3

我一直在尝试理解波束搜索算法在解码部分的自动语音识别中使用的逻辑。我尝试关注的论文是First-Pass Large Vocabulary Continuous Speech Recognition using Bi-Directional Recurrent DNNs、Lexicon-Free Conversational Speech Recognition with Neural Networks和Towards End-to-End Speech Recognition with Recurrent Neural Networks。问题是该算法背后的想法并不那么容易理解，并且论文中提供的伪代码中有很多错别字。此外，第二篇论文中的这个实现令人难以置信的难以理解，而上一篇论文中提到的这个实现不包括语言模型。

这是我在 Python 中的实现，由于缺少一些概率而失败：

class BeamSearch(object):
"""
Decoder for audio to text.

From: https://arxiv.org/pdf/1408.2873.pdf (hardcoded)
"""
def __init__(self, alphabet='" abcdefghijklmnopqrstuvwxyz'):
    # blank symbol plus alphabet
    self.alphabet = '-' + alphabet
    # index of each char
    self.char_to_index = {c: i for i, c in enumerate(self.alphabet)}

def decode(self, probs, k=100):
    """
    Decoder.

    :param probs: matrix of size Windows X AlphaLength
    :param k: beam size
    :returns: most probable prefix in A_prev
    """
    # List of prefixs, initialized with empty char
    A_prev = ['']
    # Probability of a prefix at windows time t to ending in blank
    p_b = {('', 0): 1.0}
    # Probability of a prefix at windows time t to not ending in blank
    p_nb = {('', 0): 0.0}

    # for each time window t
    for t in range(1, probs.shape[0] + 1):
        A_new = []
        # for each prefix
        for s in Z:
            for c in self.alphabet:
                if c == '-':
                    p_b[(s, t)] = probs[t-1][self.char_to_index[self.blank]] *\
                                    (p_b[(s, t-1)] +\
                                        p_nb[(s, t-1)])
                    A_new.append(s)
                else:
                    s_new = s + c
                    # repeated chars
                    if len(s) > 0 and c == s[-1]:
                        p_nb[(s_new, t)] = probs[t-1][self.char_to_index[c]] *\
                                            p_b[(s, t-1)]
                        p_nb[(s, t)] = probs[t-1][self.char_to_index[c]] *\
                                            p_b[(s, t-1)]
                    # spaces
                    elif c == ' ':
                        p_nb[(s_new, t)] = probs[t-1][self.char_to_index[c]] *\
                                           (p_b[(s, t-1)] +\
                                            p_nb[(s, t-1)])
                    else:
                        p_nb[(s_new, t)] = probs[t-1][self.char_to_index[c]] *\
                                            (p_b[(s, t-1)] +\
                                                p_nb[(s, t-1)])
                        p_nb[(s, t)] = probs[t-1][self.char_to_index[c]] *\
                                            (p_b[(s, t-1)] +\
                                                p_nb[(s, t-1)])
                    if s_new not in A_prev:
                        p_b[(s_new, t)] = probs[t-1][self.char_to_index[self.blank]] *\
                                            (p_b[(s, t-1)] +\
                                                p_nb[(s, t-1)])
                        p_nb[(s_new, t)]  = probs[t-1][self.char_to_index[c]] *\
                                                p_nb[(s, t-1)]
                    A_new.append(s_new)
        A = A_new
        s_probs = map(lambda x: (x, (p_b[(x, t)] + p_nb[(x, t)])*len(x)), A_new)
        xs = sorted(s_probs, key=lambda x: x[1], reverse=True)[:k]
        Z, best_probs = zip(*xs)
    return Z[0], best_probs[0]

任何帮助将不胜感激。

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2 回答 2

0


我使用 -inf 初始化实现了光束搜索，并遵循论文http://proceedings.mlr.press/v32/graves14.pdf中的 ctc_beam_search 算法......除了更新字符的 p_b 之外，它几乎与此相似..算法运行正常...如果存在初始化，即使此算法也可以工作..

A_prev = ['']
p_b[('',0)] = 1
p_nb[('',0)] = 0
for alphabet in alphabets:
    p_b[(alphabet,0)] = -float("inf")
    p_nb[(alphabet,0)] = -float("inf")
for t in range(1,probs.shape[0] +1):
    A_new = []
    for s in A_prev:
        if s!='':
            try:                
                p_nb[(s,t)] = p_nb[(s,t-1)]*probs[t-1][char_map[s[-1:]]]
            except:
                p_nb[(s,t)] = p_nb[(s,t-1)]*probs[t-1][char_map['<SPACE>']]*pW(s)
            if s[:-1] in A_prev:
                p_nb[(s,t)] = p_nb[(s,t)]+pr(probs[t-1],s[-1:],s[:-1],t)
        p_b[(s,t)] = (p_nb[(s,t-1)]+p_b[(s,t-1)])*probs[t-1][0]
        if s=='':
            p_nb[(s,t)] = 0
        if s not in A_new:
            A_new.append(s)
        for c in alphabets:
            s_new = s+c
            p_b[(s_new,t)] = 0
            p_nb[(s_new,t)] = pr(probs[t-1],c,s,t)
            #print s_new,' ',p_nb[(s_new,t)]
            if s_new not in A_new:
                A_new.append(s_new)
    s_probs = map(lambda x: (x,(p_b[(x, t)]+ p_nb[(x, t)])), A_new)
于 2017-05-29T07:05:51.633 回答
-1

问题是 s_new 不在 A_prev 中的块指的是对于生成的新字符串不存在的概率。使用 -float("inf") 初始化新字符串的前一个时间步，即 s_new,t-1。如果 p[(s_new,t-1)] 不存在，您可以放置​​一个 try catch 块，它将使用 -infinity。

于 2017-05-25T10:33:57.493 回答