匹配 POS 标签和单词序列

Question

我有以下两个字符串及其 POS 标签：

Sent1：“像 writer pro 或 phramology 那样的东西真的很酷。”

[('something', 'NN'), ('like', 'IN'), ('how', 'WRB'), ('writer', 'NN'), ('pro', 'NN'), ('or', 'CC'), ('phraseology', 'NN'), ('works', 'NNS'), ('would', 'MD'), ('be', 'VB'), ('really', 'RB'), ('cool', 'JJ'), ('.', '.')]

Sent2：“语法编辑器等更多选项会更好”

[('more', 'JJR'), ('options', 'NNS'), ('like', 'IN'), ('the', 'DT'), ('syntax', 'NN'), ('editor', 'NN'), ('would', 'MD'), ('be', 'VB'), ('nice', 'JJ')]

我正在寻找一种方法来检测 (return True) 是否存在以下序列："would" + be" + 形容词（无论形容词的位置如何，只要其在这些字符串中 "would" "be") 之后。在第二个字符串中，形容词 "nice" 紧跟在 "would be" 之后，但在第一个字符串中并非如此。

琐碎的情况（形容词前没有其他词；"would be nice"）在我之前的一个问题中得到解决：

我现在正在寻找一个更通用的解决方案，其中可选词可以出现在形容词之前。我是 NLTK 和 Python.

的新手

Answer 1

您似乎只需搜索连续标签 "would"，然后搜索 "be"，然后搜索标签 "JJ" 的第一个实例。像这样：

import nltk

def has_would_be_adj(S):
    # make pos tags
    pos = nltk.pos_tag(S.split())
    # Search consecutive tags for "would", "be"
    j = None  # index of found "would"
    for i, (x, y) in enumerate(zip(pos[:-1], pos[1:])):
        if x[0] == "would" and y[0] == "be":
            j = i
            break
    if j is None or len(pos) < j + 2:
        return False
    a = None  # index of found adjective
    for i, (word, tag) in enumerate(pos[j + 2:]):
        if tag == "JJ":
            a = i+j+2 #
            break
    if a is None:
        return False
    print("Found adjective {} at {}", pos[a], a)
    return True

S = "something like how writer pro or phraseology works would be really cool."
print(has_would_be_adj(S))

我敢肯定这可以写得更紧凑、更简洁，但它确实符合包装盒上的说明:)

Answer 2

勾选

from nltk.tokenize import word_tokenize
def would_be(tagged):
    return any(['would', 'be', 'JJ'] == [tagged[i][0], tagged[i+1][0], tagged[i+2][1]] for i in xrange(len(tagged) - 2))

S = "more options like the syntax editor would be nice."  
pos = nltk.pos_tag(word_tokenize(S)) 
would_be(pos)

同时检查代码

from nltk.tokenize import word_tokenize
import nltk
def checkTag(S):
    pos = nltk.pos_tag(word_tokenize(S))
    flag = 0
    for tag in pos:
        if tag[1] == 'JJ':
            flag = 1
    if flag:
        for ind,tag in enumerate(pos):
            if tag[0] == 'would' and pos[ind+1][0] == 'be':
                    return True
        return False
    return False

S = "something like how writer pro or phraseology works would be really cool."
print checkTag(S)

Answer 3

from itertools import tee,izip,dropwhile
import nltk
def check_sentence(S):
    def pairwise(iterable):
        "s -> (s0,s1), (s1,s2), (s2, s3), ..."
        a, b = tee(iterable)
        next(b, None)
        return izip(a, b)


    def consecutive_would_be(word_group):
        first, second = word_group
        (would_word, _) = first
        (be_word, _) = second
        return would_word.lower() != "would" && be_word.lower() != "be"


    for word_groups in dropwhile(consecutive_would_be, pairwise(nltk.pos_tag(nltk.word_tokenize(S))):
        first, second = word_groups
        (_, pos1) = first
        (_, pos2) = second
        if pos1 == "JJ" || pos2 == "JJ":
            return True
    return False

然后你可以像这样使用函数:

S = "more options like the syntax editor would be nice."  
check_sentence(S)

Answer 4

首先按照说明安装nltk_cli：https://github.com/alvations/nltk_cli

那么，这里是nltk_cli中的一个秘密函数，也许你会发现它有用：

alvas@ubi:~/git/nltk_cli$ cat infile.txt 
something like how writer pro or phraseology works would be really cool .
more options like the syntax editor would be nice
alvas@ubi:~/git/nltk_cli$ python senna.py --chunk2 VP+ADJP infile.txt 
would be    really cool
would be    nice

为了说明其他可能的用法：

alvas@ubi:~/git/nltk_cli$ python senna.py --chunk2 VP+VP infile.txt 
!!! NO CHUNK of VP+VP in this sentence !!!
!!! NO CHUNK of VP+VP in this sentence !!!
alvas@ubi:~/git/nltk_cli$ python senna.py --chunk2 NP+VP infile.txt 
how writer pro or phraseology works would be
the syntax editor   would be
alvas@ubi:~/git/nltk_cli$ python senna.py --chunk2 VP+NP infile.txt 
!!! NO CHUNK of VP+NP in this sentence !!!
!!! NO CHUNK of VP+NP in this sentence !!!

然后如果你想检查句子中的短语和输出 True/False，只需读取并遍历 nltk_cli 的输出并检查 if-else 条件。

Answer 5

这会有帮助吗？

s1=[('something', 'NN'), ('like', 'IN'), ('how', 'WRB'), ('writer', 'NN'), ('pro', 'NN'), ('or', 'CC'), ('phraseology', 'NN'), ('works', 'NNS'), ('would', 'MD'), ('be', 'VB'), ('really', 'RB'), ('cool', 'JJ'), ('.', '.')]

flag = True
for i,j in zip(s1[:-1],s1[1:]):
    if i[0]+" "+j[0] == "would be":
        flag = True
    if flag and (i[-1] == "JJ" or j[-1] == "JJ"):
        print "would be adjective found in the tagged string"

匹配 POS 标签和单词序列

match POS tag and sequence of words

python

nlp

nltk

pos-tagger