基于朴素贝叶斯模型的文本分类

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一、朴素贝叶斯模型

模型一：

将一个文本文档使用一个词的向量来表示。

通常文档中出现的词的个数是有限的，假设要将文档分成两类（类别0、1），分类的所有文档可能出现100个词（词典中词的个数，在实际应用中，选择训练文档中出现次数最多的n个词，n从10000到50000），那么一个特定的文档就可以用一个100维的向量来表示。每一维上要么是0，要么是1。

要计算一个文档的类别，首先要计算每个词与文档类别的关系。

运用朴素贝叶斯模型的来进行计算时，还有一个关键的假设：各个词与类别是条件独立的。写成公式如下：

=p(w₁|y)p(w₂|y)p(w₃|y)......p(w₁₀₀|y)

那么，Φ_j|y=1=p(w_j=1|y=1)，词j与类别1的关系

Φ_j|y=0=p(w_j=1|y=0)，词j与类别0的关系

建立以Φ_j|y=1，Φ_j|y=0，Φ_y=1=p(y=1)为参数的似然函数，通过极大似然法可以求出：

Φ_j|y=1=∑_i=1...m[x_jⁱ=1 and yⁱ=1] / ∑_i=1...m[yⁱ=1] ，用训练样本中类别为1且出现了词j的文档数除以所有类别为1的文档数

Φ_j|y=0=∑_i=1...m[x_jⁱ=1 and yⁱ=0] / ∑_i=1...m[yⁱ=0] ，用训练样本中类别为0且出现了词j的文档数除以所有类别为0的文档数

Φ_y=1=训练样本中类别为1的文数 / 训练样本中所有文档数

当来了一个新样本x^,，x^,=[w₁^, w₂^, w₃^, ...... w₁₀₀^, ]

计算p(y=1|x^,) = [p(x^,|y=1)p(y=1)] / p(x^,) # 将p(x)忽略掉

= [Φ_j=1|y=1w₁^, Φ_j=2|y=1w₂^, Φ_j=3|y=1w₃^,...... Φ_j=100|y=1w₁₀₀^,] #将该向量中不为0的元素求乘积

* Φ_y=1

* (1 - Φ_y=1)

比较两个值的大小，就能确定新样本的类别。

模型二：

在以上的模型中，只关注一个词是否在文档中出现，出现了就将词向量中对应的元素置为1，否则就是0。

但通常某个词可能在一个文档的不同位置多次出现，如何能更好的描述这一客观事实？

假设词典中词的总数还是100个，词的序号k={1,2,3, ... , 100},共有m个文档，第个i文档包括有n_i个词，文档xⁱ的第j个词为x_jⁱ，x_jⁱ可以是总共100个词中的任意一个。

词典中第K个词与类别1的关系：Φ_k|y=1=∑_i=1...m∑_j=1...ni[x_jⁱ=k and yⁱ=1] / ∑_i=1...m[(yⁱ=1)*n_i],

其中[x_jⁱ=k and yⁱ=1]表示第i个文档时类型1，且在该文档中的第j个词是词K。

词典中第K个词与类别0的关系：Φ_k|y=0=∑_i=1...m∑_j=1...ni[x_jⁱ=k and yⁱ=0] / ∑_i=1...m[(yⁱ=0)*n_i]，

其中[x_jⁱ=k and yⁱ=0]表示第i个文档时类型0，且在该文档中的第j个词是词K。

Φ_y=1同上一模型中一样。

通过计算得到类别1的词典向量：[Φ_1|y=1 Φ_2|y=1 Φ_3|y=1 ...... Φ_100|y=1 ]，类别0的词典向量：[Φ_1|y=0 Φ_2|y=0 Φ_3|y=0 ...... Φ_100|y=0 ]

当来了一个新样本x^,，有n个词，x^,=[w₁^, w₂^, w₃^, ...... w_n^, ]。

计算p(y=1|x^,) ={遍历x^,把其中每个词对应的Φ_k|y=1求乘机} * Φ_y=1，p(y=0|x^,) ={遍历x^,把其中每个词对应的Φ_k|y=0求乘机} * Φ_y=1 。

比较两者中，较大者，就是新样本对应的类别。

还有一点：

如果词典中的一个词在所有训练样本都没有出现，但在新文档中出现了，通过训练，计算的该词的Φ_k|y=1=0，Φ_k|y=0=0 （模型一中也类似）。

那么计算新文档的p(y=1|x^,)， p(y=0|x^,) 就都会是0.

