机器学习之隐马尔科夫模型(HMM)原理及Python实现 (大章节)
HMM
隐马尔可夫模型(hidden Markov model, HMM)是可用于标注问题的统计学模型,是生成模型。
本章节内容参考李航博士的《统计学习方法》
本章节添加了一些结论性结果的推导过程。
1. 从一个自然语言处理的例子开始
例如有三个个句子:
句子一:我/名词 看见/动词 猫/名词
句子二:猫/名词 是/动词 可爱的/形容词
句子三:我/名词 是/动词 可爱的/形容词
一般只能观察到具体的词,所以像"我 看见 猫 …"是观测集合,而词性如"名词 动词 形容词 …"是状态序列
设QQQ是所有可能的状态集合,VVV是所有可能的观测集合:
Q={q1,q2,...,qN},V={v1,v2,...,vM}Q = \{q_1, q_2, ..., q_N\}, V=\{v_1, v_2, ..., v_M\}Q={q1,q2,...,qN},V={v1,v2,...,vM}
其中, N是可能的状态数,M是可能的观测数。
例如:Q={名词,动词,形容词},V={我,看见,猫,是,可爱的},N=3,M=5Q=\{名词,动词,形容词 \},V=\{我, 看见, 猫, 是,可爱的\},N=3, M=5Q={名词,动词,形容词},V={我,看见,猫,是,可爱的},N=3,M=5
III是长度为TTT的状态序列,OOO是对应的观测序列:
I={i1,i2,...,iT},O={o1,o2,...,oT}I = \{i_1, i_2,..., i_T \}, O=\{o_1, o_2,..., o_T\}I={i1,i2,...,iT},O={o1,o2,...,oT}
例如:I=(名词,动词,名词),O=(我,看见,猫)I=(名词,动词,名词), O=(我,看见,猫)I=(名词,动词,名词),O=(我,看见,猫)
AAA是状态转移矩阵:
A=[aij]N∗N(1)A=[a_{ij}]_{N*N} \tag1A=[aij]N∗N(1)
其中,
aij=p(it+1=qj∣it=qi),i=1,2,...,N;j=1,2,...,N(2)a_{ij} = p(i_{t+1}=q_j|i_t=q_i), i=1,2,...,N; j=1,2,...,N \tag2aij=p(it+1=qj∣it=qi),i=1,2,...,N;j=1,2,...,N(2)
例如:
转态转移概率 | 名词 | 动词 | 形容词 |
---|---|---|---|
名词 | 0 | 1 | 0 |
动词 | 1/3 | 0 | 2/3 |
形容词 | 1/3 | 1/3 | 1/3 |
BBB是观测概率矩阵,也就是发射矩阵:
B=[bj(k)]N∗M(3)B=[b_j(k)]_{N*M} \tag3B=[bj(k)]N∗M(3)
其中,
bj(k)=p(ot=vk∣it=qj),k=1,2,...,M;j=1,2,...,N(4)b_j(k) = p(o_t=v_k|i_t=q_j), k=1,2,...,M; j=1,2,...,N \tag4bj(k)=p(ot=vk∣it=qj),k=1,2,...,M;j=1,2,...,N(4)
例如:
观测矩阵概率 | 我 | 看见 | 猫 | 是 | 可爱的 |
---|---|---|---|---|---|
名词 | 1 | 0 | 1 | 0 | 0 |
动词 | 0 | 1 | 0 | 1 | 0 |
形容词 | 0 | 0 | 0 | 0 | 1 |
π\piπ是初始状态概率向量:
π=(πi)(5)\pi = (\pi_i) \tag5π=(πi)(5)
其中,
πi=p(i1=qi),i=1,2,...,N(6)\pi_i = p(i_1 = q_i), i = 1,2,...,N \tag6πi=p(i1=qi),i=1,2,...,N(6)
A,BA,BA,B和π\piπ是HMM的参数,用λ\lambdaλ表示:
λ=(A,B,π)(7)\lambda = (A,B,\pi) \tag7λ=(A,B,π)(7)
例如:
名词 | 动词 | 形容词 |
---|---|---|
1 | 0 | 0 |
隐马尔可夫的三个基本问题
1.概率计算问题。给定模型λ=(A,B,π)\lambda=(A,B,\pi)λ=(A,B,π)和观测序列O=(o1,o2,...,oT)O=(o_1,o_2,...,o_T)O=(o1,o2,...,oT),计算在已知模型参数的情况下,观测序列的概率,即p(O∣λ)p(O|\lambda)p(O∣λ)。
2.学习问题。已知观测序列O=(o1,o2,...,oT)O=(o_1,o_2,...,o_T)O=(o1,o2,...,oT),估计模型参数λ=(A,B,π)\lambda=(A,B,\pi)λ=(A,B,π),使p(O∣λ)p(O|\lambda)p(O∣λ)最大。
3.预测问题,也称解码问题。已知模型λ=(A,B,π)\lambda=(A,B,\pi)λ=(A,B,π)和O=(o1,o2,...,oT)O=(o_1,o_2,...,o_T)O=(o1,o2,...,oT),求条件概率最大p(I∣O)p(I|O)p(I∣O)最大的状态序列I=(i1,i2,...,iT)I=(i_1,i_2,...,i_T)I=(i1,i2,...,iT)。
2. 概率预测问题
概率问题预测用直接计算法,计算复杂度高,可以采用动态规划形式的前向和后向算法降低计算复杂度。
为了表示方便,记:
(o1:t)=(o1,o2,...,on);(ot:T)=(ot,ot+1,...,oT)(o_{1:t} )= (o_1,o_2,...,o_n); (o_{t_:T})=(o_t,o_{t+1},...,o_T)(o1:t)=(o1,o2,...,on);(ot:T)=(ot,ot+1,...,oT)
2.1 前向算法接下来就是解前向概率p(it,o1:t∣λ)p(i_t,o_{1:t}|\lambda)p(it,o1:t∣λ):
p(it,o1:t∣λ)=∑it−1p(it−1,it,o1:t−1,ot∣λ)=∑it−1p(ot∣it−1,it,o1:t−1,λ)p(it∣it−1,o1:t−1,λ)p(it−1,o1:t−1∣λ) \begin{aligned} p(i_t,o_{1:t}|\lambda) &=\sum_{i_{t-1}} p(i_{t-1},i_t,o_{1:t-1},o_t|\lambda) \\ &=\sum_{i_{t-1}} p(o_t|i_{t-1},i_t,o_{1:t-1},\lambda)p(i_t|i_{t-1},o_{1:t-1},\lambda)p(i_{t-1},o_{1:t-1}|\lambda) \end{aligned} p(it,o1:t∣λ)=it−1∑p(it−1,it,o1:t−1,ot∣λ)=it−1∑p(ot∣it−1,it,o1:t−1,λ)p(it∣it−1,o1:t−1,λ)p(it−1,o1:t−1∣λ)
由隐马尔科夫的条件独立性假设可得:
p(ot∣it−1,it,o1:t−1,λ)=p(ot∣it,λ)p(o_t|i_{t-1},i_t,o_{1:t-1},\lambda) = p(o_t|i_t,\lambda)p(ot∣it−1,it,o1:t−1,λ)=p(ot∣it,λ)
p(it∣it−1,o1:t−1,λ)=p(it∣it−1,λ)p(i_t|i_{t-1},o_{1:t-1},\lambda)=p(i_t|i_{t-1},\lambda)p(it∣it−1,o1:t−1,λ)=p(it∣it−1,λ)
故
