NLTK的默认标记器nltk.word_tokenizer链接两个标记器,一个句子标记器,然后一个对句子操作的单词标记器。开箱就干得很好。
>>> nltk.word_tokenize("(Dr. Edwards is my friend.)")
['(', 'Dr.', 'Edwards', 'is', 'my', 'friend', '.', ')']除了让它将偏移元组返回到原始字符串而不是字符串标记之外,我还想使用相同的算法。
我所说的偏移量是指两个可以作为原始字符串索引的值。例如,在这里我会
>>> s = "(Dr. Edwards is my friend.)"
>>> s.token_spans()
[(0,1), (1,4), (5,12), (13,15), (16,18), (19,25), (25,26), (26,27)]因为s0:1是"(",s1:4是“Dr..”)以此类推。
是否有一个NLTK调用可以做到这一点,还是我必须编写自己的偏移算法?
发布于 2020-08-29 14:09:39
NLTK3.5版的TreebankWordDetokenizer支持span_tokenize()函数,因此不再需要编写自己的偏移算法:
>>> from nltk.tokenize import TreebankWordTokenizer
>>> s = '''Good muffins cost $3.88\\nin New (York). Please (buy) me\\ntwo of them.\\n(Thanks).'''
>>> expected = [(0, 4), (5, 12), (13, 17), (18, 19), (19, 23),
... (24, 26), (27, 30), (31, 32), (32, 36), (36, 37), (37, 38),
... (40, 46), (47, 48), (48, 51), (51, 52), (53, 55), (56, 59),
... (60, 62), (63, 68), (69, 70), (70, 76), (76, 77), (77, 78)]
>>> list(TreebankWordTokenizer().span_tokenize(s)) == expected
True
>>> expected = ['Good', 'muffins', 'cost', '$', '3.88', 'in',
... 'New', '(', 'York', ')', '.', 'Please', '(', 'buy', ')',
... 'me', 'two', 'of', 'them.', '(', 'Thanks', ')', '.']
>>> [s[start:end] for start, end in TreebankWordTokenizer().span_tokenize(s)] == expected
Truehttps://stackoverflow.com/questions/28678318
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