您还可以使用以下代码提取文本中每个名称实体的label ：

import nltk
for sent in nltk.sent_tokenize(sentence):
   for chunk in nltk.ne_chunk(nltk.pos_tag(nltk.word_tokenize(sent))):
      if hasattr(chunk, 'label'):
         print(chunk.label(), ' '.join(c[0] for c in chunk))

输出：

GPE WASHINGTON
GPE New York
PERSON Loretta E. Lynch
GPE Brooklyn

您可以看到Washington ， New York和Brooklyn是GPE意味着地缘政治实体

和Loretta E. Lynch是一个PERSON

使用nltk.chunk中的tree2conlltags。另外，ne_chunk还需要pos标记来标记单词标记（因此需要word_tokenize）。

from nltk import word_tokenize, pos_tag, ne_chunk
from nltk.chunk import tree2conlltags

sentence = "Mark and John are working at Google."
print(tree2conlltags(ne_chunk(pos_tag(word_tokenize(sentence))
"""[('Mark', 'NNP', 'B-PERSON'), 
    ('and', 'CC', 'O'), ('John', 'NNP', 'B-PERSON'), 
    ('are', 'VBP', 'O'), ('working', 'VBG', 'O'), 
    ('at', 'IN', 'O'), ('Google', 'NNP', 'B-ORGANIZATION'), 
    ('.', '.', 'O')] """

这将为您提供一个元组列表：[（token，pos_tag，name_entity_tag）]如果此列表不完全符合您的需求，从该列表中解析出您想要的列表当然比从nltk树更容易。

此链接的代码和详细信息;查看更多信息

您还可以继续使用以下功能提取单词：

def wordextractor(tuple1):

    #bring the tuple back to lists to work with it
    words, tags, pos = zip(*tuple1)
    words = list(words)
    pos = list(pos)
    c = list()
    i=0
    while i<= len(tuple1)-1:
        #get words with have pos B-PERSON or I-PERSON
        if pos[i] == 'B-PERSON':
            c = c+[words[i]]
        elif pos[i] == 'I-PERSON':
            c = c+[words[i]]
        i=i+1

    return c

print(wordextractor(tree2conlltags(nltk.ne_chunk(nltk.pos_tag(nltk.word_tokenize(sentence))))

编辑添加了输出文档字符串**编辑*添加了仅针对B人的输出

Tree是一个列表。块是子树，非块词是常规字符串。因此，让我们进入列表，从每个块中提取单词，然后将它们加入。

>>> chunked = nltk.ne_chunk(my_sent)
>>>
>>>  [ " ".join(w for w, t in elt) for elt in chunked if isinstance(elt, nltk.Tree) ]
['WASHINGTON', 'New York', 'Loretta E. Lynch', 'Brooklyn']

nltk.ne_chunk返回一个嵌套的nltk.tree.Tree对象，因此您必须遍历Tree对象才能到达NE。

看看带有正则表达式的命名实体识别：NLTK

>>> from nltk import ne_chunk, pos_tag, word_tokenize
>>> from nltk.tree import Tree
>>> 
>>> def get_continuous_chunks(text):
...     chunked = ne_chunk(pos_tag(word_tokenize(text)))
...     continuous_chunk = []
...     current_chunk = []
...     for i in chunked:
...             if type(i) == Tree:
...                     current_chunk.append(" ".join([token for token, pos in i.leaves()]))
...             elif current_chunk:
...                     named_entity = " ".join(current_chunk)
...                     if named_entity not in continuous_chunk:
...                             continuous_chunk.append(named_entity)
...                             current_chunk = []
...             else:
...                     continue
...     return continuous_chunk
... 
>>> my_sent = "WASHINGTON -- In the wake of a string of abuses by New York police officers in the 1990s, Loretta E. Lynch, the top federal prosecutor in Brooklyn, spoke forcefully about the pain of a broken trust that African-Americans felt and said the responsibility for repairing generations of miscommunication and mistrust fell to law enforcement."
>>> get_continuous_chunks(my_sent)
['WASHINGTON', 'New York', 'Loretta E. Lynch', 'Brooklyn']

您也可以考虑使用Spacy：

import spacy
nlp = spacy.load('en')

doc = nlp('WASHINGTON -- In the wake of a string of abuses by New York police officers in the 1990s, Loretta E. Lynch, the top federal prosecutor in Brooklyn, spoke forcefully about the pain of a broken trust that African-Americans felt and said the responsibility for repairing generations of miscommunication and mistrust fell to law enforcement.')

print([ent for ent in doc.ents])

>>> [WASHINGTON, New York, the 1990s, Loretta E. Lynch, Brooklyn, African-Americans]

当您将tree作为返回值时，我想您想选择那些标有NE子树。

这是一个简单的示例，用于将所有列表收集到列表中：

import nltk

my_sent = "WASHINGTON -- In the wake of a string of abuses by New York police officers in the 1990s, Loretta E. Lynch, the top federal prosecutor in Brooklyn, spoke forcefully about the pain of a broken trust that African-Americans felt and said the responsibility for repairing generations of miscommunication and mistrust fell to law enforcement."

parse_tree = nltk.ne_chunk(nltk.tag.pos_tag(my_sent.split()), binary=True)  # POS tagging before chunking!

named_entities = []

for t in parse_tree.subtrees():
    if t.label() == 'NE':
        named_entities.append(t)
        # named_entities.append(list(t))  # if you want to save a list of tagged words instead of a tree

print named_entities

这给出：

[Tree('NE', [('WASHINGTON', 'NNP')]), Tree('NE', [('New', 'NNP'), ('York', 'NNP')])]

或作为列表列表：

[[('WASHINGTON', 'NNP')], [('New', 'NNP'), ('York', 'NNP')]]

另请参阅： 如何导航nltk.tree.Tree？