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Python实现隐马尔可夫模型的前向后向算法的示例代码

更新时间:2020-08-17 15:54:01 作者:startmvc
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

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!