Question

我正在寻找将三元组频率存储在内存中的潜在方法，并通过以下方式动态计算unigram和bigram频率：

给出一个三元组，v，w：

count（v，w）= sum（。，v，w），即所有u

的总和

类似地，count（w）= sum（。，w）

这肯定会导致一些失踪的unigrams，例如句子开始标记，但这听起来像生成unigrams和bigrams的有效方法吗？

Answer 1

是。那可行。您可以通过使自己成为一个小型语料库来检查它，并手动进行计数以确保它是相同的。

from collections import Counter

corpus = [['the','dog','walks'], ['the','dog','runs'], ['the','cat','runs']]
corpus_with_ends = [['<s>','<s>'] + s + ['<e>'] for s in corpus]

trigram_counts = Counter(trigram for s in corpus_with_ends for trigram in zip(s,s[1:],s[2:]))

unique_bigrams = set((b,c) for a,b,c in trigram_counts)
bigram_counts = dict((bigram,sum(count for trigram,count in trigram_counts.iteritems() if trigram[1:] == bigram)) for bigram in unique_bigrams)

unique_unigrams = set((c,) for a,b,c in trigram_counts if c != '<e>')
unigram_counts = dict((unigram,sum(count for trigram,count in trigram_counts.iteritems() if trigram[2:] == unigram)) for unigram in unique_unigrams)

现在你可以检查一下：

>>> true_bigrams = [bigram for s in corpus_with_ends for bigram in zip(s[1:],s[2:])]
>>> true_bigram_counts = Counter(true_bigrams)
>>> bigram_counts == true_bigram_counts
True

>>> true_unigrams = [(unigram,) for s in corpus_with_ends for unigram in s[2:-1]]
>>> true_unigram_counts = Counter(true_unigrams)
>>> unigram_counts == true_unigram_counts
True

从三元组列表中生成unigrams和bigrams

1 个答案: