NLP-基础知识-003(词性标注)_isvbz

目标：词性标注
 
s = w1w2w3......wn 单词
 
z = (z1z2......zn)  词性
 
 
目的：argmax p(z|s)  -> Noisy Channel Model
 
= argmax p(s|z) p(z)
 
 
p(s|z) - Translation Model
 
p(z) - Language Model
 
= argmax p(w1w2...wn|z1z2....zn)p(z1z2....zn) (假设条件独立)
 
= argmax p(w1|z1) p(w2|z2) ..... p(wn|zn)p(z1)p(z2|z1)p(z3|z1z2)......
 
(马尔科夫假设)
 
= argmax Pi(i=1..n) P(wi|zi) * p(z1)p(z2|z1)p(z3|z2)......
 
=> argmax logpi(i=1...n)p(wi|zi)p(z1) pi(i=2..n)p(zj|zj-1)
= argmax sum(i=1..n) log p(wi|zi) + logp(z1) + sum(j=2..n)logp(zj|zj-1)
 
z' = argmax sum(i=1..n) log p(wi|zi) + logp(z1) + sum(j=2..n)logp(zj|zj-1)
 
 
 

"计算pi、A、B代码,traindata.txt文件数据见文章结尾"
 
tag2id, id2tag = {}, {}  # maps tag to id . tag2id: {"VB": 0, "NNP":1,..} , id2tag: {0: "VB", 1: "NNP"....}
word2id, id2word = {}, {} # maps word to id
 
for line in open('traindata.txt'):
    items = line.split('/')
    word, tag = items[0], items[1].rstrip()  # 抽取每一行里的单词和词性
    
    if word not in word2id:
        word2id[word] = len(word2id)
        id2word[len(id2word)] = word
    if tag not in tag2id:
        tag2id[tag] = len(tag2id)
        id2tag[len(id2tag)] = tag
 
M = len(word2id)  # M: 词典的大小、# of words in dictionary
N = len(tag2id)   # N: 词性的种类个数  # of tags in tag set
 
 
# 构建 pi, A, B
import numpy as np
pi = np.zeros(N)   # 每个词性出现在句子中第一个位置的概率,  N: # of tags  pi[i]: tag i出现在句子中第一个位置的概率
A = np.zeros((N, M)) # A[i][j]: 给定tag i, 出现单词j的概率。 N: # of tags M: # of words in dictionary
B = np.zeros((N,N))  # B[i][j]: 之前的状态是i, 之后转换成转态j的概率 N: # of tags
 
 
prev_tag = ""
for line in open('traindata.txt'):
    items = line.split('/')
    wordId, tagId = word2id[items[0]], tag2id[items[1].rstrip()]
    if prev_tag == "":  # 这意味着是句子的开始
        pi[tagId] += 1
        A[tagId][wordId] += 1
    else:  # 如果不是句子的开头
        A[tagId][wordId] += 1
        B[tag2id[prev_tag]][tagId] += 1
    
    if items[0] == ".":
        prev_tag = ""
    else:
        prev_tag = items[1].rstrip()
 
# normalize
pi = pi/sum(pi)
for i in range(N):
    A[i] /= sum(A[i])
    B[i] /= sum(B[i])
 
#  到此为止计算完了模型的所有的参数： pi, A, B
 
 
 

知道了pi、A、B，需要求出最优的z

维特比算法最终为一个动态规划寻找最优路径的问题，最终代码如下：

def log(v):
    if v == 0:
        return np.log(v+0.000001)
    return np.log(v)
 
def viterbi(x, pi, A, B):
    """
    x: user input string/sentence: x: "I like playing soccer"
    pi: initial probability of tags
    A: 给定tag, 每个单词出现的概率
    B: tag之间的转移概率
    """
    x = [word2id[word] for word in x.split(" ")]  # x: [4521, 412, 542 ..]
    T = len(x)
    
    dp = np.zeros((T,N))  # dp[i][j]: w1...wi, 假设wi的tag是第j个tag
    ptr = np.array([[0 for x in range(N)] for y in range(T)] ) # T*N
    # TODO: ptr = np.zeros((T,N), dtype=int)
    
    for j in range(N): # basecase for DP算法
        dp[0][j] = log(pi[j]) + log(A[j][x[0]])
    
    for i in range(1,T): # 每个单词
        for j in range(N):  # 每个词性
            # TODO: 以下几行代码可以写成一行（vectorize的操作， 会使得效率变高）
            dp[i][j] = -9999999
            for k in range(N): # 从每一个k可以到达j
                score = dp[i-1][k] + log(B[k][j]) + log(A[j][x[i]])
                if score > dp[i][j]:
                    dp[i][j] = score
                    ptr[i][j] = k
    
