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HMM算法python实现

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基础介绍，后5项为基础5元素Q = ['q0', 'q1', 'q2', 'q3']# 状态集合 States，共 N 种状态V = ['v0', 'v1']# 观测集合 Observations ， 共 M 种观测值I = [ 'i{}'.format(i) for i in range(5) ] # 某个长度为 T 的状态序列，i_t 属于QO = [ 'o{}'.format(i) for i in range(5) ] # 状态序列对应的观测值序列 ， o_t 属于 VA = [ a_ij ]# 转移概率 Transition Problity, a_ij = P( i_t+1 = q_j | i_t = q_i ) N*NB = [ bj(o_t) ]# 发射概率 Emission Problity，b_ij = P( o_t = v_k |i_t = q_j ) N*MPi = [ P_i ]# 初识状态概率 P_i = P( i_1 = q_i )基础5元素对应初始化# Q = ['盒1', '盒2', '盒3']Q = ['盒1', '盒2']V = [ '红' , '黑' ]# A = [ [ 0.2, 0.3 , 0.5 ] ,#[ 0, 0.5 , 0.5 ] ,#[ 0.4, 0.2 , 0.2 ]]A = [ [ 0.5 , 0.5 ] ,[ 0.5 , 0.5 ]]B = [ [ 0.3 , 0.7 ] ,[ 0.5 , 0.5 ] ]Pi = [ 0.5 , 0.5 ]def label_2_id(target):dt = { v:k for k,v in enumerate(V)}return [ dt[item] for item in target ]# target = label_2_id( ['红','红','黑','红'] )target = label_2_id( ['红','红'] )BruteForce暴力算法，计算复杂度：# 路径展示角度def brute_force_algorithm( target = [] ,path = '' ,prob ='' , pre = -1):ret = []path_tmp = ''prob_tmp = ''for k,v in enumerate(Q):path_tmp ='{}/{}'.format(path , v)if prob == '':prob_tmp = '{}/{},{}'.format(prob , Pi[k] , B[k][target[0]] )else:prob_tmp = '{}/{},{}'.format( prob , A[pre][k] , B[k][target[0]] )if len(target) > 1:tmp = brute_force_algorithm(target[1:] , path_tmp ,prob_tmp , pre = k )ret.extend( tmp )elif len(target) == 1:ret.append([path_tmp , prob_tmp])return ret# 总概率展示角度def brute_force_algorithm( target = [] ,path = '' ,prob = 0 , pre = -1):ret = 0for k,v in enumerate(Q):prob_tmp = probpath_tmp ='{}/{}'.format(path , v)if pre == -1 :prob_tmp += Pi[k] * B[k][target[0]]# joint 联合概率局部else:prob_tmp *= A[pre][k] * B[k][target[0]]if len(target) > 1:ret += brute_force_algorithm(target[1:] , path_tmp ,prob_tmp , pre = k )elif len(target) == 1:ret += prob_tmpreturn retForward 前向算法，时间复杂度：def forward_algorithm( target = [] ):prob = [ [ 0 for i in Q] for j in target ]for t ,o in enumerate(target):if t == 0 :for i in range( len(Q) ):prob[0][i] = Pi[i] * B[i][o]else:for id , q in enumerate(Q):for k,v in enumerate(prob[t-1]):print(v ,A[k][id] , prob , prob[t][id] )prob[t][id] += (v * A[k][id] * B[id][o])print(prob)return probBackend后向算法，计算复杂度：def backend_algorithm( target = [] ):prob = [ [ 0.0 for i in Q] for j in target ]length = len(target)for t in range( length-1 , -1 , -1):if t == length-1 :for i in range( len(Q) ):# 后向计算有点问题prob[t][i] = 1else:o = target[t+1]for id , q in enumerate(Q):if t == 0:for k,v in enumerate(prob[t+1]):prob[t][id] *= 1000prob[t][id] += ( v * A[id][k] * B[k][o] ) * 1000prob[t][id] /= 1000else:for k,v in enumerate(prob[t+1]):prob[t][id] += v * A[id][k] * B[k][o]for k,v in enumerate(prob[0]):prob[0][k] = v * Pi[k] * B[k][target[0]]return prob【HMM算法python实现】

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