• ZH奶酪:隐马尔可夫模型学习小记——forward算法+viterbi算法+forward-backward算法(Baum-welch算法)


    网上关于HMM的学习资料、博客有很多,基本都是左边摘抄一点,右边摘抄一点,这里一个图,那里一个图,公式中有的变量说不清道不明,学起来很费劲。

    经过浏览几篇博文(其实有的地方写的也比较乱),在7张4开的草稿纸上写公式、单步跟踪程序,终于还是搞清楚了HMM的原理。 

    HMM学习过程:

    1、搜索相关博客:

    隐马尔可夫模型[博客](图示比较详细,前部分还可以,后部分公式有点乱):http://www.leexiang.com/hidden-markov-model

    HMM简介、forward算法和viterbi算法[博客](含源码,算法描述不是很清晰,但是有源码可看)http://www.cnblogs.com/zhangchaoyang/articles/2219571.html

    forward-backward算法[博客](含源码,算法描述不是很清晰,但是有源码可看):http://www.cnblogs.com/zhangchaoyang/articles/2220398.html

    隐马尔科夫模型PPT—刘秉权[百度文库](算法流程、公式、参数都比较详细,有理论基础之后是很好的总结资源,但是没有具体例子,无基础的同学学习起来不是很形象。):http://wenku.baidu.com/view/2f0d944769eae009581bec04.html

    ----其他代码资源(没有理论基础,只看代码很难看懂HMM的原理)---

    UMDHMM的C语言实现:http://www.kanungo.com/software/umdhmm-v1.02.zip

    GitHub上一个UMDHMM的Python实现:https://github.com/dkyang/UMDHMM-python

    2、根据隐马尔科夫模型PPT—刘秉权[百度文库],在5张4开草稿纸上把HMM流程顺一遍,下边是整理的笔记:

     

    HMM三个算法的作用:

      forward算法:(评估)给定一HMM模型,计算一观察序列O1O2...OLEN出现的概率。

      viterbi算法:(解码)给定一HMM模型,计算一观察序列O1O2...OLEN对应的最可能的隐藏序列H1H2...HLEN及该隐藏序列出现的概率。

      forward-backward算法:(学习)给定一观察序列O1O2...OLEN,求解能够拟合这个序列的HMM模型。

    HMM三个算法之间的关系:

    forward算法中的forward变量就是forward-backforward算法中的forward变量,而backward变量与forward变量是类似的;

    forward-backward算法是为了通过类似最大似然估计的方法找到局部最优的模型参数,在迭代过程中forward变量和backward变量起了很大作用;

    viterbi算法和forward算法很相似,只是forward算法迭代过程需要的是sum,viterbi算法迭代过程需要的是max,而且viterbi算法除了输出概率,还要用逆推过程求解路径;

    当用forwar-backward算法求解出模型参数之后,用户给出一个观察序列,用viterbi算法就能求出最可能的隐藏序列以及概率了。

    首先是forward算法的Python实现:

    #-*-coding:utf-8-*-
    __author__ = 'ZhangHe'
    def forward(N,M,A,B,P,observed):
        p = 0.0
        #观察到的状态数目
        LEN = len(observed)
        #中间概率LEN*M
        Q = [([0]*N) for i in range(LEN)]
        #第一个观察到的状态,状态的初始概率乘上隐藏状态到观察状态的条件概率。
        for j in range(N):
            Q[0][j] = P[j]*B[j][observation.index(observed[0])]
        #第一个之后的状态,首先从前一天的每个状态,转移到当前状态的概率求和,然后乘上隐藏状态到观察状态的条件概率。
        for i in range(1,LEN):
            for j in range(N):
                sum = 0.0
                for k in range(N):
                    sum += Q[i-1][k]*A[k][j]
                Q[i][j] = sum * B[j][observation.index(observed[i])]
        for i in range(N):
            p += Q[LEN-1][i]
        return p
    # 3 种隐藏层状态:sun cloud rain
    hidden = []
    hidden.append('sun')
    hidden.append('cloud')
    hidden.append('rain')
    N = len(hidden)
    # 4 种观察层状态:dry dryish damp soggy
    observation = []
    observation.append('dry')
    observation.append('damp')
    observation.append('soggy')
    M = len(observation)
    # 初始状态矩阵(1*N第一天是sun,cloud,rain的概率)
    P = (0.3,0.3,0.4)
    # 状态转移矩阵A(N*N 隐藏层状态之间互相转变的概率)
    A=((0.2,0.3,0.5),(0.1,0.5,0.4),(0.6,0.1,0.3))
    # 混淆矩阵B(N*M 隐藏层状态对应的观察层状态的概率)
    B=((0.1,0.5,0.4),(0.2,0.4,0.4),(0.3,0.6,0.1))
    #假设观察到一组序列为observed,输出HMM模型(N,M,A,B,P)产生观察序列observed的概率
    observed = ['dry']
    print forward(N,M,A,B,P,observed)
    observed = ['damp']
    print forward(N,M,A,B,P,observed)
    observed = ['dry','damp']
    print forward(N,M,A,B,P,observed)
    observed = ['dry','damp','soggy']
    print forward(N,M,A,B,P,observed)

