在Python中命名为NLTK的实体识别。标识NE
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我需要将单词分类为它们的词性。像动词，名词，副词等。
我用了
nltk.word_tokenize() #to identify word in a sentence
nltk.pos_tag() #to identify the parts of speech
nltk.ne_chunk() #to identify Named entities.
>>> sentence = \"I am Jhon from America\"
>>> sent1 = nltk.word_tokenize(sentence )
>>> sent2 = nltk.pos_tag(sent1)
>>> sent3 = nltk.ne_chunk(sent2, binary=True)
>>> sent3
Tree(\'S\', [(\'I\', \'PRP\'), (\'am\', \'VBP\'), Tree(\'NE\', [(\'Jhon\', \'NNP\')]), (\'from\', \'IN\'), Tree(\'NE\', [(\'America\', \'NNP\')])])
>>> sent3[0]
(\'I\', \'PRP\')
>>> sent3[0][0]
\'I\'
>>> sent3[0][1]
\'PRP\'
>>> sent3[2]
Tree(\'NE\', [(\'Jhon\', \'NNP\')])
>>> sent3[2][0]
(\'Jhon\', \'NNP\')
>>> sent3[2][1]
Traceback (most recent call last):
File \"<pyshell#121>\", line 1, in <module>
sent3[2][1]
File \"C:\\Python26\\lib\\site-packages\\nltk\\tree.py\", line 139, in __getitem__
return list.__getitem__(self, index)
IndexError: list index out of range
没有找到相关结果
已邀请:
5 个回复
掸牛浓疗
返回树的第一个子节点，而不是节点本身 编辑：我回到家时尝试过这种方法，确实可以工作。你换
chunks = ne_chunk(postags, binary=True) for c in chunks: if hasattr(c, \'node\'): myNE.append(\' \'.join(i[0] for i in c.leaves()))佬棠
for sent in chunked_sentences: for chunk in sent: if hasattr(chunk, \"label\"): print(chunk.label())屑凉赦
O / P-\'NE \' 我认为nltk中没有任何功能可以执行上述解决方案，但作为参考，您可以在这里查看 对于循环问题，你可以做：-for i in range(len(sent3)): if \"NE\" in str(sent3[i]): print sent3[i].node目浆搽