有许多与MaltParser和/或NLTK相关的问题：
现在，在NLTK中有一个更稳定的MaltParser API版本：https://github.com/nltk/nltk/pull/944但是在解析多个句子的同时存在问题。
一次解析一句似乎很好：
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但是解析一个句子列表并没有返回DependencyGraph对象：
_path_to_maltparser = '/home/alvas/maltparser-1.8/dist/maltparser-1.8/'
_path_to_model= '/home/alvas/engmalt.linear-1.7.mco'
>>> mp = MaltParser(path_to_maltparser=_path_to_maltparser, model=_path_to_model)
>>> sent = 'I shot an elephant in my pajamas'.split()
>>> sent2 = 'Time flies like banana'.split()
>>> print(mp.parse_one(sent).tree())
(pajamas (shot I) an elephant in my)
为什么使用_path_to_maltparser = '/home/alvas/maltparser-1.8/dist/maltparser-1.8/'
_path_to_model= '/home/alvas/engmalt.linear-1.7.mco'
>>> mp = MaltParser(path_to_maltparser=_path_to_maltparser, model=_path_to_model)
>>> sent = 'I shot an elephant in my pajamas'.split()
>>> sent2 = 'Time flies like banana'.split()
>>> print(mp.parse_one(sent).tree())
(pajamas (shot I) an elephant in my)
>>> print(next(mp.parse_sents([sent,sent2])))
<listiterator object at 0x7f0a2e4d3d90>
>>> print(next(next(mp.parse_sents([sent,sent2]))))
[{u'address': 0,
u'ctag': u'TOP',
u'deps': [2],
u'feats': None,
u'lemma': None,
u'rel': u'TOP',
u'tag': u'TOP',
u'word': None},
{u'address': 1,
u'ctag': u'NN',
u'deps': [],
u'feats': u'_',
u'head': 2,
u'lemma': u'_',
u'rel': u'nn',
u'tag': u'NN',
u'word': u'I'},
{u'address': 2,
u'ctag': u'NN',
u'deps': [1, 11],
u'feats': u'_',
u'head': 0,
u'lemma': u'_',
u'rel': u'null',
u'tag': u'NN',
u'word': u'shot'},
{u'address': 3,
u'ctag': u'AT',
u'deps': [],
u'feats': u'_',
u'head': 11,
u'lemma': u'_',
u'rel': u'nn',
u'tag': u'AT',
u'word': u'an'},
{u'address': 4,
u'ctag': u'NN',
u'deps': [],
u'feats': u'_',
u'head': 11,
u'lemma': u'_',
u'rel': u'nn',
u'tag': u'NN',
u'word': u'elephant'},
{u'address': 5,
u'ctag': u'NN',
u'deps': [],
u'feats': u'_',
u'head': 11,
u'lemma': u'_',
u'rel': u'nn',
u'tag': u'NN',
u'word': u'in'},
{u'address': 6,
u'ctag': u'NN',
u'deps': [],
u'feats': u'_',
u'head': 11,
u'lemma': u'_',
u'rel': u'nn',
u'tag': u'NN',
u'word': u'my'},
{u'address': 7,
u'ctag': u'NNS',
u'deps': [],
u'feats': u'_',
u'head': 11,
u'lemma': u'_',
u'rel': u'nn',
u'tag': u'NNS',
u'word': u'pajamas'},
{u'address': 8,
u'ctag': u'NN',
u'deps': [],
u'feats': u'_',
u'head': 11,
u'lemma': u'_',
u'rel': u'nn',
u'tag': u'NN',
u'word': u'Time'},
{u'address': 9,
u'ctag': u'NNS',
u'deps': [],
u'feats': u'_',
u'head': 11,
u'lemma': u'_',
u'rel': u'nn',
u'tag': u'NNS',
u'word': u'flies'},
{u'address': 10,
u'ctag': u'NN',
u'deps': [],
u'feats': u'_',
u'head': 11,
u'lemma': u'_',
u'rel': u'nn',
u'tag': u'NN',
u'word': u'like'},
{u'address': 11,
u'ctag': u'NN',
u'deps': [3, 4, 5, 6, 7, 8, 9, 10],
u'feats': u'_',
u'head': 2,
u'lemma': u'_',
u'rel': u'dep',
u'tag': u'NN',
u'word': u'banana'}]
不会返回parse_sents()的可迭代内容？
parse_one但这不是我正在寻找的解决方案。 我的问题是如何解答为什么_path_to_maltparser = '/home/alvas/maltparser-1.8/dist/maltparser-1.8/'
_path_to_model= '/home/alvas/engmalt.linear-1.7.mco'
>>> mp = MaltParser(path_to_maltparser=_path_to_maltparser, model=_path_to_model)
>>> sent1 = 'I shot an elephant in my pajamas'.split()
>>> sent2 = 'Time flies like banana'.split()
>>> sentences = [sent1, sent2]
>>> for sent in sentences:
>>> ... print(mp.parse_one(sent).tree())
不会返回parse_sent()的可迭代内容。怎么能在NLTK代码中修复？
在@NikitaAstrakhantsev回答之后，我现在尝试输出一个解析树，但它似乎很困惑，并且在解析之前将两个句子放在一起。
parse_one()[OUT]：
# Initialize a MaltParser object with a pre-trained model.
mp = MaltParser(path_to_maltparser=path_to_maltparser, model=path_to_model)
sent = 'I shot an elephant in my pajamas'.split()
sent2 = 'Time flies like banana'.split()
# Parse a single sentence.
print(mp.parse_one(sent).tree())
print(next(next(mp.parse_sents([sent,sent2]))).tree())
从代码中看起来似乎做了一些奇怪的事情：https://github.com/nltk/nltk/blob/develop/nltk/parse/api.py#L45
为什么NLTK中的解析器抽象类在解析之前将两个句子拼凑成一个？我是否错误地呼叫了(pajamas (shot I) an elephant in my)
(shot I (banana an elephant in my pajamas Time flies like))
？如果是这样，调用parse_sents()的正确方法是什么？
答案 0 :(得分：5)
正如我在您的代码示例中看到的那样，您不能在此行中致电tree()
>>> print(next(next(mp.parse_sents([sent,sent2]))))
虽然您在所有情况下都使用tree()致电parse_one()。
否则，我不明白为什么会发生这种情况：parse_one() ParserI MaltParser的方法未在parse_sents()中覆盖，而其所做的一切只是调用{ {1}}的{1}}，请参阅the code。
更新： The line you're talking about未被调用，因为MaltParser会在parse_sents()中被覆盖并被直接调用。
我现在唯一的猜测是java lib maltparser对包含几个句子的输入文件（我的意思是this block - 运行java）没有正常工作。也许原始麦芽解析器已经改变了格式，现在它不是MaltParser。
不幸的是，我不能自己运行此代码，因为maltparser.org第二天就会停止运行。我检查了输入文件是否具有预期的格式（句子由双端线分隔），因此python包装器合并句子的可能性很小。