Question

Viterbi algorithm在“隐马尔可夫模型”中找到最可能的隐藏状态序列。我目前正在使用hhquark的以下出色代码。

import numpy as np


def viterbi_path(prior, transmat, obslik, scaled=True, ret_loglik=False):
    '''Finds the most-probable (Viterbi) path through the HMM state trellis
    Notation:
        Z[t] := Observation at time t
        Q[t] := Hidden state at time t
    Inputs:
        prior: np.array(num_hid)
            prior[i] := Pr(Q[0] == i)
        transmat: np.ndarray((num_hid,num_hid))
            transmat[i,j] := Pr(Q[t+1] == j | Q[t] == i)
        obslik: np.ndarray((num_hid,num_obs))
            obslik[i,t] := Pr(Z[t] | Q[t] == i)
        scaled: bool
            whether or not to normalize the probability trellis along the way
            doing so prevents underflow by repeated multiplications of probabilities
        ret_loglik: bool
            whether or not to return the log-likelihood of the best path
    Outputs:
        path: np.array(num_obs)
            path[t] := Q[t]
    '''
    num_hid = obslik.shape[0] # number of hidden states
    num_obs = obslik.shape[1] # number of observations (not observation *states*)

    # trellis_prob[i,t] := Pr((best sequence of length t-1 goes to state i), Z[1:(t+1)])
    trellis_prob = np.zeros((num_hid,num_obs))
    # trellis_state[i,t] := best predecessor state given that we ended up in state i at t
    trellis_state = np.zeros((num_hid,num_obs), dtype=int) # int because its elements will be used as indicies
    path = np.zeros(num_obs, dtype=int) # int because its elements will be used as indicies

    trellis_prob[:,0] = prior * obslik[:,0] # element-wise mult
    if scaled:
        scale = np.ones(num_obs) # only instantiated if necessary to save memory
        scale[0] = 1.0 / np.sum(trellis_prob[:,0])
        trellis_prob[:,0] *= scale[0]

    trellis_state[:,0] = 0 # arbitrary value since t == 0 has no predecessor
    for t in xrange(1, num_obs):
        for j in xrange(num_hid):
            trans_probs = trellis_prob[:,t-1] * transmat[:,j] # element-wise mult
            trellis_state[j,t] = trans_probs.argmax()
            trellis_prob[j,t] = trans_probs[trellis_state[j,t]] # max of trans_probs
            trellis_prob[j,t] *= obslik[j,t]
        if scaled:
            scale[t] = 1.0 / np.sum(trellis_prob[:,t])
            trellis_prob[:,t] *= scale[t]

    path[-1] = trellis_prob[:,-1].argmax()
    for t in range(num_obs-2, -1, -1):
        path[t] = trellis_state[(path[t+1]), t+1]

    if not ret_loglik:
        return path
    else:
        if scaled:
            loglik = -np.sum(np.log(scale))
        else:
            p = trellis_prob[path[-1],-1]
            loglik = np.log(p)
        return path, loglik


if __name__=='__main__':
    # Assume there are 3 observation states, 2 hidden states, and 5 observations
    priors = np.array([0.5, 0.5])
    transmat = np.array([
        [0.75, 0.25],
        [0.32, 0.68]])
    emmat = np.array([
        [0.8, 0.1, 0.1],
        [0.1, 0.2, 0.7]])
    observations = np.array([0, 1, 2, 1, 0], dtype=int)
    obslik = np.array([emmat[:,z] for z in observations]).T
    print viterbi_path(priors, transmat, obslik)                                #=> [0 1 1 1 0]
    print viterbi_path(priors, transmat, obslik, scaled=False)                  #=> [0 1 1 1 0]
    print viterbi_path(priors, transmat, obslik, ret_loglik=True)               #=> (array([0, 1, 1, 1, 0]), -7.776472586614755)
    print viterbi_path(priors, transmat, obslik, scaled=False, ret_loglik=True) #=> (array([0, 1, 1, 1, 0]), -8.0120386579275227)

但是，我真正需要的不仅是最有可能出现的序列，还有前k个最有可能出现的隐藏状态序列。

如何修改此代码以给出前k个最可能的序列？

Answer 1

换一种说法，维特比算法计算非循环加权图中其节点为（隐藏状态，时间）对的最短路径。您可以使用Yen算法找到前k个最短路径，这些路径会转换为前k个最可能的序列。这是NetworkX中Yen算法的实现。

要设置图形，我们从源节点和宿节点开始。对于所有状态i，使用权重log（prior [i] * obslik [i，0]）从源节点到节点（i，0）建立弧。对于所有状态i，所有状态j和所有时间t> 0，使用权重log（transmat [i，j] * obslik [j，t]从节点（i，t-1）到（j，t）建立弧）。令T为最后一次，使弧线从（i，T）到接收器的弧度为0。从源到接收器的每条路径都与隐藏状态序列一一对应，且隐藏状态的长度path是该序列的对数似然性。

如何为隐马尔可夫模型找到最可能的隐藏状态序列

1 个答案: