如何获得在nltk wordnet中共享特定最低常见上限的所有同音异义词？

Question

鉴于有一条来自两个常见同义词集的路径来获得最低常见的上位词，看起来应该有一些回过头来找到导致这个上位词的下位词似乎是合理的。

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)查找上位词及其距离：

>>> 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}

现在找到最小路径：

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

现在，这很无聊，因为california和alaska是WordNet层次结构中的姐妹节点。让我们过滤掉所有姐妹节点：

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

要获得american_state的子节点（我认为这是“你需要的东西”......）：

>>> 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')]

这可能看起来令人震惊但实际上，没有为alaska或california指示的上位词：

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

使用_shortest_hypernym_paths建立的连接是通过虚拟根，看看Is wordnet path similarity commutative?

Answer 2

更新的解决方案是：

alaska = wordnet.synset('Alaska.n.1')
california = wordnet.synset('California.n.1')
alaska.lowest_common_hypernyms(california)

[同义词集（ 'american_state.n.01'）]

这个旧功能是私有的，不会以这种方式工作，也许其他，但无论如何，你也可以选择x.common.hypernyms(y)来查找所有常见项目。