p(it,o1:t∣λ)=∑it−1p(ot∣it,λ)p(it∣it−1,λ)p(it−1,o1:t−1∣λ)=[∑it−1p(it−1,o1:t−1∣λ)p(it∣it−1,λ)]p(ot∣it,λ)p(i_t,o_{1:t}|\lambda)=\sum_{i_{t-1}} p(o_t|i_t,\lambda) p(i_t|i_{t-1},\lambda)p(i_{t-1},o_{1:t-1}|\lambda)=[\sum_{i_{t-1} } p(i_{t-1},o_{1:t-1}|\lambda) p(i_t|i_{t-1},\lambda)] p(o_t|i_t,\lambda)p(it,o1:t∣λ)=it−1∑p(ot∣it,λ)p(it∣it−1,λ)p(it−1,o1:t−1∣λ)=[it−1∑p(it−1,o1:t−1∣λ)p(it∣it−1,λ)]p(ot∣it,λ)
设:
αt+1(i)=p(o1:t+1,it+1=qi∣λ)(8)\alpha_{t+1}(i) = p(o_{1:t+1},i_{t+1}=q_i|\lambda) \tag8αt+1(i)=p(o1:t+1,it+1=qi∣λ)(8)
且:
p(it+1=qi∣it=qj,λ)]=ajip(i_{t+1}=q_i|i_t=q_j,\lambda)] = a_{ji}p(it+1=qi∣it=qj,λ)]=aji
p(ot+1∣it+1,λ)=bi(ot+1)p(o_{t+1}|i_{t+1},\lambda)=b_i(o_{t+1})p(ot+1∣it+1,λ)=bi(ot+1)
则:
αt+1(i)=[∑j=1Nαt(j)aji]bi(ot+1)(9)\alpha_{t+1}(i)=[\sum_{j=1}^N \alpha_t(j)a_{ji}]b_i(o_{t+1}) \tag9αt+1(i)=[j=1∑Nαt(j)aji]bi(ot+1)(9)
所以前向算法就可迭代进行。
2.2 后向算法前向算法:
1.初值α1(i)=πibi(o1)\alpha_1(i) = \pi_ib_i(o_1)α1(i)=πibi(o1)
2.递推 t=1,2,...,T−1t=1,2,...,T-1t=1,2,...,T−1
αt+1(i)=[∑j=1Nαt(j)aji]bi(ot+1)\alpha_{t+1}(i)=[\sum_{j=1}^N \alpha_t(j)a_{ji}]b_i(o_{t+1})αt+1(i)=[j=1∑Nαt(j)aji]bi(ot+1)
3.终止
p(O∣λ)=∑i=1NαT(i)p(O|\lambda) = \sum_{i=1}^N \alpha_T(i)p(O∣λ)=i=1∑NαT(i)
后向算法解决后向概率p(ot+1:T∣it,λ)p(o_{t+1:T}|i_t, \lambda)p(ot+1:T∣it,λ):
p(ot+1:T∣it,λ)=∑it+1p(it+1,ot+1,ot+2:T∣it,λ)=∑it+1p(ot+2:T∣it+1,it,ot+1,λ)p(ot+1∣it+1,it,λ)p(it+1∣it,λ) \begin{aligned} p(o_{t+1:T}|i_t, \lambda) &= \sum_{i_{t+1}} p(i_{t+1},o_{t+1},o_{t+2:T} | i_t, \lambda) \\ &= \sum_{i_{t+1}} p(o_{t+2:T}|i_{t+1}, i_t, o_{t+1}, \lambda) p(o_{t+1}|i_{t+1}, i_t, \lambda) p(i_{t+1}|i_t,\lambda)\\ \end{aligned} p(ot+1:T∣it,λ)=it+1∑p(it+1,ot+1,ot+2:T∣it,λ)=it+1∑p(ot+2:T∣it+1,it,ot+1,λ)p(ot+1∣it+1,it,λ)p(it+1∣it,λ)
由隐马尔科夫的条件独立假设得:
p(ot+2:T∣it+1,it,ot+1,λ)=p(ot+2:T∣it+1,λ)p(o_{t+2:T}|i_{t+1}, i_t, o_{t+1}, \lambda)=p(o_{t+2:T}|i_{t+1}, \lambda)p(ot+2:T∣it+1,it,ot+1,λ)=p(ot+2:T∣it+1,λ)
p(ot+1∣it+1,it,λ)=p(ot+1∣it+1,λ)p(o_{t+1}|i_{t+1}, i_t, \lambda) = p(o_{t+1}|i_{t+1}, \lambda)p(ot+1∣it+1,it,λ)=p(ot+1∣it+1,λ)
设:
βt(i)=p(ot+1:T∣it=qi,λ)(10)\beta_t(i) = p(o_{t+1:T}|i_t=q_i, \lambda) \tag{10}βt(i)=p(ot+1:T∣it=qi,λ)(10)
又:
p(it+1=qj∣it=qi,λ)=aijp(i_{t+1}=q_j|i_t=q_i,\lambda) = a_{ij}p(it+1=qj∣it=qi,λ)=aij
p(ot+1∣it+1=qj,λ)=bj(ot+1)p(o_{t+1}|i_{t+1}=q_j, \lambda) = b_j(o_{t+1})p(ot+1∣it+1=qj,λ)=bj(ot+1)
则:
βt(i)=∑j=1Naijbj(ot+1)βt+1(i)(11)\beta_t(i) = \sum_{j=1}^N a_{ij} b_j(o_{t+1}) \beta_{t+1}(i) \tag{11}βt(i)=j=1∑Naijbj(ot+1)βt+1(i)(11)
2.3 一些概率与期望值后向算法:
(1)βT(i)=1\beta_T (i) = 1βT(i)=1
(2) 对t=T-1,T-2,…,1
βt(i)=∑j=1Naijbj(ot+1)βt+1(i)\beta_t(i) = \sum_{j=1}^N a_{ij} b_j(o_{t+1}) \beta_{t+1}(i)βt(i)=j=1∑Naijbj(ot+1)βt+1(i)
(3)
p(O∣λ)=∑i=1Nπibi(o1)β1(i)p(O|\lambda) = \sum_{i=1}^N \pi_i b_i(o_1) \beta_1(i)p(O∣λ)=i=1∑Nπibi(o1)β1(i)
这两个期望值都是后面EM算法用到的中间参量
1.计算ttt时刻处于状态qiq_iqi的概率。
概率计算问题是计算p(O∣λ)p(O|\lambda)p(O∣λ),则有:
p(O∣λ)=∑itp(O,it∣λ)p(O|\lambda)=\sum_{i_t}p(O,i_t|\lambda)p(O∣λ)=it∑p(O,it∣λ)
依据隐马尔科夫的独立性假设:
p(ot+1:T∣it,o1:t,λ)=p(ot+1:T∣it,λ)p(o_{t+1:T}|i_t,o_{1:t}, \lambda) = p(o_{t+1:T}|i_t, \lambda)p(ot+1:T∣it,o1:t,λ)=p(ot+1:T∣it,λ)
所以:
p(O∣λ)=∑itp(O,it∣λ)=∑itp(ot+1:T∣it,o1:t,λ)p(it,o1:t∣λ)=∑itp(ot+1:T∣it,λ)p(it,o1:t∣λ) \begin{aligned} p(O|\lambda) &=\sum_{i_t}p(O,i_t|\lambda) \\ &=\sum_{i_t} p(o_{t+1:T}|i_t,o_{1:t}, \lambda) p(i_t,o_{1:t}|\lambda) \\ &=\sum_{i_t} p(o_{t+1:T}|i_t, \lambda) p(i_t,o_{1:t}|\lambda) \\ \end{aligned} p(O∣λ)=it∑p(O,it∣λ)=it∑p(ot+1:T∣it,o1:t,λ)p(it,o1:t∣λ)=it∑p(ot+1:T∣it,λ)p(it,o1:t∣λ)
又有:
αt(i)=p(o1:t,it=qi∣λ)(12)\alpha_t(i) = p(o_{1:t},i_t=q_i|\lambda) \tag{12}αt(i)=p(o1:t,it=qi∣λ)(12)
βt(i)=p(ot+1:T∣it=qi,λ)(13)\beta_t(i) = p(o_{t+1:T}|i_t=q_i, \lambda) \tag{13}βt(i)=p(ot+1:T∣it=qi,λ)(13)
故:
