NetworkX和Wordnet

Question

管理創建一個hypernym圖，但繼續獲得綁定方法Synset（'dog.n'）的Synset.name而不是dog.n在圖中。錯誤在哪裏？

from nltk.corpus import wordnet as wn 
import networkx as nx 
import matplotlib.pyplot as plt 

def closure_graph(synset, fn): 
    seen = set() 
    graph = nx.DiGraph() 

    def recurse(s): 
     if not s in seen: 
      seen.add(s) 
      graph.add_node(s.name) 
      for s1 in fn(s): 
       graph.add_node(s1.name) 
       graph.add_edge(s.name, s1.name) 
       recurse(s1) 

    recurse(synset) 
    return graph 


dog = wn.synsets('dog')[0] 
G = closure_graph(dog, 
         lambda s: s.hypernyms()) 
index = nx.betweenness_centrality(G) 
plt.rc('figure', figsize=(12, 7)) 
node_size = [index[n]*1000 for n in G] 
pos = nx.spring_layout(G) 
nx.draw_networkx(G, pos, node_size=node_size, edge_color='r', alpha=.3, linewidths=0) 
plt.show() 

編輯1：

成功地創建一個上位詞圖表，但保持在同義詞集合的獲得方法結合Synset.name（ 'dog.n'），而不是在圖形dog.n。錯誤在哪裏？

Answer 1

那麼打印出來的是networkx已經創建了一堆節點，每個節點都是一個功能。所以我們需要看看添加節點的命令。這發生在closure_graph。

在closure_graph的定義中，我們看到s.name被添加爲節點。這是一個函數（它可能用於實際上是在更新nltk之前的名稱）。相反，您想添加s.name()，該函數現在是一個返回名稱的函數。有4個地方發生這種情況。

from nltk.corpus import wordnet as wn 
import networkx as nx 
import matplotlib.pyplot as plt 

def closure_graph(synset, fn): 
    seen = set() 
    graph = nx.DiGraph() 

    def recurse(s): 
     if not s in seen: 
      seen.add(s) 
      graph.add_node(s.name()) 
      for s1 in fn(s): 
       graph.add_node(s1.name()) 
       graph.add_edge(s.name(), s1.name()) 
       recurse(s1) 

    recurse(synset) 
    return graph 


dog = wn.synsets('dog')[0] 
G = closure_graph(dog, 
         lambda s: s.hypernyms()) 
index = nx.betweenness_centrality(G) 
plt.rc('figure', figsize=(12, 7)) 
node_size = [index[n]*1000 for n in G] 
pos = nx.spring_layout(G) 
nx.draw_networkx(G, pos, node_size=node_size, edge_color='r', alpha=.3, linewidths=0) 
plt.show()