python模拟隐马尔可夫模型的方法是什么

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import numpy as npclass HiddenMarkov:    def __init__(self):        self.alphas = None        self.forward_P = None        self.betas = None        self.backward_P = None    # 前向算法    # Q 是状态集合，里面包含了所有可能的状态    # V 是我们的观测的集合，里面包含了所有可能的观测结果    # A 状态转移概率分布    # B 观测概率分布    # O 观测序列，依次为观测值    # PI 初始概率分布。根据这个先生成初始状态。    def forward(self, Q, V, A, B, O, PI):        # 状态序列的大小        N = len(Q)        # 观测序列的大小        M = len(O)        # 初始化前向概率alpha值        alphas = np.zeros((N, M))        # 时刻数=观测序列数        T = M        # 遍历每一个时刻，计算前向概率alpha值        for t in range(T):            # 得到序列对应的索引            indexOfO = V.index(O[t])            # 遍历状态序列            for i in range(N):                # 初始化alpha初值                if t == 0:                    # P176 公式(10.15)                    alphas[i][t] = PI[t][i] * B[i][indexOfO]                    print('alpha1(%d) = p%db%db(o1) = %f' %                          (i + 1, i, i, alphas[i][t]))                else:                    # P176 公式(10.16)                    alphas[i][t] = np.dot([alpha[t - 1] for alpha in alphas],                                          [a[i] for a in A]) * B[i][indexOfO]                    print('alpha%d(%d) = [sigma alpha%d(i)ai%d]b%d(o%d) = %f' %                          (t + 1, i + 1, t - 1, i, i, t, alphas[i][t]))        # P176 公式(10.17)        self.forward_P = np.sum([alpha[M - 1] for alpha in alphas])        self.alphas = alphas    # 后向算法    # Q 是状态集合，里面包含了所有可能的状态    # V 是我们的观测的集合，里面包含了所有可能的观测结果    # A 状态转移概率分布    # B 观测概率分布    # O 观测序列，依次为观测值    # PI 初始概率分布。根据这个先生成初始状态。    def backward(self, Q, V, A, B, O, PI):        # 状态序列的大小        N = len(Q)        # 观测序列的大小        M = len(O)        # 初始化后向概率beta值，P178 公式(10.19)        betas = np.ones((N, M))        for i in range(N):            print('beta%d(%d) = 1' % (M, i + 1))        # 对观测序列逆向遍历        for t in range(M - 2, -1, -1):            # 得到序列对应的索引            indexOfO = V.index(O[t + 1])            # 遍历状态序列            for i in range(N):                # P178 公式(10.20)                betas[i][t] = np.dot(                    np.multiply(A[i], [b[indexOfO] for b in B]),                    [beta[t + 1] for beta in betas])                realT = t + 1                realI = i + 1                print('beta%d(%d) = sigma[a%djbj(o%d)beta%d(j)] = (' %                      (realT, realI, realI, realT + 1, realT + 1),                      end='')                for j in range(N):                    print("%.2f * %.2f * %.2f + " %                          (A[i][j], B[j][indexOfO], betas[j][t + 1]),                          end='')                print("0) = %.3f" % betas[i][t])        # 取出第一个值        indexOfO = V.index(O[0])        self.betas = betas        # P178 公式(10.21)        P = np.dot(np.multiply(PI, [b[indexOfO] for b in B]),                   [beta[0] for beta in betas])        self.backward_P = P        print("P(O|lambda) = ", end="")        for i in range(N):            print("%.1f * %.1f * %.5f + " %                  (PI[0][i], B[i][indexOfO], betas[i][0]),                  end="")        print("0 = %f" % P)    # 维特比算法：动态规划解隐马尔代夫模型预测问题    # Q 是状态集合，里面包含了所有可能的状态    # V 是我们的观测的集合，里面包含了所有可能的观测结果    # A 状态转移概率分布    # B 观测概率分布    # O 观测序列，依次为观测值    # PI 初始概率分布。根据这个先生成初始状态。    def viterbi(self, Q, V, A, B, O, PI):        # 状态序列的大小        N = len(Q)        # 观测序列的大小        M = len(O)        # 初始化daltas：存当前时刻当前状态的所有单个路径的概率最大值        deltas = np.zeros((N, M))        # 初始化psis：存当前时刻当前状态所有单个路径中概率最大路径的前一时刻结点        psis = np.zeros((N, M))        # 初始化最优路径矩阵，该矩阵维度与观测序列维度相同。这是我们最后的输出。        I = np.zeros((1, M))        # 遍历观测序列        for t in range(M):            # 递推从t=2开始            realT = t + 1            # 得到序列对应的索引            indexOfO = V.index(O[t])            for i in range(N):                realI = i + 1                if t == 0:                    # P185 算法10.5 步骤(1)                    deltas[i][t] = PI[0][i] * B[i][indexOfO]                    psis[i][t] = 0                    print('delta1(%d) = pi%d * b%d(o1) = %.2f * %.2f = %.2f' %                          (realI, realI, realI, PI[0][i], B[i][indexOfO],                           deltas[i][t]))                    print('psis1(%d) = 0' % (realI))                else:                    # # P185 算法10.5 步骤(2)                    deltas[i][t] = np.max(                        np.multiply([delta[t - 1] for delta in deltas],                                    [a[i] for a in A])) * B[i][indexOfO]                    print(                        'delta%d(%d) = max[delta%d(j)aj%d]b%d(o%d) = %.2f * %.2f = %.5f'                        % (realT, realI, realT - 1, realI, realI, realT,                           np.max(                               np.multiply([delta[t - 1] for delta in deltas],                                           [a[i] for a in A])), B[i][indexOfO],                           deltas[i][t]))                    # 对于y=f(x)，argmax返回取得最大值y时的x                    psis[i][t] = np.argmax(                        np.multiply([delta[t - 1] for delta in deltas],                                    [a[i] for a in A]))                    print('psis%d(%d) = argmax[delta%d(j)aj%d] = %d' %                          (realT, realI, realT - 1, realI, psis[i][t]))        # 得到最优路径的终结点        I[0][M - 1] = np.argmax([delta[M - 1] for delta in deltas])        print('i%d = argmax[deltaT(i)] = %d' % (M, I[0][M - 1] + 1))        # 递归由后向前得到其他结点        for t in range(M - 2, -1, -1):            I[0][t] = psis[int(I[0][t + 1])][t + 1]            print('i%d = psis%d(i%d) = %d' %                  (t + 1, t + 2, t + 2, I[0][t] + 1))        # 输出最优路径        print('最优路径是：', "->".join([str(int(i + 1)) for i in I[0]]))# 习题10.1Q = [1, 2, 3]V = ['红', '白']A = [[0.5, 0.2, 0.3], [0.3, 0.5, 0.2], [0.2, 0.3, 0.5]]B = [[0.5, 0.5], [0.4, 0.6], [0.7, 0.3]]# O = ['红', '白', '红', '红', '白', '红', '白', '白']O = ['红', '白', '红', '白']    # 习题10.1的例子PI = [[0.2, 0.4, 0.4]]HMM = HiddenMarkov()HMM.forward(Q, V, A, B, O, PI)print("P(O|λ)={}".format(HMM.forward_P))# HMM.backward(Q, V, A, B, O, PI)# print("P(O|λ)={}".format(HMM.backward_P))# HMM.viterbi(Q, V, A, B, O, PI)

