关于学习朴素贝叶斯心得

朴素贝叶斯

1. 假设：基于属性相互独立。
2. 原理：基于贝叶斯定理，p（c|x）=p（x|c）*p（c）/p（x），根据贝叶斯定理，后验概率P(Y=c | X=x) = 条件概率P(X=x | Y=c) * 先验概率P(Y = c) / P(X=x)，取P(X=x | Y=c) * P(Y = c)最大的分类作为输出。
3. c表示类别（c1,c2）,x表示属性（x1，x2，x3，....）
4. P(A∣B) 表示在确定B的情况下，事件A发生的概率，而在实际情况中，我们或许更关心P(B∣A)但是只能直接获得P(A∣B) ，此时我们需要一个工具可以把P(A∣B) 和P(B∣A)相互转化， 贝叶斯定理就是这样一个公式，下面给出贝叶斯定理：
   
   P(B|A)=P(A|B)P(B)/P(A)
   对苹果分类的问题，有三个特征F = {f1, f2, f3}，两种分类C = {c1, c2}，根据贝叶斯公式有给定特征条件下，特征为ci的概率
   
   P(ci|f1f2f3)=P(f1f2f3|ci)P(ci)/P(f1f2f3)
   使得上式取得最大值的ci即为分类结果，由于对给定训练集来说，P(f1f2f3)为常数，那么就转为为求
   P(f1f2f3∣ci)P(ci)
   的最大值。
   朴素贝叶斯的假设在这里就体现了，由于特征值相互独立，那么上式可以转化为
   P(f1∣ci)P(f2∣ci)P(f3∣ci)P(ci)
   整个问题就变为求使得上式取最大值的ci，而上式中的每一项都可以从训练集中得到。
   最后讨论下Laplace校准，如果某一个特性值在训练集中出现的次数为0，那么以上我们讨论的公式就没有意义了，以为对所有的类型结果都是0。当然对训练集进行选择可以避免这种情况，但是如果避免不了就需要进行Laplace校准。其实很简单，把所有出现特征出现的次数都加上1，即为Laplace校准。

 #构造训练集data <- matrix(c("sunny","hot","high","weak","no",                 "sunny","hot","high","strong","no",                 "overcast","hot","high","weak","yes",                 "rain","mild","high","weak","yes",                 "rain","cool","normal","weak","yes",                 "rain","cool","normal","strong","no",                 "overcast","cool","normal","strong","yes",                 "sunny","mild","high","weak","no",                 "sunny","cool","normal","weak","yes",                 "rain","mild","normal","weak","yes",                 "sunny","mild","normal","strong","yes",                 "overcast","mild","high","strong","yes",                 "overcast","hot","normal","weak","yes",                 "rain","mild","high","strong","no"), byrow = TRUE,               dimnames = list(day = c(),                                  

#byrow项控制排列元素时是否按行进行，dimnames给定行和列的名称.               condition = c("outlook","temperature",                 "humidity","wind","playtennis")), nrow=14, ncol=5);#计算先验概率show(data)prior.yes = sum(data[,5] == "yes") / length(data[,5]);prior.no  = sum(data[,5] == "no")  / length(data[,5]);#模型naive.bayes.prediction <- function(condition.vec) {# Calculate unnormlized posterior probability for playtennis = yes.playtennis.yes <-sum((data[,1] == condition.vec[1]) & (data[,5] == "yes")) / sum(data[,5] == "yes") * # P(outlook = f_1 | playtennis = yes)sum((data[,2] == condition.vec[2]) & (data[,5] == "yes")) / sum(data[,5] == "yes") * # P(temperature = f_2 | playtennis = yes)sum((data[,3] == condition.vec[3]) & (data[,5] == "yes")) / sum(data[,5] == "yes") * # P(humidity = f_3 | playtennis = yes)sum((data[,4] == condition.vec[4]) & (data[,5] == "yes")) / sum(data[,5] == "yes") * # P(wind = f_4 | playtennis = yes)      prior.yes; # P(playtennis = yes)# Calculate unnormlized posterior probability for playtennis = no.playtennis.no <-sum((data[,1] == condition.vec[1]) & (data[,5] == "no"))  / sum(data[,5] == "no")  * # P(outlook = f_1 | playtennis = no)sum((data[,2] == condition.vec[2]) & (data[,5] == "no"))  / sum(data[,5] == "no")  * # P(temperature = f_2 | playtennis = no)sum((data[,3] == condition.vec[3]) & (data[,5] == "no"))  / sum(data[,5] == "no")  * # P(humidity = f_3 | playtennis = no)sum((data[,4] == condition.vec[4]) & (data[,5] == "no"))  / sum(data[,5] == "no")  * # P(wind = f_4 | playtennis = no)prior.no; # P(playtennis = no)return(list(post.pr.yes = playtennis.yes,post.pr.no  = playtennis.no,prediction  = ifelse(playtennis.yes >= playtennis.no, "yes", "no")));}#预测naive.bayes.prediction(c("rain",     "hot",  "high",   "strong"));naive.bayes.prediction(c("sunny",    "mild", "normal", "weak"));naive.bayes.prediction(c("overcast", "mild", "normal", "weak"));

naive.bayes.prediction(c("rain",     "hot",  "high",   "strong"));
## $post.pr.yes
## [1] 0.005291005
## 
## $post.pr.no
## [1] 0.02742857
## 
## $prediction
## [1] "no"
naive.bayes.prediction(c("sunny",    "mild", "normal", "weak"));
## $post.pr.yes
## [1] 0.02821869
## 
## $post.pr.no
## [1] 0.006857143
## 
## $prediction
## [1] "yes"
naive.bayes.prediction(c("overcast", "mild", "normal", "weak"));
## $post.pr.yes
## [1] 0.05643739
## 
## $post.pr.no
## [1] 0
## 
## $prediction
## [1] "yes"