Python实现隐马尔可夫模型的前向后向算法的示例代码
本篇文章对隐马尔可夫模型的前向和后向算法进行了Python实现,并且每种算法都给出了循环和递归两种方式的实现。
前向算法Python实现
循环方式
import numpy as np def hmm_forward(Q, V, A, B, pi, T, O, p): """ :param Q: 状态集合 :param V: 观测集合 :param A: 状态转移概率矩阵 :param B: 观测概率矩阵 :param pi: 初始概率分布 :param T: 观测序列和状态序列的长度 :param O: 观测序列 :param p: 存储各个状态的前向概率的列表,初始为空 """ for t in range(T): # 计算初值 if t == 0: for i in range(len(Q)): p.append(pi[i] * B[i, V[O[0]]]) # 初值计算完毕后,进行下一时刻的递推运算 else: alpha_t_ = 0 alpha_t_t = [] for i in range(len(Q)): for j in range(len(Q)): alpha_t_ += p[j] * A[j, i] alpha_t_t.append(alpha_t_ * B[i, V[O[t]]]) alpha_t_ = 0 p = alpha_t_t return sum(p) # 《统计学习方法》书上例10.2 Q = [1, 2, 3] V = {'红':0, '白':1} A = np.array([[0.5, 0.2, 0.3], [0.3, 0.5, 0.2], [0.2, 0.3, 0.5]]) B = np.array([[0.5, 0.5], [0.4, 0.6], [0.7, 0.3]]) pi = [0.2, 0.4, 0.4] T = 3 O = ['红', '白', '红'] p = [] print(hmm_forward(Q, V, A, B, pi, T, O, p)) # 0.130218
递归方式
import numpy as np def hmm_forward_(Q, V, A, B, pi, T, O, p, T_final): """ :param T_final:递归的终止条件 """ if T == 0: for i in range(len(Q)): p.append(pi[i] * B[i, V[O[0]]]) else: alpha_t_ = 0 alpha_t_t = [] for i in range(len(Q)): for j in range(len(Q)): alpha_t_ += p[j] * A[j, i] alpha_t_t.append(alpha_t_ * B[i, V[O[T]]]) alpha_t_ = 0 p = alpha_t_t if T >= T_final: return sum(p) return hmm_forward_(Q, V, A, B, pi, T+1, O, p, T_final) Q = [1, 2, 3] V = {'红':0, '白':1} A = np.array([[0.5, 0.2, 0.3], [0.3, 0.5, 0.2], [0.2, 0.3, 0.5]]) B = np.array([[0.5, 0.5], [0.4, 0.6], [0.7, 0.3]]) pi = [0.2, 0.4, 0.4] T = 0 O = ['红', '白', '红'] p = [] T_final = 2 # T的长度是3,T的取值是(0时刻, 1时刻, 2时刻) print(hmm_forward_(Q, V, A, B, pi, T, O, p, T_final))
后向算法Python实现
循环方式
import numpy as np def hmm_backward(Q, V, A, B, pi, T, O, beta_t, T_final): for t in range(T, -1, -1): if t == T_final: beta_t = beta_t else: beta_t_ = 0 beta_t_t = [] for i in range(len(Q)): for j in range(len(Q)): beta_t_ += A[i, j] * B[j, V[O[t + 1]]] * beta_t[j] beta_t_t.append(beta_t_) beta_t_ = 0 beta_t = beta_t_t if t == 0: p=[] for i in range(len(Q)): p.append(pi[i] * B[i, V[O[0]]] * beta_t[i]) beta_t = p return sum(beta_t) # 《统计学习方法》课后题10.1 Q = [1, 2, 3] V = {'红':0, '白':1} A = np.array([[0.5, 0.2, 0.3], [0.3, 0.5, 0.2], [0.2, 0.3, 0.5]]) B = np.array([[0.5, 0.5], [0.4, 0.6], [0.7, 0.3]]) pi = [0.2, 0.4, 0.4] T = 3 O = ['红', '白', '红', '白'] beta_t = [1, 1, 1] T_final = 3 print(hmm_backward_(Q, V, A, B, pi, T, O, beta_t, T_final)) # 0.06009
递归方式
import numpy as np def hmm_backward(Q, V, A, B, pi, T, O, beta_t, T_final): if T == T_final: beta_t = beta_t else: beta_t_ = 0 beta_t_t = [] for i in range(len(Q)): for j in range(len(Q)): beta_t_ += A[i, j] * B[j, V[O[T+1]]] * beta_t[j] beta_t_t.append(beta_t_) beta_t_ = 0 beta_t = beta_t_t if T == 0: p=[] for i in range(len(Q)): p.append(pi[i] * B[i, V[O[0]]] * beta_t[i]) beta_t = p return sum(beta_t) return hmm_backward(Q, V, A, B, pi, T-1, O, beta_t, T_final) jpgQ = [1, 2, 3] V = {'红':0, '白':1} A = np.array([[0.5, 0.2, 0.3], [0.3, 0.5, 0.2], [0.2, 0.3, 0.5]]) B = np.array([[0.5, 0.5], [0.4, 0.6], [0.7, 0.3]]) pi = [0.2, 0.4, 0.4] T = 3 O = ['红', '白', '红', '白'] beta_t = [1, 1, 1] T_final = 3 print(hmm_backward_(Q, V, A, B, pi, T, O, beta_t, T_final)) # 0.06009
这里我有个问题不理解,这道题的正确答案应该是0.061328,我计算出的答案和实际有一点偏差,我跟踪了代码的计算过程,发现在第一次循环完成后,计算结果是正确的,第二次循环后的结果就出现了偏差,我怀疑是小数部分的精度造成,希望有人能给出一个更好的解答,如果是代码的问题也欢迎指正。
以上所述是小编给大家介绍的Python实现隐马尔可夫模型的前向后向算法,希望对大家有所帮助!
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