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)
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 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。
来源：CSDN
作者：润°
链接：https://blog.csdn.net/weixin_43352637/article/details/103773612