前向算法求解隐马模型的第一个基本问题

隐马模型的介绍实现以及一个简单的实例

http://blog.csdn.net/liugallup/article/details/51519008

实现代码

# -*- coding: utf-8 -*-
# HMM Forward algorithm

# input  Matrix A,B vector pi

import numpy as np
#按行存储
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]])
O = np.array([0, 1, 0])  # T=3
# O=np.array([1 ,0, 1])#T=3
#初始状态向量
pi = np.array([0.2, 0.4, 0.4])

N = 3  # N kind state
M = 2  # M kind of observation
T = 3

# initialize:

Aerfa = np.zeros((3, 3), np.float)
for i in range(N):
    Aerfa[0, i] = pi[i] * B[i, O[0]]

# Recursion:
for t in range(T - 1):
    for i in range(N):
        for j in range(N):
            Aerfa[t + 1, i] += Aerfa[t, j] * A[j, i]
        Aerfa[t + 1, i] *= B[i, O[t + 1]]

        # compute P(O|lamda) and termination
P = 0
for i in range(N):
    P += Aerfa[T - 1, i]  # begin with 0 so T-1

print P

# backward

# initialize:
Beta = np.zeros((T, N), np.float)
print Beta
for i in range(N):
    Beta[T - 1, i] = 1

# recursion:
for t in range(T - 2, -1, -1):
    for i in range(N):
        for j in range(N):
            Beta[t, i] += A[i, j] * B[j, O[t + 1]] * Beta[t + 1, j]


            # termination:
P_back = 0
for i in range(N):
    P_back += pi[i] * B[i, O[0]] * Beta[0, i]

print P_back