Viterbi算法的Python实现
我正在做一个Python项目,其中我想使用Viterbi算法。有人知道Viterbi算法的完整Python实现吗?维基百科上的这篇文章的正确性似乎在谈话页面上受到质疑。有人有指针吗?嗯,我可以发我的。这并不漂亮,但请让我知道,如果你需要澄清。我最近写这篇文章是专门为词性标注写的Viterbi算法的Python实现,python,viterbi,Python,Viterbi,我正在做一个Python项目,其中我想使用Viterbi算法。有人知道Viterbi算法的完整Python实现吗?维基百科上的这篇文章的正确性似乎在谈话页面上受到质疑。有人有指针吗?嗯,我可以发我的。这并不漂亮,但请让我知道,如果你需要澄清。我最近写这篇文章是专门为词性标注写的 class Trellis: trell = [] def __init__(self, hmm, words): self.trell = [] temp = {}
class Trellis:
trell = []
def __init__(self, hmm, words):
self.trell = []
temp = {}
for label in hmm.labels:
temp[label] = [0,None]
for word in words:
self.trell.append([word,copy.deepcopy(temp)])
self.fill_in(hmm)
def fill_in(self,hmm):
for i in range(len(self.trell)):
for token in self.trell[i][1]:
word = self.trell[i][0]
if i == 0:
self.trell[i][1][token][0] = hmm.e(token,word)
else:
max = None
guess = None
c = None
for k in self.trell[i-1][1]:
c = self.trell[i-1][1][k][0] + hmm.t(k,token)
if max == None or c > max:
max = c
guess = k
max += hmm.e(token,word)
self.trell[i][1][token][0] = max
self.trell[i][1][token][1] = guess
def return_max(self):
tokens = []
token = None
for i in range(len(self.trell)-1,-1,-1):
if token == None:
max = None
guess = None
for k in self.trell[i][1]:
if max == None or self.trell[i][1][k][0] > max:
max = self.trell[i][1][k][0]
token = self.trell[i][1][k][1]
guess = k
tokens.append(guess)
else:
tokens.append(token)
token = self.trell[i][1][token][1]
tokens.reverse()
return tokens
我在的示例存储库中发现了以下内容。你在找这样的东西吗
def viterbi_segment(text, P):
"""Find the best segmentation of the string of characters, given the
UnigramTextModel P."""
# best[i] = best probability for text[0:i]
# words[i] = best word ending at position i
n = len(text)
words = [''] + list(text)
best = [1.0] + [0.0] * n
## Fill in the vectors best, words via dynamic programming
for i in range(n+1):
for j in range(0, i):
w = text[j:i]
if P[w] * best[i - len(w)] >= best[i]:
best[i] = P[w] * best[i - len(w)]
words[i] = w
## Now recover the sequence of best words
sequence = []; i = len(words)-1
while i > 0:
sequence[0:0] = [words[i]]
i = i - len(words[i])
## Return sequence of best words and overall probability
return sequence, best[-1]
我刚刚纠正了的伪实现。从最初的(不正确的)版本开始,我花了一段时间才弄清楚哪里出了问题,但我最终还是成功了,部分原因是Kevin Murphy在中实现了
viterbi_path.m
在具有变量的HMM对象的上下文中,如图所示:
hmm = HMM()
hmm.priors = np.array([0.5, 0.5]) # pi = prior probs
hmm.transition = np.array([[0.75, 0.25], # A = transition probs. / 2 states
[0.32, 0.68]])
hmm.emission = np.array([[0.8, 0.1, 0.1], # B = emission (observation) probs. / 3 obs modes
[0.1, 0.2, 0.7]])
运行Viterbi(最佳路径)算法的Python函数如下:
def viterbi (self,observations):
"""Return the best path, given an HMM model and a sequence of observations"""
# A - initialise stuff
nSamples = len(observations[0])
nStates = self.transition.shape[0] # number of states
c = np.zeros(nSamples) #scale factors (necessary to prevent underflow)
viterbi = np.zeros((nStates,nSamples)) # initialise viterbi table
psi = np.zeros((nStates,nSamples)) # initialise the best path table
best_path = np.zeros(nSamples); # this will be your output
# B- appoint initial values for viterbi and best path (bp) tables - Eq (32a-32b)
viterbi[:,0] = self.priors.T * self.emission[:,observations(0)]
c[0] = 1.0/np.sum(viterbi[:,0])
viterbi[:,0] = c[0] * viterbi[:,0] # apply the scaling factor
psi[0] = 0;
# C- Do the iterations for viterbi and psi for time>0 until T
for t in range(1,nSamples): # loop through time
for s in range (0,nStates): # loop through the states @(t-1)
trans_p = viterbi[:,t-1] * self.transition[:,s]
psi[s,t], viterbi[s,t] = max(enumerate(trans_p), key=operator.itemgetter(1))
viterbi[s,t] = viterbi[s,t]*self.emission[s,observations(t)]
c[t] = 1.0/np.sum(viterbi[:,t]) # scaling factor
viterbi[:,t] = c[t] * viterbi[:,t]
# D - Back-tracking
best_path[nSamples-1] = viterbi[:,nSamples-1].argmax() # last state
for t in range(nSamples-1,0,-1): # states of (last-1)th to 0th time step
best_path[t-1] = psi[best_path[t],t]
return best_path
这是一个老问题,但其他答案都不是我所需要的,因为我的应用程序没有特定的观察状态 在@Rhubarb之后,我还重新实现了Kevin Murphy的(请参见
viterbi_path.m
),但我将它保持在原来的位置。我还包括了一个简单的测试用例
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)
请注意,此实现不直接使用发射概率,而是使用变量obslik
。通常,emissions[i,j]:=Pr(观测状态==j |隐藏状态==i)
对于特定观测状态i
,使排放。形状==(num\u hidden\u states,num\u obs\u states)
然而,给定一个序列
观测值[t]:=在时间t
的观测值,维特比算法所需要的只是每个隐藏状态的观测值的可能性。因此,obslik[i,t]:=Pr(观测值[t]|隐藏状态==i)
。观察到的状态的实际值是不必要的。我已经修改了@Rhubarb对于已知边际概率的条件的答案(例如,通过计算向前-向后算法)
这是我的。它是直接从原文中改写的。它使用
numpy
来转换它们的ndarray
,但在其他方面它是一个纯python3实现
import numpy as np
def viterbi(y, A, B, Pi=None):
"""
Return the MAP estimate of state trajectory of Hidden Markov Model.