    输出结果:

    0.21
    0.51
    0.1074
    0.030162

    其中前两个结果和手工计算的一样;

    后两个结果没有手工计算,但是在调试程序过程中单步跟踪运行代码,运行过程与手工计算过程相同。

    然后是Viterbi算法的Python实现:

    def viterbi(N,M,A,B,P,hidden,observed):
        sta = []
        LEN = len(observed)
        Q = [([0]*N) for i in range(LEN)]
        path = [([0]*N) for i in range(LEN)]
        #第一天计算,状态的初始概率*隐藏状态到观察状态的条件概率
        for j in range(N):
            Q[0][j]=P[j]*B[j][observation.index(observed[0])]
            path[0][j] = -1
        # 第一天以后的计算
        # 前一天的每个状态转移到当前状态的概率最大值
        # *
        # 隐藏状态到观察状态的条件概率
        for i in range(1,LEN):
            for j in range(N):
                max = 0.0
                index = 0
                for k in range(N):
                    if(Q[i-1][k]*A[k][j] > max):
                        max = Q[i-1][k]*A[k][j]
                        index = k
                Q[i][j] = max * B[j][observation.index(observed[i])]
                path[i][j] = index
        #找到最后一天天气呈现哪种观察状态的概率最大
        max = 0.0
        idx = 0
        for i in range(N):
            if(Q[LEN-1][i]>max):
                max = Q[LEN-1][i]
                idx = i
        print "最可能隐藏序列的概率:"+str(max)
        sta.append(hidden[idx])
        #逆推回去找到每天出现哪个隐藏状态的概率最大
        for i in range(LEN-1,0,-1):
            idx = path[i][idx]
            sta.append(hidden[idx])
        sta.reverse()
        return sta;
    # 3 种隐藏层状态:sun cloud rain
    hidden = []
    hidden.append('sun')
    hidden.append('cloud')
    hidden.append('rain')
    N = len(hidden)
    # 4 种观察层状态:dry dryish damp soggy
    observation = []
    observation.append('dry')
    observation.append('damp')
    observation.append('soggy')
    M = len(observation)
    # 初始状态矩阵(1*N第一天是sun,cloud,rain的概率)
    P = (0.3,0.3,0.4)
    # 状态转移矩阵A(N*N 隐藏层状态之间互相转变的概率)
    A=((0.2,0.3,0.5),(0.1,0.5,0.4),(0.6,0.1,0.3))
    # 混淆矩阵B(N*M 隐藏层状态对应的观察层状态的概率)
    B=((0.1,0.5,0.4),(0.2,0.4,0.4),(0.3,0.6,0.1))
    #假设观察到一组序列为observed,输出HMM模型(N,M,A,B,P)产生观察序列observed的概率
    observed = ['dry','damp','soggy']
    print viterbi(N,M,A,B,P,hidden,observed)

    输出:

    最可能隐藏序列的概率:0.005184
    ['rain', 'rain', 'sun']

