《统计学习方法》 李航著 第十章 隐马尔可夫模型
我是小白一个；本文代码转载地址文末有注释；有问题请多指教
先看书，看完书，代码就看懂了。程序只是将算法翻译成机器认识的罢了
import numpy as np
class HiddenMarkov:
def __init__(self):
self.alphas = None
self.forward_P = None
self.betas = None
self.backward_P = None
# 前向算法
# Q 是状态集合，里面包含了所有可能的状态
# V 是我们的观测的集合，里面包含了所有可能的观测结果
# A 状态转移概率分布
# B 观测概率分布
# O 观测序列，依次为观测值
# PI 初始概率分布。根据这个先生成初始状态。
def forward(self, Q, V, A, B, O, PI):
# 状态序列的大小
N = len(Q)
# 观测序列的大小
M = len(O)
# 初始化前向概率alpha值
alphas = np.zeros((N, M))
# 时刻数=观测序列数
T = M
# 遍历每一个时刻，计算前向概率alpha值
for t in range(T):
# 得到序列对应的索引
indexOfO = V.index(O[t])
# 遍历状态序列
for i in range(N):
# 初始化alpha初值
if t == 0:
# P176 公式(10.15)
alphas[i][t] = PI[t][i] * B[i][indexOfO]
print('alpha1(%d) = p%db%db(o1) = %f' %
(i + 1, i, i, alphas[i][t]))
else:
# P176 公式(10.16)
alphas[i][t] = np.dot([alpha[t - 1] for alpha in alphas],
[a[i] for a in A]) * B[i][indexOfO]
print('alpha%d(%d) = [sigma alpha%d(i)ai%d]b%d(o%d) = %f' %
(t + 1, i + 1, t - 1, i, i, t, alphas[i][t]))
# P176 公式(10.17)
self.forward_P = np.sum([alpha[M - 1] for alpha in alphas])
self.alphas = alphas
# 后向算法
# Q 是状态集合，里面包含了所有可能的状态
# V 是我们的观测的集合，里面包含了所有可能的观测结果
# A 状态转移概率分布
# B 观测概率分布
# O 观测序列，依次为观测值
# PI 初始概率分布。根据这个先生成初始状态。
def backward(self, Q, V, A, B, O, PI):
# 状态序列的大小
N = len(Q)
# 观测序列的大小
M = len(O)
# 初始化后向概率beta值，P178 公式(10.19)
betas = np.ones((N, M))
for i in range(N):
print('beta%d(%d) = 1' % (M, i + 1))
# 对观测序列逆向遍历
for t in range(M - 2, -1, -1):
# 得到序列对应的索引
indexOfO = V.index(O[t + 1])
# 遍历状态序列
for i in range(N):
# P178 公式(10.20)
betas[i][t] = np.dot(
np.multiply(A[i], [b[indexOfO] for b in B]),
[beta[t + 1] for beta in betas])
realT = t + 1
realI = i + 1
print('beta%d(%d) = sigma[a%djbj(o%d)beta%d(j)] = (' %
(realT, realI, realI, realT + 1, realT + 1),
end='')
for j in range(N):
print("%.2f * %.2f * %.2f + " %
(A[i][j], B[j][indexOfO], betas[j][t + 1]),
end='')
print("0) = %.3f" % betas[i][t])
# 取出第一个值
indexOfO = V.index(O[0])
self.betas = betas
# P178 公式(10.21)
P = np.dot(np.multiply(PI, [b[indexOfO] for b in B]),
[beta[0] for beta in betas])
self.backward_P = P
print("P(O|lambda) = ", end="")
for i in range(N):
print("%.1f * %.1f * %.5f + " %
(PI[0][i], B[i][indexOfO], betas[i][0]),
end="")
print("0 = %f" % P)
# 维特比算法：动态规划解隐马尔代夫模型预测问题
# Q 是状态集合，里面包含了所有可能的状态
# V 是我们的观测的集合，里面包含了所有可能的观测结果
# A 状态转移概率分布
# B 观测概率分布
# O 观测序列，依次为观测值
# PI 初始概率分布。根据这个先生成初始状态。
def viterbi(self, Q, V, A, B, O, PI):
# 状态序列的大小
N = len(Q)
# 观测序列的大小
M = len(O)
# 初始化daltas：存当前时刻当前状态的所有单个路径的概率最大值
deltas = np.zeros((N, M))
# 初始化psis：存当前时刻当前状态所有单个路径中概率最大路径的前一时刻结点
psis = np.zeros((N, M))
# 初始化最优路径矩阵，该矩阵维度与观测序列维度相同。这是我们最后的输出。
I = np.zeros((1, M))
# 遍历观测序列
for t in range(M):
# 递推从t=2开始
realT = t + 1
# 得到序列对应的索引
indexOfO = V.index(O[t])
for i in range(N):
realI = i + 1
if t == 0:
# P185 算法10.5 步骤(1)
deltas[i][t] = PI[0][i] * B[i][indexOfO]
psis[i][t] = 0
print('delta1(%d) = pi%d * b%d(o1) = %.2f * %.2f = %.2f' %
(realI, realI, realI, PI[0][i], B[i][indexOfO],
deltas[i][t]))
print('psis1(%d) = 0' % (realI))
else:
# # P185 算法10.5 步骤(2)
deltas[i][t] = np.max(
np.multiply([delta[t - 1] for delta in deltas],
[a[i] for a in A])) * B[i][indexOfO]
print(
'delta%d(%d) = max[delta%d(j)aj%d]b%d(o%d) = %.2f * %.2f = %.5f'
% (realT, realI, realT - 1, realI, realI, realT,
np.max(
np.multiply([delta[t - 1] for delta in deltas],
[a[i] for a in A])), B[i][indexOfO],
deltas[i][t]))
# 对于y=f(x)，argmax返回取得最大值y时的x
psis[i][t] = np.argmax(
np.multiply([delta[t - 1] for delta in deltas],
[a[i] for a in A]))
print('psis%d(%d) = argmax[delta%d(j)aj%d] = %d' %
(realT, realI, realT - 1, realI, psis[i][t]))
# 得到最优路径的终结点
I[0][M - 1] = np.argmax([delta[M - 1] for delta in deltas])
print('i%d = argmax[deltaT(i)] = %d' % (M, I[0][M - 1] + 1))
# 递归由后向前得到其他结点
for t in range(M - 2, -1, -1):
I[0][t] = psis[int(I[0][t + 1])][t + 1]
print('i%d = psis%d(i%d) = %d' %
(t + 1, t + 2, t + 2, I[0][t] + 1))
# 输出最优路径
print('最优路径是：', "->".join([str(int(i + 1)) for i in I[0]]))
# 习题10.1
Q = [1, 2, 3]
V = ['红', '白']
A = [[0.5, 0.2, 0.3], [0.3, 0.5, 0.2], [0.2, 0.3, 0.5]]
B = [[0.5, 0.5], [0.4, 0.6], [0.7, 0.3]]
# O = ['红', '白', '红', '红', '白', '红', '白', '白']
O = ['红', '白', '红', '白'] # 习题10.1的例子
PI = [[0.2, 0.4, 0.4]]
HMM = HiddenMarkov()
HMM.forward(Q, V, A, B, O, PI)
print("P(O|λ)={}".format(HMM.forward_P))
# HMM.backward(Q, V, A, B, O, PI)
# print("P(O|λ)={}".format(HMM.backward_P))
# HMM.viterbi(Q, V, A, B, O, PI)
结果
前向算法
alpha1(1) = p0b0b(o1) = 0.100000
alpha1(2) = p1b1b(o1) = 0.160000
alpha1(3) = p2b2b(o1) = 0.280000
alpha2(1) = [sigma alpha0(i)ai0]b0(o1) = 0.077000
alpha2(2) = [sigma alpha0(i)ai1]b1(o1) = 0.110400
alpha2(3) = [sigma alpha0(i)ai2]b2(o1) = 0.060600
alpha3(1) = [sigma alpha1(i)ai0]b0(o2) = 0.041870
alpha3(2) = [sigma alpha1(i)ai1]b1(o2) = 0.035512
alpha3(3) = [sigma alpha1(i)ai2]b2(o2) = 0.052836
alpha4(1) = [sigma alpha2(i)ai0]b0(o3) = 0.021078
alpha4(2) = [sigma alpha2(i)ai1]b1(o3) = 0.025188