Parameters
----------
y : array (T,)
Observation state sequence. int dtype.
A : array (K, K)
State transition matrix. See HiddenMarkovModel.state_transition for
details.
B : array (K, M)
Emission matrix. See HiddenMarkovModel.emission for details.
Pi: optional, (K,)
Initial state probabilities: Pi[i] is the probability x[0] == i. If
None, uniform initial distribution is assumed (Pi[:] == 1/K).
Returns
-------
x : array (T,)
Maximum a posteriori probability estimate of hidden state trajectory,
conditioned on observation sequence y under the model parameters A, B,
Pi.
T1: array (K, T)
the probability of the most likely path so far
T2: array (K, T)
the x_j-1 of the most likely path so far
"""
# Cardinality of the state space
K = A.shape[0]
# Initialize the priors with default (uniform dist) if not given by caller
Pi = Pi if Pi is not None else np.full(K, 1 / K)
T = len(y)
T1 = np.empty((K, T), 'd')
T2 = np.empty((K, T), 'B')
# Initilaize the tracking tables from first observation
T1[:, 0] = Pi * B[:, y[0]]
T2[:, 0] = 0
# Iterate throught the observations updating the tracking tables
for i in range(1, T):
T1[:, i] = np.max(T1[:, i - 1] * A.T * B[np.newaxis, :, y[i]].T, 1)
T2[:, i] = np.argmax(T1[:, i - 1] * A.T, 1)
# Build the output, optimal model trajectory
x = np.empty(T, 'B')
x[-1] = np.argmax(T1[:, T - 1])
for i in reversed(range(1, T)):
x[i - 1] = T2[x[i], i]
return x, T1, T2
我有点困惑为什么这比NLTK帖子高,它们的实现是否不正确?OP您觉得我的完全未记录的代码令人满意吗?可能是因为它比NLTK代码更容易适应您的需要。@placeybordeaux此函数
hmm.t(k,token)
做什么?我试图复制代码,但我不知道hmm.t(k,token)
做什么。你能提供一个例子吗?@Mohammed-hmm可以追溯到这里,但我非常确定hmm.t(k,token)
是从状态k转换到token的概率,hmm.e(token,word)
是发出给定token的单词的概率。查看NLTK代码也可能会有所帮助。老实说,我的帖子不太好看,也不太可读。在回答中错误地发表评论,观察结果(0)是错误的,对吗?应该是观测值[0]和观测值[t]?我不明白在执行psi[最佳路径[t],t]
时,你怎么不会出错,因为最佳路径是float类型,你只能用int索引?@MikeVella我补充道:bp=np.zeros(nSamples)。astype(int)
在包含4个可能类别的12个值的np.array上运行此算法时,我会出错。您会遇到什么错误?你是如何调用这个函数的?
import numpy as np
def viterbi(y, A, B, Pi=None):
"""
Return the MAP estimate of state trajectory of Hidden Markov Model.
Parameters
----------
y : array (T,)
Observation state sequence. int dtype.
A : array (K, K)
State transition matrix. See HiddenMarkovModel.state_transition for
details.
B : array (K, M)
Emission matrix. See HiddenMarkovModel.emission for details.
Pi: optional, (K,)
Initial state probabilities: Pi[i] is the probability x[0] == i. If
None, uniform initial distribution is assumed (Pi[:] == 1/K).
Returns
-------
x : array (T,)
Maximum a posteriori probability estimate of hidden state trajectory,
conditioned on observation sequence y under the model parameters A, B,
Pi.
T1: array (K, T)
the probability of the most likely path so far
T2: array (K, T)
the x_j-1 of the most likely path so far
"""
# Cardinality of the state space
K = A.shape[0]
# Initialize the priors with default (uniform dist) if not given by caller
Pi = Pi if Pi is not None else np.full(K, 1 / K)
T = len(y)
T1 = np.empty((K, T), 'd')
T2 = np.empty((K, T), 'B')
# Initilaize the tracking tables from first observation
T1[:, 0] = Pi * B[:, y[0]]
T2[:, 0] = 0
# Iterate throught the observations updating the tracking tables
for i in range(1, T):
T1[:, i] = np.max(T1[:, i - 1] * A.T * B[np.newaxis, :, y[i]].T, 1)
T2[:, i] = np.argmax(T1[:, i - 1] * A.T, 1)
# Build the output, optimal model trajectory
x = np.empty(T, 'B')
x[-1] = np.argmax(T1[:, T - 1])
for i in reversed(range(1, T)):
x[i - 1] = T2[x[i], i]
return x, T1, T2