     GITHUB上一个Python实现的完整HMM:

    import numpy as np
    
    DELTA = 0.001
    
    class HMM:
    
    
        def __init__(self, pi, A, B):
            self.pi = pi
            self.A = A
            self.B = B
            self.M = B.shape[1]
            self.N = A.shape[0]
            
        def forward(self,obs):
            T = len(obs)
            N = self.N
            
            alpha = np.zeros([N,T])
            alpha[:,0] = self.pi[:] * self.B[:,obs[0]-1]                                                                                                      
        
            for t in xrange(1,T):
                for n in xrange(0,N):
                    alpha[n,t] = np.sum(alpha[:,t-1] * self.A[:,n]) * self.B[n,obs[t]-1]
                         
            prob = np.sum(alpha[:,T-1])
            return prob, alpha
            
        def forward_with_scale(self, obs):
            """see scaling chapter in "A tutorial on hidden Markov models and 
            selected applications in speech recognition." 
            """
            T = len(obs)
            N = self.N
            alpha = np.zeros([N,T])
            scale = np.zeros(T)
    
            alpha[:,0] = self.pi[:] * self.B[:,obs[0]-1]
            scale[0] = np.sum(alpha[:,0])
            alpha[:,0] /= scale[0]
    
            for t in xrange(1,T):
                for n in xrange(0,N):
                    alpha[n,t] = np.sum(alpha[:,t-1] * self.A[:,n]) * self.B[n,obs[t]-1]
                scale[t] = np.sum(alpha[:,t])
                alpha[:,t] /= scale[t]
    
            logprob = np.sum(np.log(scale[:]))
            return logprob, alpha, scale    
            
        def backward(self, obs):
            T = len(obs)
            N = self.N
            beta = np.zeros([N,T])
            
            beta[:,T-1] = 1
            for t in reversed(xrange(0,T-1)):
                for n in xrange(0,N):
                    beta[n,t] = np.sum(self.B[:,obs[t+1]-1] * self.A[n,:] * beta[:,t+1])
                    
            prob = np.sum(beta[:,0])
            return prob, beta
    
        def backward_with_scale(self, obs, scale):
            T = len(obs)
            N = self.N
            beta = np.zeros([N,T])
    
            beta[:,T-1] = 1 / scale[T-1]
            for t in reversed(xrange(0,T-1)):
                for n in xrange(0,N):
                    beta[n,t] = np.sum(self.B[:,obs[t+1]-1] * self.A[n,:] * beta[:,t+1])
                    beta[n,t] /= scale[t]
            
            return beta
    
        def viterbi(self, obs):
            T = len(obs)
            N = self.N
            psi = np.zeros([N,T]) # reverse pointer
            delta = np.zeros([N,T])
            q = np.zeros(T)
            temp = np.zeros(N)        
            
            delta[:,0] = self.pi[:] * self.B[:,obs[0]-1]    
            
            for t in xrange(1,T):
                for n in xrange(0,N):
                    temp = delta[:,t-1] * self.A[:,n]    
                    max_ind = argmax(temp)
                    psi[n,t] = max_ind
                    delta[n,t] = self.B[n,obs[t]-1] * temp[max_ind]
    
            max_ind = argmax(delta[:,T-1])
            q[T-1] = max_ind
            prob = delta[:,T-1][max_ind]
    
            for t in reversed(xrange(1,T-1)):
                q[t] = psi[q[t+1],t+1]    
                
            return prob, q, delta    
            
        def viterbi_log(self, obs):
            
            T = len(obs)
            N = self.N
            psi = np.zeros([N,T])
            delta = np.zeros([N,T])
            pi = np.zeros(self.pi.shape)
            A = np.zeros(self.A.shape)
            biot = np.zeros([N,T])
    
            pi = np.log(self.pi)        
            A = np.log(self.A)
            biot = np.log(self.B[:,obs[:]-1])
    
            delta[:,0] = pi[:] + biot[:,0]
    
            for t in xrange(1,T):
                for n in xrange(0,N):
                    temp = delta[:,t-1] + self.A[:,n]    
                    max_ind = argmax(temp)
                    psi[n,t] = max_ind
                    delta[n,t] = temp[max_ind] + biot[n,t]   
    
            max_ind = argmax(delta[:,T-1])
            q[T-1] = max_ind            
            logprob = delta[max_ind,T-1]      
                    
            for t in reversed(xrange(1,T-1)):
                q[t] = psi[q[t+1],t+1]    
    
            return logprob, q, delta
    
        def baum_welch(self, obs):
            T = len(obs)
            M = self.M
            N = self.N        
            alpha = np.zeros([N,T])
            beta = np.zeros([N,T])
            scale = np.zeros(T)
            gamma = np.zeros([N,T])
            xi = np.zeros([N,N,T-1])
        
            # caculate initial parameters
            logprobprev, alpha, scale = self.forward_with_scale(obs)
            beta = self.backward_with_scale(obs, scale)            
            gamma = self.compute_gamma(alpha, beta)    
            xi = self.compute_xi(obs, alpha, beta)    
            logprobinit = logprobprev        
            
            # start interative 
            while True:
                # E-step
                self.pi = 0.001 + 0.999*gamma[:,0]
                for i in xrange(N):
                    denominator = np.sum(gamma[i,0:T-1])
                    for j in xrange(N): 
                        numerator = np.sum(xi[i,j,0:T-1])
                        self.A[i,j] = numerator / denominator
                                       
                self.A = 0.001 + 0.999*self.A
                for j in xrange(0,N):
                    denominator = np.sum(gamma[j,:])
                    for k in xrange(0,M):
                        numerator = 0.0
                        for t in xrange(0,T):
                            if obs[t]-1 == k:
                                numerator += gamma[j,t]
                        self.B[j,k] = numerator / denominator
                self.B = 0.001 + 0.999*self.B
    
                # M-step
                logprobcur, alpha, scale = self.forward_with_scale(obs)
                beta = self.backward_with_scale(obs, scale)            
                gamma = self.compute_gamma(alpha, beta)    
                xi = self.compute_xi(obs, alpha, beta)    
    
                delta = logprobcur - logprobprev
                logprobprev = logprobcur
                # print "delta is ", delta
                if delta <= DELTA:
                    break     
                    
            logprobfinal = logprobcur
            return logprobinit, logprobfinal                
                
        def compute_gamma(self, alpha, beta):
            gamma = np.zeros(alpha.shape)
            gamma = alpha[:,:] * beta[:,:]
            gamma = gamma / np.sum(gamma,0)
            return gamma
                
        def compute_xi(self, obs, alpha, beta):
            T = len(obs)
            N = self.N
            xi = np.zeros((N, N, T-1))
                
            for t in xrange(0,T-1):        
                for i in xrange(0,N):
                    for j in xrange(0,N):
                        xi[i,j,t] = alpha[i,t] * self.A[i,j] * 
                                    self.B[j,obs[t+1]-1] * beta[j,t+1]
                xi[:,:,t] /= np.sum(np.sum(xi[:,:,t],1),0)    
            return xi
    
    def read_hmm(hmmfile):
        fhmm = open(hmmfile,'r') 
    
        M = int(fhmm.readline().split(' ')[1])
        N = int(fhmm.readline().split(' ')[1]) 
        
        A = np.array([])
        fhmm.readline()
        for i in xrange(N):
            line = fhmm.readline()
            if i == 0:
                A = np.array(map(float,line.split(',')))
            else:
                A = np.vstack((A,map(float,line.split(','))))
            
        B = np.array([])
        fhmm.readline()
        for i in xrange(N):
            line = fhmm.readline()
            if i == 0:
                B = np.array(map(float,line.split(',')))
            else:
                B = np.vstack((B,map(float,line.split(','))))
        
        fhmm.readline()
        line = fhmm.readline()
        pi = np.array(map(float,line.split(',')))
        
        fhmm.close()
        return M, N, pi, A, B 
        
    def read_sequence(seqfile):
        fseq = open(seqfile,'r') 
        
        T = int(fseq.readline().split(' ')[1])
        line = fseq.readline()
        obs = np.array(map(int,line.split(',')))
        
        fseq.close()
        return T, obs
  • 相关阅读:
    是否可以从一个静态(static)方法内部发出对非静态 (non-static)方法的调用?
    是否可以继承 String 类?
    如何实现字符串的反转及替换?
    String s = new String(“xyz”);创建了几个字符串对象?
    Serial 与 Parallel GC 之间的不同之处?
    SVG是什么?
    阐述静态变量和实例变量的区别?
    Java 中 ++ 操作符是线程安全的吗?
    什么是 REST / RESTful 以及它的用途是什么?
    WebSQL是HTML 5规范的一部分吗?
  • 原文地址:https://www.cnblogs.com/CheeseZH/p/4229910.html
Copyright © 2020-2023  润新知