给定“the”，“开始”的概率是多少？

【问题标题】：What is the Probability of ‘begining’ given ‘the’?给定“the”，“开始”的概率是多少？
【发布时间】：2014-05-07 23:19:37
【问题描述】：

Using an NLTK Conditional Frequency Distribution and the nltk.bigrams function, train a bigram model on the Genesis:

text = nltk.corpus.genesis.words('english-kjv.txt')
bigrams = nltk.bigrams(text)
cfd = nltk.ConditionalFreqDist(bigrams)
Answer the following questions

What is the Probability of ‘begining’ given ‘the’?
What is the probability of ‘the’?

注意：您作为答案给出的概率必须是可从该语料库计算得出的概率。

您好，有什么可以帮帮我的吗？这是在 nltk 书中。当我得到它时，我得到了 78%，这没有意义。我试图在 Python 中计算它。

【问题讨论】：

零，“开始”不是这样拼写的 :)
我的天才！ ..那么呢？我还是 78

标签： python nltk corpus tagged-corpus

【解决方案1】：

probability of 'beginning' intersect 'the' 之间有一些区别

p('beginning','the')

和probability of 'beginning' given 'the':

p('beginning'|'the') = p('beginning','the') / p('the')

尝试：

from collections import Counter

import nltk

text = nltk.corpus.genesis.words('english-kjv.txt')
bigrams = nltk.bigrams(text)
cfd_bigrams = Counter(bigrams)
cfd_unigrams = Counter(list(text))

print "p('said','unto') =", cfd_bigrams[u'said', u'unto'] / float(sum(cfd_bigrams.values()))

print "p('said'|'unto') =", (cfd_bigrams[u'said', u'unto'] / float(sum(cfd_bigrams.values()))) / cfd_unigrams[u'unto']

print "p('beginning','the') =", cfd_bigrams[u'beginning', u'the']

[出]：

p('said','unto') = 0.00397649844738
p('said'|'unto') = 6.73982787691e-06
p('beginning','the') = 0

【讨论】：