• 结果

1. 前向算法

alpha1(1) = p0b0b(o1) = 0.100000alpha1(2) = p1b1b(o1) = 0.160000alpha1(3) = p2b2b(o1) = 0.280000alpha2(1) = [sigma alpha0(i)ai0]b0(o1) = 0.077000alpha2(2) = [sigma alpha0(i)ai1]b1(o1) = 0.110400alpha2(3) = [sigma alpha0(i)ai2]b2(o1) = 0.060600alpha3(1) = [sigma alpha1(i)ai0]b0(o2) = 0.041870alpha3(2) = [sigma alpha1(i)ai1]b1(o2) = 0.035512alpha3(3) = [sigma alpha1(i)ai2]b2(o2) = 0.052836alpha4(1) = [sigma alpha2(i)ai0]b0(o3) = 0.021078alpha4(2) = [sigma alpha2(i)ai1]b1(o3) = 0.025188alpha4(3) = [sigma alpha2(i)ai2]b2(o3) = 0.013824P(O|λ)=0.06009079999999999

2. 后向算法

beta4(1) = 1beta4(2) = 1beta4(3) = 1beta3(1) = sigma[a1jbj(o4)beta4(j)] = (0.50 * 0.50 * 1.00 + 0.20 * 0.60 * 1.00 + 0.30 * 0.30 * 1.00 + 0) = 0.460beta3(2) = sigma[a2jbj(o4)beta4(j)] = (0.30 * 0.50 * 1.00 + 0.50 * 0.60 * 1.00 + 0.20 * 0.30 * 1.00 + 0) = 0.510beta3(3) = sigma[a3jbj(o4)beta4(j)] = (0.20 * 0.50 * 1.00 + 0.30 * 0.60 * 1.00 + 0.50 * 0.30 * 1.00 + 0) = 0.430beta2(1) = sigma[a1jbj(o3)beta3(j)] = (0.50 * 0.50 * 0.46 + 0.20 * 0.40 * 0.51 + 0.30 * 0.70 * 0.43 + 0) = 0.246beta2(2) = sigma[a2jbj(o3)beta3(j)] = (0.30 * 0.50 * 0.46 + 0.50 * 0.40 * 0.51 + 0.20 * 0.70 * 0.43 + 0) = 0.231beta2(3) = sigma[a3jbj(o3)beta3(j)] = (0.20 * 0.50 * 0.46 + 0.30 * 0.40 * 0.51 + 0.50 * 0.70 * 0.43 + 0) = 0.258beta1(1) = sigma[a1jbj(o2)beta2(j)] = (0.50 * 0.50 * 0.25 + 0.20 * 0.60 * 0.23 + 0.30 * 0.30 * 0.26 + 0) = 0.112beta1(2) = sigma[a2jbj(o2)beta2(j)] = (0.30 * 0.50 * 0.25 + 0.50 * 0.60 * 0.23 + 0.20 * 0.30 * 0.26 + 0) = 0.122beta1(3) = sigma[a3jbj(o2)beta2(j)] = (0.20 * 0.50 * 0.25 + 0.30 * 0.60 * 0.23 + 0.50 * 0.30 * 0.26 + 0) = 0.105P(O|lambda) = 0.2 * 0.5 * 0.11246 + 0.4 * 0.4 * 0.12174 + 0.4 * 0.7 * 0.10488 + 0 = 0.060091P(O|λ)=[0.0600908]

3. 维特比算法

delta1(1) = pi1 * b1(o1) = 0.20 * 0.50 = 0.10psis1(1) = 0delta1(2) = pi2 * b2(o1) = 0.40 * 0.40 = 0.16psis1(2) = 0delta1(3) = pi3 * b3(o1) = 0.40 * 0.70 = 0.28psis1(3) = 0delta2(1) = max[delta1(j)aj1]b1(o2) = 0.06 * 0.50 = 0.02800psis2(1) = argmax[delta1(j)aj1] = 2delta2(2) = max[delta1(j)aj2]b2(o2) = 0.08 * 0.60 = 0.05040psis2(2) = argmax[delta1(j)aj2] = 2delta2(3) = max[delta1(j)aj3]b3(o2) = 0.14 * 0.30 = 0.04200psis2(3) = argmax[delta1(j)aj3] = 2delta3(1) = max[delta2(j)aj1]b1(o3) = 0.02 * 0.50 = 0.00756psis3(1) = argmax[delta2(j)aj1] = 1delta3(2) = max[delta2(j)aj2]b2(o3) = 0.03 * 0.40 = 0.01008psis3(2) = argmax[delta2(j)aj2] = 1delta3(3) = max[delta2(j)aj3]b3(o3) = 0.02 * 0.70 = 0.01470psis3(3) = argmax[delta2(j)aj3] = 2delta4(1) = max[delta3(j)aj1]b1(o4) = 0.00 * 0.50 = 0.00189psis4(1) = argmax[delta3(j)aj1] = 0delta4(2) = max[delta3(j)aj2]b2(o4) = 0.01 * 0.60 = 0.00302psis4(2) = argmax[delta3(j)aj2] = 1delta4(3) = max[delta3(j)aj3]b3(o4) = 0.01 * 0.30 = 0.00220psis4(3) = argmax[delta3(j)aj3] = 2i4 = argmax[deltaT(i)] = 2i3 = psis4(i4) = 2i2 = psis3(i3) = 2i1 = psis2(i2) = 3最优路径是： 3->2->2->2

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