词性标注-隐马尔可夫模型

发布时间：2023-02-02 09:00

词性标注-隐马尔可夫模型

今天刚学完条件随机场，然后想找个例子实战一下，写写代码。于是关注到了词性标注，因为在学习的过程中，很多博客、视频、书等等都常常提到词性标注作为讲解的例子。然后我突然想到前面学HMM的过程中虽然利用盒子与球模型实现了HMM的算法，但是还没有用一个更广泛，更实际的例子去实现它，于是决定先用HMM实现词性标注。

流程很简单，先利用现有的语料库训练HMM模型，这是学习问题，然后利用viterbi算法做任意输入的句子做词性标注，即解码问题

我们先来看监督式学习的情况

语料来自于https://github.com/junman/POS-tagging，利用并进一步封装了其中读取文件的代码以忽略与模型无关的细节

如下是dataset.py的文件内容，只关注模型的读者直接使用即可

import numpy as np

def read_dataset():
    state_list = ['Ag', 'a', 'ad', 'an', 'Bg', 'b', 'c', 'Dg',
                'd', 'e', 'f', 'h', 'i', 'j', 'k', 'l',
                'Mg', 'm', 'Ng', 'n', 'nr', 'ns', 'nt', 'nx',
                'nz', 'o', 'p', 'q', 'Rg', 'r', 's','na',
                'Tg', 't','u', 'Vg', 'v', 'vd', 'vn','vvn',
                'w', 'Yg', 'y', 'z']

    data = []
    word_count = {}

    with open('corpus_POS.txt') as filess:
        for line in filess:  #一次处理全部的行？？列表可能会影响效率？？
            vocabs=[]
            classifies=[]
            line=line.strip()
            if not line:
                continue
            words=line.split(" ")#分解为多个单词
            for word in words:
                position=word.index('/')  #如果是[中国人民/n]
                if '[' in word and ']' in word:
                    vocabs.append(word[1:position])
                    classifies.append(word[position+1:-1])
                    if word[1:position] in word_count.keys():
                        word_count[word[1:position]] +=1
                    else:
                        word_count[word[1:position]] = 0
                    break
                if  '[' in word:
                    vocabs.append(word[1:position])
                    classifies.append(word[position+1:])
                    if word[1:position] in word_count.keys():
                        word_count[word[1:position]] +=1
                    else:
                        word_count[word[1:position]] = 0
                    break
                if ']' in  word:
                    vocabs.append(word[:position])
                    classifies.append(word[position+1:-1])
                    if word[:position] in word_count.keys():
                        word_count[word[:position]] +=1
                    else:
                        word_count[word[:position]] = 0
                    break
                vocabs.append(word[:position])
                classifies.append(word[position+1:])
                if word[:position] in word_count.keys():
                    word_count[word[:position]] +=1
                else:
                    word_count[word[:position]] = 0

            if  len(vocabs)!=len(classifies):
                print('词汇数量与类别数量不一致')
                break  #不一致退出程序
            else:
                data.append((vocabs,classifies))

    # 我们需要将词汇和词性转化成数字编码
    observation_list = word_count.keys()

    word_index = { word : i for i,word in enumerate(observation_list)}
    state_index = { pos : i for i,pos in enumerate(state_list)}

    # 遍历所有句子
    pad_length = 30
    X_data = []
    y_data = []
    for vocabs,classifies in data:
        digit_vocabs = []
        digit_classifies = []
        for i in range(min(len(vocabs),pad_length)):
            digit_vocabs.append(word_index[vocabs[i]])
        for i in range(min(len(vocabs),pad_length)):
            digit_classifies.append(state_index[classifies[i]])
        if (len(digit_vocabs)<pad_length):
            for i in range(pad_length-len(digit_vocabs)):
                digit_vocabs.append(len(observation_list))
                digit_classifies.append(len(state_list))
        X_data.append(digit_vocabs)
        y_data.append(digit_classifies)

