Question

我有一个句子列表列表，用单词标记，然后用pos标记，所以结果显然是一个包含元素的列表：

[(w1,pos_tag1)(w2,pos_tag2)]
[(w3,pos_tag3),(w4,pos_tag4),(w5,pos_tag5)]
[(w6,pos_tag6),(w7,pos_tag7)]

我需要得到一个pos_tags列表，其顺序与所有句子中的顺序相同。我尝试的是列表上的迭代

tags = [x [1] for x in in element in list]

但这不起作用。如何在这些列表中包含所有标签？

感谢

Answer 1

您可以使用zip(*list)惯用法解包元组列表，请参阅Unpacking a list / tuple of pairs into two lists / tuples

>>> from nltk import pos_tag
>>> tagged_sent = pos_tag('The quick brown fox jumps over the lazy dog'.split()) 
>>> tagged_sent
[('The', 'DT'), ('quick', 'JJ'), ('brown', 'NN'), ('fox', 'NN'), ('jumps', 'VBZ'), ('over', 'IN'), ('the', 'DT'), ('lazy', 'JJ'), ('dog', 'NN')]
>>> words, tags = zip(*tagged_sent)
>>> tags
('DT', 'JJ', 'NN', 'NN', 'VBZ', 'IN', 'DT', 'JJ', 'NN')

Answer 2

假设您有一个单词 - 标签对列表列表：

tagged_sentences = [[(w1, t1), (w2, t2), ...], [(w5, t5), ...],...]

您可以执行以下操作来获取标记列表的列表：

> tags = [[tag for word, tag in sent] for sent in tagged_sentences]
# tags = [[x[1] for x in sent] for sent in tagged_sentences]
[[t1, t2, ...], [t5, ...], ...]

如果你想平整列表，即获得所有句子中所有标签的平面列表：

> tags = [tag for sent in tagged_sentences for word, tag in sent]
# tags = [x[1] for sent in tagged_sentences for x in sent]
[t1, t2, ..., t5, ...]

仅从（word，pos_tag）的元组获取POS标签列表

2 个答案: