转换矩阵中的多个 ngram，概率不会增加到 1

问题描述

我正在尝试找到一种方法，使用 python 和 numpy 为给定文本使用 unigrams、bigrams 和 trigrams 制作转换矩阵。每行的概率应等于 1。我首先使用 bigrams 进行了此操作，并且效果很好：

distinct_words = list(word_dict.keys())
dwc = len(distinct_words)

matrix = np.zeros((dwc,dwc),dtype=np.float)
for i in range(len(distinct_words)):
    word = distinct_words[i]
    first_word_idx = i
    total = 0
    for bigram,count in ngrams.items():
        word_1,word_2 = bigram.split(" ")
        if word_1 == word:
            total += count
    for bigram,word_2 = bigram.split(" ")
        if word_1 == word:
            second_word_idx = index_dict[word_2]
            matrix[first_word_idx,second_word_idx] = count / total

但现在我想添加 unigrams 和 trigrams 并加权它们的概率（trigrams * .6，bigrams * .2，unigrams *.2）。我不认为我的 python 很简洁，这是一个问题，但我也不知道如何使用多个 n-gram（和权重，虽然说实话权重是次要的），这样我仍然可以获得所有的概率从任何给定的行加起来为一。

distinct_words = list(word_dict.keys())
dwc = len(distinct_words)

matrix = np.zeros((dwc,dtype=np.float)
for i in range(len(distinct_words)):
  word = distinct_words[i]
  first_word_index = i 
  bi_total = 0
  tri_total=0
  tri_prob = 0
  bi_prob = 0
  uni_prob = word_dict[word] / len(distinct_words)
  if i < len(distinct_words)-1:
    for trigram,count in trigrams.items():
      word_1,word_2,word_3 = trigram.split()
      if word_1 + word_2 == word + distinct_words[i+1]:
        tri_total += count
    for trigram,word_3 = trigram.split()
      if word_1 + word_2 == word + distinct_words[i+1]:
        second_word_index = index_dict[word_2]
        tri_prob = count/bigrams[word_1 + " " + word_2]
  for bigram,count in bigrams.items():
    word_1,word_2 = bigram.split(" ")
    if word_1 == word:
      bi_total += count
  for bigram,word_2 = bigram.split(" ")
    if word_1 == word:
      second_word_index = index_dict[word_2]
      bi_prob = count / bi_total
      matrix[first_word_index,second_word_index] = (tri_prob * .4) + (bi_prob * .2) + (word_dict[word]/len(word_dict) *.2)

我正在阅读 this lecture 以了解如何设置我的概率矩阵，这似乎很有道理，但我不确定我哪里出错了。

如果有帮助，我的 n_grams 来自 this - 它只是生成一个 n_gram 字典作为字符串及其计数。

def get_ngram(words,n):
    word_dict = {}
    for i,word in enumerate(words):
        if i > (n-2):
            n_gram = []
            for num in range(n):
                index = i - num
                n_gram.append(words[index])
            if len(n_gram) > 1:
                formatted_gram = ""
                for word in reversed(n_gram):
                    formatted_gram += word + " "
            else:
                formatted_gram = n_gram[0]
            stripped = formatted_gram.strip() if formatted_gram else n_gram[0]
            if stripped in word_dict:
                word_dict[stripped] += 1
            else:
                word_dict[stripped] = 1
    return word_dict

解决方法

让我们尝试以最有效的方式在纯 Python 中完成，仅依赖列表和字典推导式。

假设我们有一个由 3 个单词“a”、“b”和“c”组成的玩具文本：

np.random.seed(42)
text = " ".join([np.random.choice(list("abc")) for _ in range(100)])
text
'c a c c a a c b c c c c a c b a b b b b a a b b a a a c c c b c b b c 
 b c c a c a c c a a c b a b b b a b a b c c a c c b a b b b b b b b a 
 c b b b b b b c c b c a b a a b c a b a a a a c a a a c a a'

然后要制作 unigrams、bigrams 和 trigrams，您可以按以下步骤操作：

unigrams = text.split()
unigram_counts = dict()
for unigram in unigrams:
    unigram_counts[unigram] = unigram_counts.get(unigram,0) +1

bigrams = ["".join(bigram) for bigram in zip(unigrams[:-1],unigrams[1:])]
bigram_counts = dict()
for bigram in bigrams:
    bigram_counts[bigram] = bigram_counts.get(bigram,0) +1

trigrams = ["".join(trigram) for trigram in zip(unigrams[:-2],unigrams[1:-1],unigrams[2:])]
trigram_counts = dict()
for trigram in trigrams:
    trigram_counts[trigram] = trigram_counts.get(trigram,0) +1

要合并权重并标准化：

weights = [.2,.2,.6]
dics = [unigram_counts,bigram_counts,trigram_counts]
weighted_counts = {k:v*w for d,w in zip(dics,weights) for k,v in d.items()}
#desired output
freqs = {k:v/sum(weighted_counts.values()) for k,v in weighted_counts.items()}

我们有什么：

pprint(freqs)

---

{'a': 0.06693711967545637,'aa': 0.02434077079107505,'aaa': 0.024340770791075043,...

最后，健全性检查：

print(sum(freqs.values()))

---

0.999999999999999

可以进一步自定义此代码以包含您的标记化规则，例如，或通过一次循环遍历不同的 gram 使其更短。