基本概念：用于分类的朴素贝叶斯算法

Question

我想我或多或少地了解朴素贝叶斯，但对于简单的二进制文本分类测试的实现，我有几个问题。

假设文档D_i 是词汇表x_1, x_2, ...x_n 的某个子集

有两个类 c_i 任何文档都可以落入，我想为某些输入文档 D 计算 P(c_i|D)，它与 P(D|c_i)P(c_i) 成正比

我有三个问题

1. P(c_i) 是 #docs in c_i/ #total docs 或 #words in c_i/ #total words
2. 应该P(x_j|c_i) 是#times x_j appears in D/ #times x_j appears in c_i
3. 假设训练集中不存在 x_j，我是否给它一个概率为 1，这样它就不会改变计算？

例如，假设我有一个训练集：

training = [("hello world", "good")
            ("bye world", "bad")]

所以课程会有

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是好的概率

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

Answer 1

由于这个问题太宽泛，所以我只能有限地回答：-

第一个：- P(c_i) 是 #docs in c_i/ #total docs 或 #words in c_i/ #total words

P(c_i) = #c_i/#total docs

第二个：- P(x_j|c_i) 应该是#times x_j 出现在 D/#times x_j 出现在 c_i。
在 @larsmans 注意到之后..

It is exactly occurrence of word in a document
by total number of words in that class in whole dataset.

3rd:-假设训练集中不存在一个 x_j，我是否给它一个概率为 1，这样它就不会改变计算？

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.