p(O,it=qi∣λ)=p(ot+1:T∣it=qi,λ)p(it=qi,o1:t∣λ)=αt(i)βt(i)p(O,i_t=q_i|\lambda) = p(o_{t+1:T}|i_t=q_i, \lambda) p(i_t=q_i,o_{1:t}|\lambda) = \alpha_t(i) \beta_t(i)p(O,it=qi∣λ)=p(ot+1:T∣it=qi,λ)p(it=qi,o1:t∣λ)=αt(i)βt(i)
p(O∣λ)=∑itαt(i)βt(i)p(O|\lambda) = \sum_{i_t} \alpha_t(i) \beta_t(i)p(O∣λ)=it∑αt(i)βt(i)
设:
γt(i)=p(it=qi∣O,λ)\gamma_t(i) = p(i_t=q_i|O,\lambda)γt(i)=p(it=qi∣O,λ)
于是可以得到:
γt(i)=p(it=qi∣O,λ)=p(it=qi,O∣λ)p(O∣λ)=αt(i)βt(i)∑j=1Nαt(j)βt(j)(14)\gamma_t(i) = p(i_t=q_i|O,\lambda) = \frac {p(i_t=q_i,O|\lambda)}{p(O|\lambda)} = \frac {\alpha_t(i) \beta_t(i)}{\sum_{j=1}^N \alpha_t(j) \beta_t(j)} \tag{14}γt(i)=p(it=qi∣O,λ)=p(O∣λ)p(it=qi,O∣λ)=∑j=1Nαt(j)βt(j)αt(i)βt(i)(14)
2.计算计算ttt时刻处于状态qiq_iqi且计算t+1t+1t+1时刻处于状态qjq_jqj的概率
p(O∣λ)=∑it∑it+1p(O,it,it+1∣λ)=∑it∑it+1p(o1:t,ot+1,ot+2:T,it,it+1∣λ)=∑it∑it+1p(ot+2:T∣o1:t,ot+1,it,it+1,λ)p(ot+1∣o1:t,it,it+1,λ)p(it+1∣it,o1:t,λ)p(it,o1:t∣λ) \begin{aligned} p(O|\lambda) &=\sum_{i_t} \sum_{i_{t+1}} p(O,i_t, i_{t+1}|\lambda) \\ &=\sum_{i_t} \sum_{i_{t+1}} p(o_{1:t},o_{t+1},o_{t+2:T},i_t, i_{t+1}|\lambda) \\ &=\sum_{i_t} \sum_{i_{t+1}} p(o_{t+2:T}|o_{1:t},o_{t+1},i_t, i_{t+1},\lambda)p(o_{t+1}|o_{1:t},i_t,i_{t+1},\lambda) p(i_{t+1}|i_t,o_{1:t},\lambda) p(i_t,o_{1:t}|\lambda) \\ \end{aligned} p(O∣λ)=it∑it+1∑p(O,it,it+1∣λ)=it∑it+1∑p(o1:t,ot+1,ot+2:T,it,it+1∣λ)=it∑it+1∑p(ot+2:T∣o1:t,ot+1,it,it+1,λ)p(ot+1∣o1:t,it,it+1,λ)p(it+1∣it,o1:t,λ)p(it,o1:t∣λ)
由隐马尔科夫的独立性假设可得:
p(O∣λ)=∑it∑it+1p(ot+2:T∣it+1,λ)p(ot+1∣it+1,λ)p(it+1∣it,λ)p(it,o1:t∣λ)p(O|\lambda) = \sum_{i_t} \sum_{i_{t+1}} p(o_{t+2:T}| i_{t+1},\lambda)p(o_{t+1}|i_{t+1},\lambda) p(i_{t+1}|i_t,\lambda) p(i_t,o_{1:t}|\lambda)p(O∣λ)=it∑it+1∑p(ot+2:T∣it+1,λ)p(ot+1∣it+1,λ)p(it+1∣it,λ)p(it,o1:t∣λ)
设:
ξt(i,j)=p(it=qi,it+1=qj∣O,λ)\xi_t(i,j)=p(i_t=q_i,i_{t+1}=q_j|O,\lambda)ξt(i,j)=p(it=qi,it+1=qj∣O,λ)
又有公式(2)(4)(12)(13)
得:
ξt(i,j)=p(it=qi,it+1=qj∣O,λ)p(O∣λ)=αt(i)aijbj(ot+1)βt+1(j)∑i=1N∑j=1Nαt(i)aijbj(ot+1)βt+1(j)(15)\xi_t(i,j) = \frac {p(i_t=q_i,i_{t+1}=q_j|O,\lambda)}{p(O|\lambda)} =\frac {\alpha_t(i) a_{ij} b_j(o_{t+1}) \beta_{t+1}(j)} {\sum_{i=1}^N \sum_{j=1}^N \alpha_t(i) a_{ij} b_j(o_{t+1}) \beta_{t+1}(j)} \tag{15}ξt(i,j)=p(O∣λ)p(it=qi,it+1=qj∣O,λ)=∑i=1N∑j=1Nαt(i)aijbj(ot+1)βt+1(j)αt(i)aijbj(ot+1)βt+1(j)(15)
3. 学习问题
3.1 监督学习如果有标记好状态序列的样本,那就太好办了,直接将接个矩阵统计的各个维度定义后进行统计就可以了。统计过程中注意概率之和为一的约束。
3.2 无监督学习如果没有标记状态序列的样本,可以用Baum-Welch算法(EM算法)实现。
已知:包含SSS个长度为TTT的观测序列的观测序列{O1,O2,...,OS}\{O_1,O_2,...,O_S \}{O1,O2,...,OS}
目标:学习隐马尔可夫模型的参数λ=(A,B,π)\lambda=(A,B,\pi)λ=(A,B,π)
记观测数据OOO,隐数据III,那么隐马尔可夫模型可以表示为:
p(O∣λ)=∑Ip(O∣I,λ)p(I∣λ)p(O|\lambda) = \sum_I p(O|I,\lambda) p(I|\lambda)p(O∣λ)=I∑p(O∣I,λ)p(I∣λ)
E步:
因为对λ\lambdaλ而言,1/p(O∣λ‾)1/p(O| \overline \lambda)1/p(O∣λ)是常数项,所以
Q(λ,λ‾)=EI[logp(O,I∣λ)∣O,λ‾]=∑Ilogp(O,I∣λ)p(I∣O,λ‾)=∑Ilogp(O,I∣λ)p(I,O∣λ‾)p(O∣λ‾)=∑Ilogp(O,I∣λ)p(I,O∣λ‾) \begin{aligned} Q(\lambda,\overline \lambda) &= E_I[\log p(O,I|\lambda)|O, \overline \lambda] \\ &= \sum_I \log p(O,I|\lambda) p(I|O,\overline \lambda) \\ &= \sum_I \log p(O,I|\lambda) \frac {p(I,O|\overline \lambda)}{p(O| \overline \lambda)} \\ &= \sum_I \log p(O,I|\lambda) p(I,O|\overline \lambda) \\ \end{aligned} Q(λ,λ)=EI[logp(O,I∣λ)∣O,λ]=I∑logp(O,I∣λ)p(I∣O,λ)=I∑logp(O,I∣λ)p(O∣λ)p(I,O∣λ)=I∑logp(O,I∣λ)p(I,O∣λ)
将概率计算问题2.1小姐中前向算法的递归公式展开就可以得到:
p(O,I∣λ)=πi1bi1(o1)ai1i2bi2(o2)...aiT−1iTbiT(oT)=πi1[∏t=1T−1aitit+1][∏t=1Tbit(ot)]p(O,I|\lambda) = \pi_{i_1} b_{i_1}(o_1) a_{i_1i_2} b_{i_2}(o_2) ... a_{i_{T-1}i_T} b_{iT}(o_T) = \pi_{i_1} [\prod_{t=1}^{T-1} a_{i_ti_{t+1}}][\prod_{t=1}^T b_{it}(o_t)]p(O,I∣λ)=πi1bi1(o1)ai1i2bi2(o2)...aiT−1iTbiT(oT)=πi1[t=1∏T−1aitit+1][t=1∏Tbit(ot)]