    X_data = np.array(X_data)
    y_data = np.array(y_data)
    return X_data,y_data,word_index,state_index

然后就是利用我在白板推导系列Pytorch-隐马尔可夫模型-学习问题以及白板推导系列Pytorch-隐马尔可夫模型-解码问题两篇博客中实现的极大似然估计定义的模型代码和viterbi解码的代码，训练数据传进去，跑一跑，解个码就结束了

是这样吗？

我起初以为是这样的，结果现实狠狠的给了我一巴掌。

我这才知道之前写的代码存在多大的问题。

如果你看过我上面两篇文章，你用里面的代码去训练模型，你会发现你似乎永远看不到代码运行结束？这是为什么？监督式学习的代码之前在盒子和球模型不是表现挺好吗？模型参数什么的预测都挺准。但是有一个问题，我们当时用的盒子与球模型只有四个状态，两种观测，所以我们直接把A，B写成矩阵，一个一个算，模型仍然很快训练完成，因为整个模型只有 $4 + 4 * 4 + 4 * 2 = 28$ 个参数。

但是现在不一样了。

给定的语料中，状态（词性）有44个，观测（单词）有46571个，总共有 $44 + 44 * 44 + 44 * 46571$ 总共2051104个参数。另外，我们定义的发射矩阵维度是44x46571，但其中大部分元素都是0，用这么庞大的空间存储这么一点点有效的数据，我亏的慌。

所以看到github上的代码后，我改啊改，改啊改，终于把它改成了库里面的形状

修改前

def train(self,data):
    # data:Sx2xT
    # [[[i1,i2,...],[o1,o2,...]],[],[]]
    self.pi = np.zeros(shape=(self.N,))
    self.A = np.zeros(shape=(self.N,self.N))
    self.B = np.zeros(shape=(self.N,self.M))

    S = len(data)
    self.T = len(data[0][0])
    # 求 pi
    for i in range(self.N):
        for j in range(S):
            self.pi[i] += data[j][0][0]==i
        self.pi[i] = self.pi[i]/S
    # 求 A
    for i in range(self.N):
        for j in range(self.N):
            fenzi = 0
            fenmu = 0
            for k in range(S):
                for t in range(self.T-1):
                    fenzi += data[k][0][t]==i and data[k][0][t+1]==j
                    fenmu += data[k][0][t]==i
            self.A[i][j] = fenzi/fenmu
    
    # 求 B
    for j in range(self.N):
        for k in range(self.M):
            fenzi = 0
            fenmu = 0
            for i in range(S):
                for t in range(self.T):
                    fenzi += data[i][0][t]==j and data[i][1][t]==k
                    fenmu += data[i][0][t]==j
            self.B[j][k] = fenzi/fenmu
    return self.pi,self.A,self.B

修改后的监督学习方法如下

def train(self,O,I):
    self.pi = np.zeros(shape=(self.N,))
    self.A = [{} for i in range(self.N)]
    self.B = [{} for i in range(self.N)]

    S = I.shape[0]
    self.T = I.shape[1]
    state_count = {}
    for i in range(self.N):
        state_count[i] = (I==i).sum()
        if state_count[i]==0:
            state_count[i] = 0.001
        self.pi[i] = (I[:,0]==i).sum()/S

    for j in range(S):
        for t in range(self.T):
            if t<self.T-1:
                self.A[I[j,t]][I[j,t+1]] = self.A[t].get(t+1,0)+1
            self.B[I[j,t]][O[j,t]] = self.B[I[j,t]].get(O[j,t],0)+1
    for i in range(self.N):
        for state in self.A[i].keys():
            self.A[i][state] = self.A[i].get(state)/state_count[i]
        for observation in self.B[i].keys():
            self.B[i][observation] = self.B[i].get(observation)/state_count[i]
    return self.pi,self.A,self.B