如何克服这个问题？可以将计算Φ_k|y=1和Φ_k|y=0的公式做一点修改：

Φ_k|y=1={∑_i=1...m∑_j=1...ni[x_jⁱ=k and yⁱ=1] + 1} / {∑_i=1...m[(yⁱ=1)*n_i] + 100 }

Φ_k|y=0={∑_i=1...m∑_j=1...ni[x_jⁱ=k and yⁱ=0] + 1} / {∑_i=1...m[(yⁱ=0)*n_i] + 100}

二、实现

使用R编程语言，基于模型二的实现代码如下：

## = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
## initialize
## = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
library(Matrix)
numTrainDocs <- 700   # number of documents, 400, 100, 50
numTokens <- 2500      # number of words totally

trDoc <- read.table(file="train-features.txt")
trLabel <- read.table(file="train-labels.txt")
x <- sparseMatrix(i=trDoc$V1, j=trDoc$V2, x=trDoc$V3, dims=c(numTrainDocs,numTokens))
x <- as.matrix(x)
y <- matrix(data=trLabel$V1,ncol=1)


## = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
## training
## = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =

## 1. calculate the phi of y, i.e. p(y=1)
phiOfy1 <- colSums(y) / dim(y)[1]
phiOfy0 <- 1 - phiOfy1

## 2. calculate the phi of k to given y=1 or 0, i.e. p(Xj=k|y=1)  p(Xj=k|y=0)
## # # #
# The number of parameters is numTokens, which is same to the vocabulary length.
# i.e. Each word in vocabulary is a feature. Each feature need a parameter.
phiArray1 <- numeric(numTokens)  # y=1, parameter vector
phiArray0 <- numeric(numTokens)  # y=0, parameter vector
sum1 <- sum(x[y==1,])
sum0 <- sum(x[y==0,])
for (i in 1:numTokens) {  # calculte each parameter, one by one
    phiArray1[i] <- ( sum(x[y==1,i]) + 1 )  /  ( sum1 + numTokens )
    phiArray0[i] <- ( sum(x[y==0,i]) + 1 )  /  ( sum0 + numTokens )
}


## = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
## test
## = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
xTestDoc <- read.table(file="test-features.txt")
yTest <- read.table(file="test-labels.txt")

xTest <- sparseMatrix(i=xTestDoc$V1, j=xTestDoc$V2, x=xTestDoc$V3)
xTest <- as.matrix(xTest)

numTestDoc <- dim(xTest)[1]
numTestTokens <- dim(xTest)[2]

testDocProbility1 <- xTest %*% as.matrix(log(phiArray1)) + log(phiOfy1)
testDocProbility0 <- xTest %*% as.matrix(log(phiArray0)) + log(phiOfy0)

testLabel <- (testDocProbility1 > testDocProbility0)
testLabel[testLabel == TRUE] <- 1

classErrorNum <- sum(abs(yTest - as.vector(testLabel)))
classErrorRatio <- classErrorNum / numTestDoc

View Code

三、测试结果

实验的数据中，词共有2500个（词典中词个数），测试四次，训练文档个数分别是700,400,100,50，测试集有260个文档.

训练文档数	错分类文档数	错分率
700	5	1.92%
400	6	2.30%
100	6	2.30%
50	7	2.69%

随着训练文档数增加，错分率下降。总体来说，朴素贝叶斯进行文档分类的错分率还是相对较低的。