alpha4(3) = [sigma alpha2(i)ai2]b2(o3) = 0.013824
P(O|λ)=0.06009079999999999
后向算法
beta4(1) = 1
beta4(2) = 1
beta4(3) = 1
beta3(1) = sigma[a1jbj(o4)beta4(j)] = (0.50 * 0.50 * 1.00 + 0.20 * 0.60 * 1.00 + 0.30 * 0.30 * 1.00 + 0) = 0.460
beta3(2) = sigma[a2jbj(o4)beta4(j)] = (0.30 * 0.50 * 1.00 + 0.50 * 0.60 * 1.00 + 0.20 * 0.30 * 1.00 + 0) = 0.510
beta3(3) = sigma[a3jbj(o4)beta4(j)] = (0.20 * 0.50 * 1.00 + 0.30 * 0.60 * 1.00 + 0.50 * 0.30 * 1.00 + 0) = 0.430
beta2(1) = sigma[a1jbj(o3)beta3(j)] = (0.50 * 0.50 * 0.46 + 0.20 * 0.40 * 0.51 + 0.30 * 0.70 * 0.43 + 0) = 0.246
beta2(2) = sigma[a2jbj(o3)beta3(j)] = (0.30 * 0.50 * 0.46 + 0.50 * 0.40 * 0.51 + 0.20 * 0.70 * 0.43 + 0) = 0.231
beta2(3) = sigma[a3jbj(o3)beta3(j)] = (0.20 * 0.50 * 0.46 + 0.30 * 0.40 * 0.51 + 0.50 * 0.70 * 0.43 + 0) = 0.258
beta1(1) = sigma[a1jbj(o2)beta2(j)] = (0.50 * 0.50 * 0.25 + 0.20 * 0.60 * 0.23 + 0.30 * 0.30 * 0.26 + 0) = 0.112
beta1(2) = sigma[a2jbj(o2)beta2(j)] = (0.30 * 0.50 * 0.25 + 0.50 * 0.60 * 0.23 + 0.20 * 0.30 * 0.26 + 0) = 0.122
beta1(3) = sigma[a3jbj(o2)beta2(j)] = (0.20 * 0.50 * 0.25 + 0.30 * 0.60 * 0.23 + 0.50 * 0.30 * 0.26 + 0) = 0.105
P(O|lambda) = 0.2 * 0.5 * 0.11246 + 0.4 * 0.4 * 0.12174 + 0.4 * 0.7 * 0.10488 + 0 = 0.060091
P(O|λ)=[0.0600908]
维特比算法
delta1(1) = pi1 * b1(o1) = 0.20 * 0.50 = 0.10
psis1(1) = 0
delta1(2) = pi2 * b2(o1) = 0.40 * 0.40 = 0.16
psis1(2) = 0
delta1(3) = pi3 * b3(o1) = 0.40 * 0.70 = 0.28
psis1(3) = 0
delta2(1) = max[delta1(j)aj1]b1(o2) = 0.06 * 0.50 = 0.02800
psis2(1) = argmax[delta1(j)aj1] = 2
delta2(2) = max[delta1(j)aj2]b2(o2) = 0.08 * 0.60 = 0.05040
psis2(2) = argmax[delta1(j)aj2] = 2
delta2(3) = max[delta1(j)aj3]b3(o2) = 0.14 * 0.30 = 0.04200
psis2(3) = argmax[delta1(j)aj3] = 2
delta3(1) = max[delta2(j)aj1]b1(o3) = 0.02 * 0.50 = 0.00756
psis3(1) = argmax[delta2(j)aj1] = 1
delta3(2) = max[delta2(j)aj2]b2(o3) = 0.03 * 0.40 = 0.01008
psis3(2) = argmax[delta2(j)aj2] = 1
delta3(3) = max[delta2(j)aj3]b3(o3) = 0.02 * 0.70 = 0.01470
psis3(3) = argmax[delta2(j)aj3] = 2
delta4(1) = max[delta3(j)aj1]b1(o4) = 0.00 * 0.50 = 0.00189
psis4(1) = argmax[delta3(j)aj1] = 0
delta4(2) = max[delta3(j)aj2]b2(o4) = 0.01 * 0.60 = 0.00302
psis4(2) = argmax[delta3(j)aj2] = 1
delta4(3) = max[delta3(j)aj3]b3(o4) = 0.01 * 0.30 = 0.00220
psis4(3) = argmax[delta3(j)aj3] = 2
i4 = argmax[deltaT(i)] = 2
i3 = psis4(i4) = 2
i2 = psis3(i3) = 2
i1 = psis2(i2) = 3
最优路径是： 3->2->2->2
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