NLTK 分块并遍历结果树

Question

我正在使用 NLTK RegexpParser 从标记的标记中提取名词组和动词组。

如何遍历生成的树以仅查找 NP 或 V 组的块？

from nltk.chunk import RegexpParser

grammar = '''
NP: {<DT>?<JJ>*<NN>*}
V: {<V.*>}'''
chunker = RegexpParser(grammar)
token = [] ## Some tokens from my POS tagger
chunked = chunker.parse(tokens)
print chunked

#How do I walk the tree?
#for chunk in chunked:
#    if chunk.??? == 'NP':
#         print chunk

（S （NP 运营商/NN） 为/IN 组织-/JJ 和/CC 细胞培养/JJ 为/IN (NP/DT准备/NN) 的/IN （NP 植入物/NNS） 和/CC （NP 植入/NN） (含V/VBG) （NP/DT 运营商/NN） ./.)

Answer 1

令牌中的小错误

from nltk.chunk import RegexpParser
grammar = '''
NP: {<DT>?<JJ>*<NN>*}
V: {<V.*>}'''
chunker = RegexpParser(grammar)
token = [] ## Some tokens from my POS tagger
//chunked = chunker.parse(tokens) // token defined in the previous line but used tokens in chunker.parse(tokens)
chunked = chunker.parse(token) // Change in this line
print chunked

Answer 2

Savino 的答案很好，但值得注意的是，子树也可以通过索引访问，例如

for n in range(len(chunked)):
    do_something_with_subtree(chunked[n])

Answer 3

def preprocess(sent):
sent = nltk.word_tokenize(sent)
sent = nltk.pos_tag(sent)
return sent



pattern = 'NP: {<JJ>*<NNP.*>*}'
cp =   nltk.RegexpParser(pattern)
exp = []
for line in lines:
    line = preprocess(line)
    cs = cp.parse(line)
    for n in cs:
        if isinstance(n, nltk.tree.Tree):
            if n.label() == 'NP':
                if len(n.leaves()) > 1:
                    req = ''
                    for leaf in n.leaves():
                        req += leaf[0]+' '
                    exp.append(req)
print(exp)   

Answer 4

这应该有效：

for n in chunked:
    if isinstance(n, nltk.tree.Tree):               
        if n.label() == 'NP':
            do_something_with_subtree(n)
        else:
            do_something_with_leaf(n)