如何獲得在ntlk wordnet中共享特定最低公共上位詞的所有下位詞？

Question

由於有兩個常見的同義詞集獲得最低的共同上位詞路徑，它似乎是合理的應該是要想辦法往回走，發現導致該上位詞

from nltk.corpus import wordnet as wn 
alaska = wn.synset('Alaska.n.1') 
california = wn.synset('California.n.1') 
common_hypernym = alaska.lowest_common_hypernyms(california)[0] 

common_hypernym 
Synset('american_state.n.01') 

common_hypernym.do_something_awesome() 
['Alabama.n.1', 'Alaska.n.1', ...] #all 50 american states 

Answer 1

使用Synset1._shortest_path_distance(Synset2)找到上位詞的下義詞和他們的距離：

>>> from nltk.corpus import wordnet as wn 
>>> alaska = wn.synset('Alaska.n.1') 
>>> california = wn.synset('California.n.1') 

>>> alaska._shortest_hypernym_paths(california) 
{Synset('district.n.01'): 4, Synset('location.n.01'): 6, Synset('region.n.03'): 5, Synset('physical_entity.n.01'): 8, Synset('entity.n.01'): 9, Synset('state.n.01'): 2, Synset('administrative_district.n.01'): 3, Synset('object.n.01'): 7, Synset('alaska.n.01'): 0, Synset('*ROOT*'): 10, Synset('american_state.n.01'): 1} 

現在找到最小路徑：

>>> paths = alaska._shortest_hypernym_paths(california) 
>>> min(paths, key=paths.get) 
Synset('alaska.n.01') 

現在，這是無聊，因爲california和alaska是WordNet層次結構中的姐妹節點。讓我們過濾掉所有的姐妹節點：

>>> paths = {k:v for k,v in paths.items() if v > 0} 
>>> min(paths, key=paths.get) 
Synset('american_state.n.01') 

爲了得到american_state的子節點（我認爲這是「東西真棒」你需要...）：

>>> min(paths, key=paths.get).hyponyms() 
[Synset('free_state.n.02'), Synset('slave_state.n.01')] 
>>> list(min(paths, key=paths.get).closure(lambda s:s.hyponyms())) 
[Synset('free_state.n.02'), Synset('slave_state.n.01')] 

這可能看起來令人震驚但實際上，有沒有爲alaska或california表示上位：

>>> alaska.hypernyms() 
[] 
>>> california.hypernyms() 
[] 

並採用_shortest_hypernym_paths是連接製成由我ans虛擬根，看看Is wordnet path similarity commutative?