于是:
Q(λ,λ‾)=∑Ilogπi1p(O,I∣λ‾)+∑I(∑t=1T−1aitit+1)p(O,I∣λ‾)+∑I(∑t=1Tbit(ot))p(O,I∣λ‾)(16)Q(\lambda, \overline \lambda) = \sum_I \log \pi_{i_1} p(O, I| \overline \lambda) + \sum_I (\sum_{t=1}^{T-1} a_{i_ti_{t+1}}) p(O, I| \overline \lambda) + \sum_I (\sum_{t=1}^T b_{it}(o_t)) p(O, I| \overline \lambda) \tag{16}Q(λ,λ)=I∑logπi1p(O,I∣λ)+I∑(t=1∑T−1aitit+1)p(O,I∣λ)+I∑(t=1∑Tbit(ot))p(O,I∣λ)(16)
特此说明隐变量
隐马尔可夫模型的隐变量就是观测序列对应的状态序列,所以隐变量可以用(14)式的变量表示
后面在M步中更新模型参数的时候也用到了(15)式,是不是就说明隐变量是两个,其实不是的,这儿只是为了表示的方便和算法的方便。
也就是在E步中,用γ\gammaγ和ξ\xiξ表示隐变量,只是为了编程和表示的便利,这两个变量在E步中信息是重复的。
M步:
1.求解πi\pi_iπi
由(15)式可得:
L(πi1)=∑Ilogπi1p(O,I∣λ‾)=∑iNlogπi1p(O,i1=i∣λ‾)L(\pi_{i_1}) = \sum_I \log \pi_{i_1} p(O, I| \overline \lambda) = \sum_{i}^N \log \pi_{i_1} p(O, i_1=i| \overline \lambda)L(πi1)=I∑logπi1p(O,I∣λ)=i∑Nlogπi1p(O,i1=i∣λ)
又因为πi\pi_iπi满足约束条件∑i=1Nπi1=1\sum_{i=1}^N \pi_{i_1}=1∑i=1Nπi1=1,利用拉格朗日乘子法,写出拉格朗日函数:
∑i=1Nlogπip(O,i1=i∣λ‾)+γ(∑i=1Nπi−1)\sum_{i=1}^N \log \pi_{i} p(O, i_1=i| \overline \lambda) + \gamma(\sum_{i=1}^N \pi_{i} - 1)i=1∑Nlogπip(O,i1=i∣λ)+γ(i=1∑Nπi−1)
对其求偏导并且令其结果为0得:
∂∂πi[∑i=1Nlogπip(O,i=i∣λ‾)+γ(∑i1=1Nπi−1)]=0(17)\frac {\partial} {\partial \pi_i} [\sum_{i=1}^N \log \pi_{i} p(O, i=i| \overline \lambda) + \gamma(\sum_{i_1=1}^N \pi_{i} - 1)]=0 \tag{17}∂πi∂[i=1∑Nlogπip(O,i=i∣λ)+γ(i1=1∑Nπi−1)]=0(17)
得:
p(O,i1=i∣λ‾)+γπi=0p(O, i_1=i| \overline \lambda) + \gamma \pi_i=0p(O,i1=i∣λ)+γπi=0
得到:
πi=p(O,i1=i∣λ‾)−λ\pi_i = \frac {p(O, i_1=i| \overline \lambda)} {-\lambda}πi=−λp(O,i1=i∣λ)
带入∑i=1Nπi1=1\sum_{i=1}^N \pi_{i_1}=1∑i=1Nπi1=1的:
−λ=∑i=1Np(O,i1=i∣λ‾)=p(o∣λ‾)-\lambda = \sum_{i=1}^N p(O, i_1=i| \overline \lambda) = p(o|\overline \lambda)−λ=i=1∑Np(O,i1=i∣λ)=p(o∣λ)
求得并有公式(14):
πi=p(O,i1=i∣λ‾)p(o∣λ‾)=γ1(i)(18)\pi_i = \frac {p(O, i_1=i| \overline \lambda)} {p(o|\overline \lambda)} = \gamma_1(i) \tag{18}πi=p(o∣λ)p(O,i1=i∣λ)=γ1(i)(18)
2.求解aija_{ij}aij:
L(aij)=∑I(∑t=1T−1aitit+1)p(O,I∣λ‾)=∑i=1N(∑t=1T−1aitit+1)(∑j=1Np(O,it=i,it+1=j∣λ‾))=∑i=1N∑j=1N∑t=1T−1aijp(O,it=i,it+1=j∣λ‾)L(a_{ij})=\sum_I (\sum_{t=1}^{T-1} a_{i_ti_{t+1}}) p(O, I| \overline \lambda) = \sum_{i=1}^N (\sum_{t=1}^{T-1} a_{i_ti_{t+1}}) ( \sum_{j=1}^N p(O, i_t=i, i_{t+1}=j| \overline \lambda) ) \\ = \sum_{i=1}^N \sum_{j=1}^N \sum_{t=1}^{T-1} a_{ij} p(O, i_t=i, i_{t+1}=j| \overline \lambda) L(aij)=I∑(t=1∑T−1aitit+1)p(O,I∣λ)=i=1∑N(t=1∑T−1aitit+1)(j=1∑Np(O,it=i,it+1=j∣λ))=i=1∑Nj=1∑Nt=1∑T−1aijp(O,it=i,it+1=j∣λ)
应用约束条件∑j=1Naij=1\sum_{j=1}^N a_{ij} = 1∑j=1Naij=1,用拉格朗日乘子法可以求出:
∑i=1N∑j=1N∑t=1T−1aijp(O,it=i,it+1=j∣λ‾)+λ(∑j=1Naij−1)\sum_{i=1}^N \sum_{j=1}^N \sum_{t=1}^{T-1} a_{ij} p(O, i_t=i, i_{t+1}=j| \overline \lambda) + \lambda(\sum_{j=1}^N a_{ij} - 1)i=1∑Nj=1∑Nt=1∑T−1aijp(O,it=i,it+1=j∣λ)+λ(j=1∑Naij−1)
对上式求骗到并等于0得到:
∂∂aij[∑i=1N∑j=1N∑t=1T−1aijp(O,it=i,it+1=j∣λ‾)+λ(∑j=1Naij−1)]=0\frac {\partial}{\partial a_{ij}} [\sum_{i=1}^N \sum_{j=1}^N \sum_{t=1}^{T-1} a_{ij} p(O, i_t=i, i_{t+1}=j| \overline \lambda) + \lambda(\sum_{j=1}^N a_{ij} - 1)] = 0∂aij∂[i=1∑Nj=1∑Nt=1∑T−1aijp(O,it=i,it+1=j∣λ)+λ(j=1∑Naij−1)]=0
得到:
∑t=1T−1p(O,it=i,it+1=j∣λ‾)+λaij=0\sum_{t=1}^{T-1} p(O, i_t=i, i_{t+1}=j| \overline \lambda) + \lambda a_{ij} = 0 t=1∑T−1p(O,it=i,it+1=j∣λ)+λaij=0
所以:
aij=∑t=1T−1p(O,it=i,it+1=j∣λ‾)−λa_{ij} = \frac {\sum_{t=1}^{T-1} p(O, i_t=i, i_{t+1}=j| \overline \lambda)}{- \lambda} aij=−λ∑t=1T−1p(O,it=i,it+1=j∣λ)
将上式带入∑j=1Naij=1\sum_{j=1}^N a_{ij} = 1∑j=1Naij=1:
