词性标注器的代码实现

╰半橙微兮° 2022-03-14 14:06 181阅读 0赞

##### 1.遍历语料库，整理好words和tags #####

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

##### 2.初始化A、B矩阵 #####

关于A、B矩阵的简单理解，可以参考我之前的文章[通过丢硬币的案例理解HMM所解决的问题][HMM]  
π代表每种词性出现的概率。

# 构建 pi, A, B
    # N: # of tags M: # of words in dictionary
    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

##### 3.计算A、B矩阵及π的值 #####

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 : 模型的参数

##### 4.用viterbi算法实现计算句子的词性 #####

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]])

##### 5.校验 #####

最后我们可以拿一个句子看一下词性标注的情况

x = "Social Security number , passport number and details about the services provided for the payment"
    viterbi(x, pi, A, B)

打印结果如下  
![在这里插入图片描述][2019030418295468.png]

相关代码及训练数据及已上传至[github][]，有疑问的同学请题Issue或在下方留言。

[HMM]: https://blog.csdn.net/lt326030434/article/details/88096841
[2019030418295468.png]: /images/20220313/009f9c151bf544298162afdeb5b12173.png
[github]: https://github.com/Maginaaa/PosTagging