修改后的代码仍然是极大似然估计，但它不再是套用通过数学推导出来的公式，而是直接根据频率估计概率的思想编码，二者表现形式可能没有多大区别，但行为模式不同。

读者对比之后可以发现，状态转移矩阵A和发射矩阵B都从ndarray变成了字典的列表。并且可以看到不再出现

for k in range(self.M)

代码上的细节读者可慢慢体会。

读者可前往https://download.csdn.net/download/qq_41335232/45109996直接下载完整压缩包，或者自行复制本文代码创建文件

最后我把剩下的代码贴上来(dataset.py前面已经贴过了)，然后语料corpus_POS.txt读者可前往https://github.com/junman/POS-tagging下载

# 监督学习算法
import numpy as np
from dataset import read_dataset

class SupervisedModel:
    def __init__(self,n_states) -> None:
        self.N = n_states+1

    def train(self,O,I):
        self.pi = np.zeros(shape=(self.N,))
        self.A = [{} for i in range(self.N)]
        self.B = [{} for i in range(self.N)]

        S = I.shape[0]
        self.T = I.shape[1]
        state_count = {}
        for i in range(self.N):
            state_count[i] = (I==i).sum()
            if state_count[i]==0:
                state_count[i] = 0.001
            self.pi[i] = (I[:,0]==i).sum()/S

        for j in range(S):
            for t in range(self.T):
                if t<self.T-1:
                    self.A[I[j,t]][I[j,t+1]] = self.A[t].get(t+1,0)+1
                self.B[I[j,t]][O[j,t]] = self.B[I[j,t]].get(O[j,t],0)+1
        for i in range(self.N):
            for state in self.A[i].keys():
                self.A[i][state] = self.A[i].get(state)/state_count[i]
            for observation in self.B[i].keys():
                self.B[i][observation] = self.B[i].get(observation)/state_count[i]
        return self.pi,self.A,self.B

    def decode(self,O):
        T = len(O)
        delta = np.zeros(shape=(T,self.N))
        fi = np.zeros(shape=(T,self.N),dtype=int)
        # 初始化
        for i in range(self.N):
            delta[0][i] = self.B[i].get(O[0],0)*self.pi[i]
        # 前向计算
        for t in range(0,T-1):
            for i in range(self.N):
                max_val = 0
                max_index = 0
                for j in range(self.N):
                    p = self.A[j].get(i,0)*delta[t][j]
                    if p>max_val:
                        max_index = j
                        max_val = p
                delta[t+1][i] = self.B[i].get(O[t+1],0)*max_val
                fi[t+1][i] = max_index
        #回溯
        I = []
        index = delta[T-1].argmax()
        I.append(index)
        for t in reversed(range(1,T)):
            index = fi[t,index]
            I.insert(0,index)
        return I
    
X_data,y_data,word_index,state_index = read_dataset()

def word2index(word_index,words):
        digit_words = []
        for word in words:
            digit_words.append(word_index[word])
        return digit_words
    
def index2state(state_index,indexes):
    state_list = list(state_index.keys())
    states = []
    for index in indexes:
        states.append(state_list[index])
    return states

model = SupervisedModel(len(state_index.keys()))
model.train(X_data,y_data)

words = ["我","要","吃饭"]
words = word2index(word_index,words)
posids = model.decode(words)
print(index2state(state_index,posids))

至于无监督的词性标注，我目前还没想到要怎么优化baum-welch的代码，但从理论上说，直接使用之前的代码也是可以训练的，但是肯定需要很长的训练时间。并且效果也不怎么好。

读者可利用hmmlearn库简单验证

import numpy as np
from dataset import read_dataset
from hmmlearn import hmm

X_data,y_data,word_index,state_index = read_dataset()
n_states = len(state_index.keys())
model = hmm.MultinomialHMM(n_components=n_states, n_iter=20, tol=0.001)
model.fit(X_data)

# decode 测试
words = ["我","要","吃饭"]
model.decode(np.array([word2index(words)]).T,algorithm="viterbi")

词性标注-隐马尔可夫模型

词性标注-隐马尔可夫模型

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