−λ=∑j=1N∑t=1T−1p(O,it=i,it+1=j∣λ‾)=∑t=1T−1p(O,it=i∣λ‾)- \lambda = \sum_{j=1}^N \sum_{t=1}^{T-1} p(O, i_t=i, i_{t+1}=j| \overline \lambda) = \sum_{t=1}^{T-1} p(O, i_t=i| \overline \lambda) −λ=j=1∑Nt=1∑T−1p(O,it=i,it+1=j∣λ)=t=1∑T−1p(O,it=i∣λ)
故得:
aij=∑t=1T−1p(O,it=i,it+1=j∣λ‾)∑t=1T−1p(O,it=i∣λ‾)=∑t=1T−1p(O,it=i,it+1=j∣λ‾)/p(o∣λ‾)∑t=1T−1p(O,it=i∣λ‾)/p(o∣λ‾)a_{ij} = \frac {\sum_{t=1}^{T-1} p(O, i_t=i, i_{t+1}=j| \overline \lambda)}{\sum_{t=1}^{T-1} p(O, i_t=i| \overline \lambda) } = \frac {\sum_{t=1}^{T-1} p(O, i_t=i, i_{t+1}=j| \overline \lambda) / p(o|\overline \lambda)} {\sum_{t=1}^{T-1} p(O, i_t=i| \overline \lambda) / p(o|\overline \lambda) } aij=∑t=1T−1p(O,it=i∣λ)∑t=1T−1p(O,it=i,it+1=j∣λ)=∑t=1T−1p(O,it=i∣λ)/p(o∣λ)∑t=1T−1p(O,it=i,it+1=j∣λ)/p(o∣λ)
将(14)和(15)带入的:
aij=∑t=1T−1ξt(i,j)∑t=1T−1γt(i)(19)a_{ij} = \frac {\sum_{t=1}^{T-1} \xi_t(i,j)} {\sum_{t=1}^{T-1} \gamma_t(i) } \tag{19}aij=∑t=1T−1γt(i)∑t=1T−1ξt(i,j)(19)
3.求解bjkb_j{k}bjk:
L(bjk)=∑I(∑t=1Tbit(ot))p(O,I∣λ‾)=∑j=1N∑t=1Tbj(ot)p(O,it=j∣λ‾)L(b_j{k}) = \sum_I (\sum_{t=1}^T b_{it}(o_t)) p(O, I| \overline \lambda) = \sum_{j=1}^N \sum_{t=1}^T b_{j}(o_t) p(O, i_t=j| \overline \lambda) L(bjk)=I∑(t=1∑Tbit(ot))p(O,I∣λ)=j=1∑Nt=1∑Tbj(ot)p(O,it=j∣λ)
在约束条件∑k=1Mbj(k)=1\sum_{k=1}^M b_j(k) = 1∑k=1Mbj(k)=1的拉格朗日乘子法:
∑j=1N∑t=1Tbj(ot)p(O,it=j∣λ‾)+λ(∑k=1Mbj(k)−1)\sum_{j=1}^N \sum_{t=1}^T b_{j}(o_t) p(O, i_t=j| \overline \lambda) + \lambda(\sum_{k=1}^M b_j(k) - 1)j=1∑Nt=1∑Tbj(ot)p(O,it=j∣λ)+λ(k=1∑Mbj(k)−1)
对其求偏导得:
∂∂bj(k)[∑j=1N∑t=1Tbj(ot)p(O,it=j∣λ‾)+λ(∑k=1Mbj(k)−1)]=0\frac {\partial}{\partial b_j(k)} [\sum_{j=1}^N \sum_{t=1}^T b_{j}(o_t) p(O, i_t=j| \overline \lambda) + \lambda(\sum_{k=1}^M b_j(k) - 1)] = 0∂bj(k)∂[j=1∑Nt=1∑Tbj(ot)p(O,it=j∣λ)+λ(k=1∑Mbj(k)−1)]=0
因为只有在ot=vko_t=v_kot=vk时偏导才不会等于0,以I(ot=vk)I(o_t=v_k)I(ot=vk)表示,则:
∑t=1Tp(O,it=j∣λ‾)I(ot=vk)+λbj(ot)I(ot=vk)=0\sum_{t=1}^T p(O, i_t=j| \overline \lambda) I(o_t=v_k) + \lambda b_{j}(o_t)I(o_t=v_k) = 0t=1∑Tp(O,it=j∣λ)I(ot=vk)+λbj(ot)I(ot=vk)=0
bj(ot)I(ot=vk)b_{j}(o_t)I(o_t=v_k)bj(ot)I(ot=vk)可以写作bj(k)b_{j}(k)bj(k),故:
bj(k)=∑t=1Tp(O,it=j∣λ‾)I(ot=vk)−λb_{j}(k) = \frac {\sum_{t=1}^T p(O, i_t=j| \overline \lambda) I(o_t=v_k)} {- \lambda}bj(k)=−λ∑t=1Tp(O,it=j∣λ)I(ot=vk)
将上式带入∑k=1Mbj(k)=1\sum_{k=1}^M b_j(k) = 1∑k=1Mbj(k)=1得:
−λ=∑k=1M∑t=1Tp(O,it=j∣λ‾)I(ot=vk)=∑t=1Tp(O,it=j∣λ‾)- \lambda = \sum_{k=1}^M \sum_{t=1}^T p(O, i_t=j| \overline \lambda) I(o_t=v_k) = \sum_{t=1}^T p(O, i_t=j| \overline \lambda)−λ=k=1∑Mt=1∑Tp(O,it=j∣λ)I(ot=vk)=t=1∑Tp(O,it=j∣λ)
得到:
bj(k)=∑t=1Tp(O,it=j∣λ‾)I(ot=vk)∑t=1Tp(O,it=j∣λ‾)b_{j}(k) = \frac {\sum_{t=1}^T p(O, i_t=j| \overline \lambda) I(o_t=v_k)} {\sum_{t=1}^T p(O, i_t=j| \overline \lambda)}bj(k)=∑t=1Tp(O,it=j∣λ)∑t=1Tp(O,it=j∣λ)I(ot=vk)
又有(14)式可得:
bj(k)=∑t=1,ot=vkTγt(j)∑t=1Tγt(j)(20)b_{j}(k) = \frac {\sum_{t=1,o_t=v_k}^T \gamma_t(j)} {\sum_{t=1}^T \gamma_t(j)} \tag{20}bj(k)=∑t=1Tγt(j)∑t=1,ot=vkTγt(j)(20)
EM算法总结:
E步:γt(i)=p(it=qi∣O,λ)=p(it=qi,O∣λ)p(O∣λ)=αt(i)βt(i)∑j=1Nαt(j)βt(j)\gamma_t(i) = p(i_t=q_i|O,\lambda) = \frac {p(i_t=q_i,O|\lambda)}{p(O|\lambda)} = \frac {\alpha_t(i) \beta_t(i)}{\sum_{j=1}^N \alpha_t(j) \beta_t(j)} γt(i)=p(it=qi∣O,λ)=p(O∣λ)p(it=qi,O∣λ)=∑j=1Nαt(j)βt(j)αt(i)βt(i)
ξt(i,j)=p(it=qi,it+1=qj∣O,λ)p(O∣λ)=αt(i)aijbj(ot+1)βt+1(j)∑i=1N∑j=1Nαt(i)aijbj(ot+1)βt+1(j)\xi_t(i,j) = \frac {p(i_t=q_i,i_{t+1}=q_j|O,\lambda)}{p(O|\lambda)} =\frac {\alpha_t(i) a_{ij} b_j(o_{t+1}) \beta_{t+1}(j)} {\sum_{i=1}^N \sum_{j=1}^N \alpha_t(i) a_{ij} b_j(o_{t+1}) \beta_{t+1}(j)} ξt(i,j)=p(O∣λ)p(it=qi,it+1=qj∣O,λ)=∑i=1N∑j=1Nαt(i)aijbj(ot+1)βt+1(j)αt(i)aijbj(ot+1)βt+1(j)
