Python實現(xiàn)隱馬爾可夫模型的前向后向算法的示例代碼

更新時間：2019年12月31日 11:56:52 作者：潤°

這篇文章主要介紹了Python實現(xiàn)隱馬爾可夫模型的前向后向算法,本文通過實例代碼給大家講解的非常詳細，具有一定的參考借鑒價值，需要的朋友可以參考下

本篇文章對隱馬爾可夫模型的前向和后向算法進行了Python實現(xiàn)，并且每種算法都給出了循環(huán)和遞歸兩種方式的實現(xiàn)。

前向算法Python實現(xiàn)

循環(huán)方式

import numpy as np
def hmm_forward(Q, V, A, B, pi, T, O, p):
  """
  :param Q: 狀態(tài)集合
  :param V: 觀測集合
  :param A: 狀態(tài)轉(zhuǎn)移概率矩陣
  :param B: 觀測概率矩陣
  :param pi: 初始概率分布
  :param T: 觀測序列和狀態(tài)序列的長度
  :param O: 觀測序列
  :param p: 存儲各個狀態(tài)的前向概率的列表，初始為空
  """
  for t in range(T):
    # 計算初值
    if t == 0:
      for i in range(len(Q)):
        p.append(pi[i] * B[i, V[O[0]]])
    # 初值計算完畢后，進行下一時刻的遞推運算
    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)
# 《統(tǒng)計學習方法》書上例10.2
Q = [1, 2, 3]
V = {'紅':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 = ['紅', '白', '紅']
p = []
print(hmm_forward(Q, V, A, B, pi, T, O, p)) # 0.130218

遞歸方式

import numpy as np
def hmm_forward_(Q, V, A, B, pi, T, O, p, T_final):
  """
  :param T_final:遞歸的終止條件
  """
  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 = {'紅':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 = ['紅', '白', '紅']
p = []
T_final = 2 # T的長度是3，T的取值是(0時刻, 1時刻, 2時刻)
print(hmm_forward_(Q, V, A, B, pi, T, O, p, T_final))

后向算法Python實現(xiàn)

循環(huán)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)
# 《統(tǒng)計學習方法》課后題10.1
Q = [1, 2, 3]
V = {'紅':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 = ['紅', '白', '紅', '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

遞歸方式

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 = {'紅':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 = ['紅', '白', '紅', '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

這里我有個問題不理解，這道題的正確答案應該是0.061328，我計算出的答案和實際有一點偏差，我跟蹤了代碼的計算過程，發(fā)現(xiàn)在第一次循環(huán)完成后，計算結(jié)果是正確的，第二次循環(huán)后的結(jié)果就出現(xiàn)了偏差，我懷疑是小數(shù)部分的精度造成，希望有人能給出一個更好的解答，如果是代碼的問題也歡迎指正。

以上所述是小編給大家介紹的Python實現(xiàn)隱馬爾可夫模型的前向后向算法，希望對大家有所幫助！

您可能感興趣的文章: