基本概念：朴素贝叶斯算法进行分类
我认为我或多或少地理解朴素贝叶斯，但是我对于其简单的二进制文本分类tast的实现有几个问题。
假设文档D_i是词汇表x_1, x_2, ...x_n一些子集
有两个类c_i任何文档都可以落在上面，我想为一些与P(D|c_i)P(c_i)成比例的输入文档D计算P(c_i|D) P(D|c_i)P(c_i)
我有三个问题
P(c_i)是#docs in c_i/ #total docs #words in c_i/ #total words或#words in c_i/ #total words P(x_j|c_i)是#times x_j appears in D/ #times x_j appears in c_i x_j ，我是否给它一个1的概率以便它不改变计算？ 例如，让我们说我有一套训练集：
training = [("hello world", "good")
("bye world", "bad")]
所以课堂会有
good_class = {"hello": 1, "world": 1}
bad_class = {"bye":1, "world:1"}
all = {"hello": 1, "world": 2, "bye":1}
所以现在如果我想计算一个测试字符串好的概率
test1 = ["hello", "again"]
p_good = sum(good_class.values())/sum(all.values())
p_hello_good = good_class["hello"]/all["hello"]
p_again_good = 1 # because "again" doesn't exist in our training set
p_test1_good = p_good * p_hello_good * p_again_good
由于这个问题太广泛，所以我只能以一种限制的方式回答：
第1位： - P（c_i）在c_i / #total文档中为#docs或在c_i /＃总词中为#words
P(c_i) = #c_i/#total docs
第二： -如果P（x_j | c_i）是＃x_j出现在D / #times x_j出现在c_i中。
@larsmans注意到后..
It is exactly occurrence of word in a document
by total number of words in that class in whole dataset.
第三：假设训练集中不存在x_j，我是否给它一个1的概率，以便它不会改变计算？
For That we have laplace correction or Additive smoothing. It is applied on
p(x_j|c_i)=(#times x_j appears in D+1)/ (#times x_j +|V|) which will neutralize
the effect not occurring features.
上一篇: Basic concepts: Naive Bayes algorithm for classification