M步:
πi=p(O,i1=i∣λ‾)p(o∣λ‾)=γ1(i)\pi_i = \frac {p(O, i_1=i| \overline \lambda)} {p(o|\overline \lambda)} = \gamma_1(i) πi=p(o∣λ)p(O,i1=i∣λ)=γ1(i)aij=∑t=1T−1ξt(i,j)∑t=1T−1γt(i)a_{ij} = \frac {\sum_{t=1}^{T-1} \xi_t(i,j)} {\sum_{t=1}^{T-1} \gamma_t(i) } aij=∑t=1T−1γt(i)∑t=1T−1ξt(i,j)
bj(k)=∑t=1,ot=vkTγt(j)∑t=1Tγt(j)b_{j}(k) = \frac {\sum_{t=1,o_t=v_k}^T \gamma_t(j)} {\sum_{t=1}^T \gamma_t(j)} bj(k)=∑t=1Tγt(j)∑t=1,ot=vkTγt(j)
4. 预测问题(解码问题)
用维特比算法进行求解:
已知:模型λ=(A,B,π)\lambda=(A,B,\pi)λ=(A,B,π)和O=(o1,o2,...,oT)O=(o_1,o_2,...,o_T)O=(o1,o2,...,oT)
求:条件概率最大p(I∣O,λ)p(I|O,\lambda)p(I∣O,λ)最大的状态序列I=(i1,i2,...,iT)I=(i_1,i_2,...,i_T)I=(i1,i2,...,iT)
因为p(O)p(O)p(O)是一个定值,所以:
maxIp(I∣O,λ)=maxIp(I,O∣λ)/p(O∣λ)=maxIp(I,O∣λ)\max_I p(I|O,\lambda) = \max_I p(I, O|\lambda) / p(O|\lambda) = \max_I p(I, O|\lambda)Imaxp(I∣O,λ)=Imaxp(I,O∣λ)/p(O∣λ)=Imaxp(I,O∣λ)
定义在时刻ttt状态为iii的所有单个路径(i1,i2,...,it)(i_1,i_2,...,i_t)(i1,i2,...,it)中概率最大值为:
δt(i)=maxi1,i2,...,it−1p(it=i,it−1:i1,ot:1∣λ)\delta_t(i) = \max_{i_1,i_2,...,i_{t-1}} p(i_t=i, i_{t-1:i_1},o_{t:1}|\lambda)δt(i)=i1,i2,...,it−1maxp(it=i,it−1:i1,ot:1∣λ)
递推推导:
p(it+1=i,it:1,ot+1:1∣λ)=p(it+1=i,it,it−1:1,ot+1,ot:1∣λ)=p(ot+1∣it+1=i,it,ot:1,λ)p(it+1=i∣it,it−1:1,ot:1,λ)p(it,it−1:1,ot:1∣λ)=p(ot+1∣it+1=i,λ)p(it+1=i∣it,λ)p(it,it−1:1,ot:1∣λ) \begin{aligned} &p(i_{t+1}=i,i_{t:1},o_{t+1:1}| \lambda) \\ &=p(i_{t+1}=i,i_t,i_{t-1:1},o_{t+1},o_{t:1}| \lambda) \\ &= p(o_{t+1}|i_{t+1}=i,i_t,o_{t:1},\lambda) p(i_{t+1}=i|i_t,i_{t-1:1},o_{t:1}, \lambda) p(i_t,i_{t-1:1},o_{t:1}|\lambda) \\ &= p(o_{t+1}|i_{t+1}=i,\lambda) p(i_{t+1}=i|i_t,\lambda) p(i_t,i_{t-1:1},o_{t:1}|\lambda) \\ \end{aligned} p(it+1=i,it:1,ot+1:1∣λ)=p(it+1=i,it,it−1:1,ot+1,ot:1∣λ)=p(ot+1∣it+1=i,it,ot:1,λ)p(it+1=i∣it,it−1:1,ot:1,λ)p(it,it−1:1,ot:1∣λ)=p(ot+1∣it+1=i,λ)p(it+1=i∣it,λ)p(it,it−1:1,ot:1∣λ)
故:
δt+1(i)=maxi1,i2,...,it−1p(it+1=i,it:1,ot+1:1∣λ)=max1≤j≤N[δt(j)aji]bi(ot+1)(21)\delta_{t+1}(i) = \max_{i_1,i_2,...,i_{t-1}} p(i_{t+1}=i,i_{t:1},o_{t+1:1}| \lambda) = \max_{1 \le j \le N} [\delta _t(j) a_{ji}] b_i(o_{t+1}) \tag{21}δt+1(i)=i1,i2,...,it−1maxp(it+1=i,it:1,ot+1:1∣λ)=1≤j≤Nmax[δt(j)aji]bi(ot+1)(21)
定义在时刻ttt状态为iii的所有单个路径(i1,i2,...,it−1)(i_1,i_2,...,i_{t-1})(i1,i2,...,it−1)中概率最大的第t−1t-1t−1个节点为:
ψt(i)=argmax1≤j≤N[δt−1(j)aji](22)\psi_t(i) = \arg \max_{1 \le j \le N}[\delta _{t-1}(j) a_{ji}] \tag{22}ψt(i)=arg1≤j≤Nmax[δt−1(j)aji](22)
5. python实现模型
5.1 参数对应关系下面说一下上面公式中出现的参数和下面模型之中的名称的对应关系(公式中的符号将会和代码一致):
:param N: NNN 表示状态数
:param M: MMM 表示观测数
:param V: VVV 表示观测集合,维度(M,)(M,)(M,)
:param A: AAA 对应于状态转移矩阵,维度(N,N)(N, N)(N,N)
:param B: BBB对应于观测概率矩阵(发射矩阵),维度(N,M)(N, M)(N,M)
:param pi: π\piπ 对应于初始状态向量,维度(N,)(N,)(N,)
:param S: SSS 表示输入句子数量
:param T: TTT 表示每个句子的个数
:param gamma: γ\gammaγ 隐变量,表示状态的概率矩阵,维度(S,N,T)(S,N,T)(S,N,T)
:param xi: ξ\xiξ 隐变量,表示状态的概率矩阵,维度(S,N,N,T)(S,N,N,T)(S,N,N,T)
:param alpha: α\alphaα 前向算法结果,维度(N,T)(N,T)(N,T)
:param beta: β\betaβ 后向算法结果,维度(N,T)(N,T)(N,T)
:param delta: δ\deltaδ 维特比算法中存储概率最大值,维度(N,T)(N,T)(N,T)
:param psi: ψ\psiψ 维特比算法中存储概率最大值索引,维度(N,T)(N,T)(N,T)
:param I: III 输出的状态向量,维度(T,)(T,)(T,)
小技巧:为了避免连续乘积带来内存溢出,一般先用对数进行计算,最后再用指数运算还原。
logsumexp() # http://bayesjumping.net/log-sum-exp-trick/
log∑iexp(xi)=b+log∑iexp(xi−b)\log\sum_i \exp(x_i) = b + \log \sum_i \exp(x_i-b)logi∑exp(xi)=b+logi∑exp(xi−b)
def logSumExp(ns): max = np.max(ns) ds = ns - max sumOfExp = np.exp(ds).sum() return max + np.log(sumOfExp)5.2 python实现HMM
