Question

我一直试图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。我试过的论文是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。问题是算法背后的想法并不那么容易理解，并且在论文中提供的伪代码中存在很多拼写错误。此外，第二篇论文中的this implementation令人难以置信，并且this one，从上一篇文章中提到，并不包含语言模型。

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]

任何帮助都将非常感激。

Answer 1

我使用-inf初始化实现了波束搜索，并且还遵循了论文http://proceedings.mlr.press/v32/graves14.pdf中的ctc_beam_search算法......除了更新p_b中的字符外，它几乎与此类似。算法运行正常...如果初始化存在，这个算法也可以工作..

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)

Answer 2

问题是s_new中不在A_prev中的块指的是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长，即s_new，t-1。你可以设置一个try catch块，如果p [（s_new，t-1）]不存在，它将使用-infinity。

用于解码RNN输出的波束搜索算法

2 个答案: