypernyms of a synset to file
    for hyper in synset.closure(lambda s: s.hypernyms()):
        edges.add((synset.name(),hyper.name()))
# also write transitive closure for all instances of a synset
for instance in synset.instance_hyponyms():
    for hyper in instance.closure(lambda s: s.instance_hypernyms()):
        edges.add((instance.name(),hyper.name()))
        for h in hyper.closure(lambda s: s.hypernyms()):
            edges.add((instance.name(),h.name()))

# dataframe of nouns
nouns = pandas.DataFrame(list(edges),columns=['id1','id2'])

#collect edges
edge_list = pd.DataFrame(columns=['child','parent'])
child_list = []
parent_list = []

for row in range(nouns.shape[0]):
  if row > 0 and row % 100000 == 0:
    print(f'row {row}/{nouns.shape[0]}')

  w1 = nouns.iloc[row]['id1']
  w2 = nouns.iloc[row]['id2']

  ss_w1 = wn.synset(w1)
  ss_w2 = wn.synset(w2)

  level_w1 = min([len(path) for path in ss_w1.hypernym_paths()])
  level_w2 = min([len(path) for path in ss_w2.hypernym_paths()])

  child_list.append(f'{w1}.L{level_w1}')
  parent_list.append(f'{w2}.L{level_w2}')

这给了我一个看起来像这样的数据框：

this link

我在每个名词后面加上了相关的等级，以“ L [level]”的形式出现在边缘集中。但是出了点问题。如果我们查看最后一个条目（行743085），则会将单词“ bulimia.n.02”分配给级别9，即，它距根节点9步之遥，而其父节点显然是“ entity.n.01”这是根节点本身。

我做错了什么？