import numpy as np class MyHMM(object): def __init__(self, N=None, A=None, B=None, pi=None): """ HMM模型: >隐马尔可夫的三个基本问题 >1.概率计算问题。给定模型$\lambda=(A,B,\pi)$和观测序列$O=(o_1,o_2,...,o_T)$,计算在已知模型参数的情况下,观测序列的概率, 即$p(O|\lambda)$。用前向算法或后向算法。 >2.学习问题。已知观测序列$O=(o_1,o_2,...,o_T)$,估计模型参数$\lambda=(A,B,\pi)$,使$p(O|\lambda)$最大。用BW算法。 >3.预测问题,也称解码问题。已知模型$\lambda=(A,B,\pi)$和$O=(o_1,o_2,...,o_T)$,求条件概率最大$p(I|O)$最大的状态序列 $I=(i_1,i_2,...,i_T)$。用维特比算法解码。 :param N: $N$ 表示状态数 :param M: $M$ 表示观测数 :param V: $V$ 表示观测集合,维度$(M,)$ :param A: $A$ 对应于状态转移矩阵,维度$(N, N)$ :param B: $B$对应于观测概率矩阵(发射矩阵),维度$(N, M)$ :param pi: $pi$ 对应于初始状态向量,维度$(N,)$ :param S: $S$ 表示输入句子数量 :param T: $T$ 表示每个句子的个数 :param gamma: $gamma$ 隐变量,表示状态的概率矩阵,维度$(S,N,T)$ :param xi: $xi$ 隐变量,表示状态的概率矩阵,维度$(S,N,N,T)$ :param alpha: $alpha$ 前向算法结果,维度$(N,T)$ :param beta: $beta$ 后向算法结果,维度$(N,T)$ :param delta: $delta$ 维特比算法中存储概率最大值,维度$(N,T)$ :param psi: $psi$ 维特比算法中存储概率最大值索引,维度$(N,T)$ :param I: $I$ 输出的状态向量,维度$(T,)$ """ self.N = N # 状态数 self.params = { 'A': A, 'B': B, 'pi': pi, 'gamma': None, 'xi': None } self.M = None # 观测数 self.S = None # 句子个数 self.T = None # 每个句子的长度 self.V = None # 观测集合 self.eps = np.finfo(float).eps np.random.seed(2) def _init_params(self): """ 初始化模型参数 :return: """ def generate_random_n_data(N): ret = np.random.rand(N) return ret / np.sum(ret) pi = generate_random_n_data(self.N) A = np.array([generate_random_n_data(self.N) for _ in range(self.N)]) B = np.array([generate_random_n_data(self.M) for _ in range(self.N)]) gamma = np.zeros((self.S, self.N, self.T)) xi = np.zeros((self.S, self.N, self.N, self.T)) self.params = { 'A': A, 'B': B, 'pi': pi, 'gamma': gamma, 'xi': xi } def logSumExp(self, ns, axis=None): max = np.max(ns) ds = ns - max sumOfExp = np.exp(ds).sum(axis=axis) return max + np.log(sumOfExp) def _forward(self, O_s): """ 前向算法,公式参考博客公式(9) :param O_s: 单个序列,维度(N,) :return: """ A = self.params['A'] B = self.params['B'] pi = self.params['pi'] T = len(O_s) log_alpha = np.zeros((self.N, T)) for i in range(self.N): log_alpha[i, 0] = np.log(pi[i] + self.eps) + np.log(B[i, O_s[0]]) for t in range(1, T): for i in range(self.N): log_alpha[i, t] = self.logSumExp(np.array([log_alpha[_i, t-1] + np.log(A[_i, i] + self.eps) + np.log(B[i, O_s[t]]) for _i in range(self.N)])) return log_alpha def _backward(self, O_s): """ 后向算法,参考博客公式(11) :param O_s: 单个序列,维度(N,) :return: """ A = self.params['A'] B = self.params['B'] pi = self.params['pi'] T = len(O_s) log_beta = np.zeros((self.N, T)) for i in range(self.N): log_beta[i, T-1] = 0 for t in range(T-2, -1, -1): for i in range(self.N): log_beta[i, t] = self.logSumExp(np.array([ log_beta[_i, t+1] + np.log(A[i, _i] + self.eps) + np.log(B[_i, O_s[t+1]] + self.eps) for _i in range(self.N)])) return log_beta def _E_step(self, O): """ BW算法的E_step 计算隐变量,参考博客公式(9)(11) :param O: :return: """ A = self.params['A'] B = self.params['B'] pi = self.params['pi'] # 对S个句子依次执行 for s in range(self.S): O_s = O[s] log_alpha = self._forward(O_s) log_beta = self._backward(O_s) # 前向算法得到的最大似然 log_likelihood = self.logSumExp(log_alpha[:, self.T -1]) # log p(O|lambda) # # 后向算法得到的最大似然 (两个结果应该是相等的) # log_likelihood = self.logSumExp(np.array([np.log(pi[_i] + self.eps) + np.log(B[_i, 0] + self.eps) + log_beta[_i, 0] for _i in range(self.N)])) for i in range(self.N): self.params['gamma'][s, i, self.T-1] = log_alpha[i, self.T-1] + log_beta[i, self.T-1] - log_likelihood for t in range(self.T - 1): for i in range(self.N): self.params['gamma'][s, i, t] = log_alpha[i, t] + log_beta[i, t] - log_likelihood for j in range(self.N): self.params['xi'][s, i, j, t] = log_alpha[i, t] + np.log(A[i, j] + self.eps) + np.log(B[j, O_s[t + 1]] + self.eps) + log_beta[j, t+1] - log_likelihood def _M_step(self, O): """ BW算法的M_step。