    # decoding: 把最好的tag sequence 打印出来
    best_seq = [0]*T  # best_seq = [1,5,2,23,4,...]  
    # step1: 找出对应于最后一个单词的词性
    best_seq[T-1] = np.argmax(dp[T-1])
    
    # step2: 通过从后到前的循环来依次求出每个单词的词性
    for i in range(T-2, -1, -1): # T-2, T-1,... 1, 0
        best_seq[i] = ptr[i+1][best_seq[i+1]]
        
    # 到目前为止, best_seq存放了对应于x的 词性序列
    for i in range(len(best_seq)):
        print (id2tag[best_seq[i]])
    
 

最终验证，输入一句话，可以得出对应的词性:

x = "Social Security number , passport number and details about the services provided for the payment"
print(viterbi(x, pi, A, B))
 
NNP
NNP
NN
,
NN
NN
CC
NNS
IN
DT
NNS
VBN
IN
DT
NN

traindata.txt 部分训练语料如下所示：
 
Newsweek/NNP
,/,
trying/VBG
to/TO
keep/VB
pace/NN
with/IN
rival/JJ
Time/NNP
magazine/NN
,/,
announced/VBD
new/JJ
advertising/NN
rates/NNS
for/IN
1990/CD
and/CC
said/VBD
it/PRP
will/MD
introduce/VB
a/DT
new/JJ
incentive/NN
plan/NN
for/IN
advertisers/NNS
./.
The/DT
new/JJ
ad/NN
plan/NN
from/IN
Newsweek/NNP
,/,
a/DT
unit/NN
of/IN
the/DT
Washington/NNP
Post/NNP
Co./NNP
,/,
is/VBZ
the/DT
second/JJ
incentive/NN
plan/NN
the/DT
magazine/NN
has/VBZ
offered/VBN
advertisers/NNS
in/IN
three/CD
years/NNS
./.
Plans/NNS
that/WDT
give/VBP
advertisers/NNS
discounts/NNS
for/IN
maintaining/VBG
or/CC
increasing/VBG
ad/NN
spending/NN
have/VBP
become/VBN
permanent/JJ
fixtures/NNS
at/IN
the/DT
news/NN
weeklies/NNS
and/CC
underscore/VBP
the/DT
fierce/JJ
competition/NN
between/IN
Newsweek/NNP
,/,
Time/NNP
Warner/NNP
Inc./NNP
's/POS
Time/NNP
magazine/NN
,/,
and/CC
Mortimer/NNP
B./NNP
Zuckerman/NNP
's/POS
U.S./NNP
News/NNP
&/CC
World/NNP
Report/NNP
./.
Alan/NNP
Spoon/NNP
,/,
recently/RB
named/VBN
Newsweek/NNP
president/NN
,/,
said/VBD
Newsweek/NNP
's/POS
ad/NN
rates/NNS
would/MD
increase/VB
5/CD
%/NN
in/IN
January/NNP
./.
A/DT
full/JJ
,/,
four-color/JJ
page/NN
in/IN
Newsweek/NNP
will/MD
cost/VB
$/$
100,980/CD
./.
In/IN
mid-October/NNP
,/,
Time/NNP
magazine/NN
lowered/VBD
its/PRP$
guaranteed/VBN
circulation/NN
rate/NN
base/NN
for/IN
1990/CD
while/IN
not/RB
increasing/VBG
ad/NN
page/NN
rates/NNS
;/:
with/IN
a/DT
lower/JJR
circulation/NN
base/NN
,/,
Time/NNP
's/POS
ad/NN
rate/NN
will/MD
be/VB
effectively/RB
7.5/CD
%/NN
higher/JJR
per/IN
subscriber/NN
;/:
a/DT
full/JJ
page/NN
in/IN
Time/NNP
costs/VBZ
about/IN
$/$
120,000/CD
./.
U.S./NNP
News/NNP
has/VBZ
yet/RB
to/TO
announce/VB
its/PRP$
1990/CD
ad/NN
rates/NNS
./.
Newsweek/NNP
said/VBD
it/PRP
will/MD
introduce/VB
the/DT
Circulation/NNP
Credit/NNP
Plan/NNP
,/,
which/WDT
awards/VBZ
space/NN
credits/NNS
to/TO
advertisers/NNS
on/IN
``/``
renewal/NN
advertising/NN
./.
''/''
The/DT
magazine/NN
will/MD
reward/VB
with/IN
``/``
page/NN
bonuses/NNS
''/''
advertisers/NNS
who/WP
in/IN
1990/CD
meet/VBP
or/CC
exceed/VBP
their/PRP$
1989/CD
spending/NN
,/,
as/RB
long/RB
as/IN
they/PRP
spent/VBD
$/$
325,000/CD
in/IN
1989/CD
and/CC
$/$
340,000/CD
in/IN
1990/CD
./.