Python实现隐马尔可夫模型的前向后向算法的示例代码

Python  /  管理员 发布于 7年前   167

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 npdef 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10.2Q = [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 = 3O = ['红', '白', '红']p = []print(hmm_forward(Q, V, A, B, pi, T, O, p)) # 0.130218

递归方式

import numpy as npdef 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 = 0O = ['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 npdef 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10.1Q = [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 = 3O = ['红', '白', '红', '白']beta_t = [1, 1, 1]T_final = 3print(hmm_backward_(Q, V, A, B, pi, T, O, beta_t, T_final)) # 0.06009

递归方式

import numpy as npdef 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 = 3O = ['红', '白', '红', '白']beta_t = [1, 1, 1]T_final = 3print(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！