参考博客公式(18)(19)(20) :param O: :return: """ gamma = self.params['gamma'] xi = self.params['xi'] count_gamma = np.zeros((self.S, self.N, self.M)) count_xi = np.zeros((self.S, self.N, self.N)) for s in range(self.S): O_s = O[s, :] for i in range(self.N): for k in range(self.M): if not (O_s == k).any(): count_gamma[s, i, k] = np.log(self.eps) else: count_gamma[s, i, k] = self.logSumExp(gamma[s, i, O_s == k]) for j in range(self.N): count_xi[s, i, j] = self.logSumExp(xi[s, i, j, :]) self.params['pi'] = np.exp(self.logSumExp(gamma[:, :, 0], axis=0) - np.log(self.S + self.eps)) np.testing.assert_almost_equal(self.params['pi'].sum(), 1) for i in range(self.N): for k in range(self.M): self.params['B'][i, k] = np.exp(self.logSumExp(count_gamma[:, i, k]) - self.logSumExp( count_gamma[:, i, :] )) for j in range(self.N): self.params['A'][i, j] = np.exp(self.logSumExp(count_xi[:, i, j]) - self.logSumExp( count_xi[:, i, :] )) np.testing.assert_almost_equal(self.params['A'][i, :].sum(), 1) np.testing.assert_almost_equal(self.params['B'][i, :].sum(), 1) def fit(self, O, V=(0,1,2,3,4), max_iter=20): O = np.array(O) self.S, self.T = O.shape self.M = len(V) self.V = V print(self.S, self.T) self._init_params() for i in range(max_iter): self._E_step(O) self._M_step(O) def decode(self, O_s): """ 用维特比算法解码。参考公式(21)(22) :param O_s: :return: """ O_s = np.array(O_s) if len(O_s.shape) != 1: raise ('只容许一个序列进行解码.') T = len(O_s) delta = np.zeros((self.N, self.T)) psi = np.zeros((self.N, self.T)) for i in range(self.N): psi[i, 0] = 0 delta[i, 0] = np.log(self.params['pi'][i] + self.eps) + np.log(self.params['B'][i, O_s[0]]) for t in range(1, T): for i in range(self.N): seq_prob = [delta[j, t-1] + np.log(self.params['A'][j, i] + self.eps) + np.log(self.params['B'][i, O_s[t]]) for j in range(self.N)] delta[i, t] = np.max(seq_prob) psi[i, t] = np.argmax(seq_prob) pointer = np.argmax(delta[:, -1]) I = [pointer] for t in reversed(range(1, T)): pointer = int(psi[int(pointer), t]) I.append(pointer) I.reverse() return I5.3 模型测试
def generate_data(): O = [['我', '看见', '猫'], ['猫', '是', '可爱的'], ['我', '是', '可爱的']] word2index = {} index2word = {} for sentence in O: for word in sentence: if word not in word2index.keys(): word2index[word] = len(word2index) index2word[len(index2word)] = word print(word2index) print(index2word) O_input = [] for sentence in O: O_input.append([word2index[word] for word in sentence]) print(O_input) return O_input def run_my_model(): O_input = generate_data() N = 3 # 隐变量的维度设为3,表示有3种词性 my = MyHMM(N=N) my.fit(O_input) print('A:') print(my.params['A']) print('B:') print(my.params['B']) print('pi:') print(my.params['pi']) I = my.decode(O_s=(2, 1, 0)) print("I:") print(I)
打印的结果为:
A:
[[0.33333528 0.33332093 0.33334378]
[0.43652988 0.25000742 0.3134627 ]
[0.2500279 0.4999721 0.25 ]]
B:
[[9.91856630e-17 1.12533719e-04 1.06467892e-01 3.70235380e-05 8.93382551e-01]
[7.40229993e-17 3.33285967e-01 1.91726675e-07 6.66712274e-01 1.56719927e-06]
[5.31681136e-01 1.48466699e-11 4.68318638e-01 1.95615241e-11 2.25912305e-07]]
pi:
[1.16786216e-17 3.50652002e-23 1.00000000e+00]
I:
[2, 1, 2]
参考资料:
《统计学习方法》李航著
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