机器学习与R笔记之线性回归

普通最小二乘法OLS
y=ax+b     f(e)=sum((y-y')^2) 
b=cor(x,y)/var(x)  #证明省略 协方差/方差
a=mean(y)-b*mean(x)
person相关系数
R=cor(x,y)
多元线性回归
Y=XB+E
YBE为向量X为带偏执矩阵--最小二次法求系数
B=solve((t(x)%*%x))%*%t(x)%*%y
估计B的函数
reg <- function(y, x) {
  x <- as.matrix(x)
  x <- cbind(Intercept = 1, x) #Intercept = 1命名为Intercept且该列全为1
  b <- solve(t(x) %*% x) %*% t(x) %*% y
  colnames(b) <- "estimate"  #estimate估计
  print(b)
}
reg(y = launch$distress_ct, x = launch[2:4])
model <- lm(distress_ct ~ temperature + field_check_pressure + flight_num , data = launch)
对比结果一致
利用直方图看分布
hist(insurance$expenses)
利用table统计出现个数
table(insurance$region)
改进的散点图矩阵--数值是相关系数-椭圆越拉伸相关性越强-红色的线为拟合曲线
library(psych)
pairs.panels(insurance[c("age", "bmi", "children", "expenses")])
线性回归模型lm()默认stats包里
复相关系数R--皮尔逊相关
决定系数R方-判定模型介绍因变量的值的程度-越接近1越好=复相关系数的平方
R方=cor(pre,real)^2
提升模型的性能
1添加非线性关系x^2 
2转换数值型转换为二进制指标ifelse函数
insurance$bmi30 <- ifelse(insurance$bmi >= 30, 1, 0)
3加入相互作用的影响bmi30*smoker 乘法关系 或者bmi30:smoker 冒号显示之间有相互作用
lm(expenses ~ age + bmi30*smoker, data = insurance)
lm(expenses ~ age + bmi30:smoker, data = insurance)
4全部放在一起
ins_model2 <- lm(expenses ~ age + age2 + children + bmi + sex +

                   bmi30*smoker + region, data = insurance)

launch <- read.csv("challenger.csv")# estimate beta manuallyb <- cov(launch$temperature, launch$distress_ct) / var(launch$temperature)b# estimate alpha manuallya <- mean(launch$distress_ct) - b * mean(launch$temperature)a# calculate the correlation of launch datar <- cov(launch$temperature, launch$distress_ct) /       (sd(launch$temperature) * sd(launch$distress_ct))rcor(launch$temperature, launch$distress_ct)# computing the slope using correlationr * (sd(launch$distress_ct) / sd(launch$temperature))# confirming the regression line using the lm function (not in text)model <- lm(distress_ct ~ temperature, data = launch)modelsummary(model)# creating a simple multiple regression functionreg <- function(y, x) {  x <- as.matrix(x)  x <- cbind(Intercept = 1, x)  b <- solve(t(x) %*% x) %*% t(x) %*% y  colnames(b) <- "estimate"  print(b)}# examine the launch datastr(launch)# test regression model with simple linear regressionreg(y = launch$distress_ct, x = launch[2])# use regression model with multiple regressionreg(y = launch$distress_ct, x = launch[2:4])# confirming the multiple regression result using the lm function (not in text)#model <- lm(distress_ct ~ temperature + pressure + launch_id, data = launch)model <- lm(distress_ct ~ temperature + field_check_pressure + flight_num , data = launch)model## Example: Predicting Medical Expenses ----## Step 2: Exploring and preparing the data ----insurance <- read.csv("insurance.csv", stringsAsFactors = TRUE)str(insurance)# summarize the charges variablesummary(insurance$expenses)# histogram of insurance charges利用直方图看分布hist(insurance$expenses)# table of region利用table统计出现个数table(insurance$region)# exploring relationships among features: correlation matrixcor(insurance[c("age", "bmi", "children", "expenses")])# visualing relationships among features: scatterplot matrix两两散点图矩阵pairs(insurance[c("age", "bmi", "children", "expenses")])# more informative scatterplot matrix改进的散点图矩阵library(psych)pairs.panels(insurance[c("age", "bmi", "children", "expenses")])## Step 3: Training a model on the data ----ins_model <- lm(expenses ~ age + children + bmi + sex + smoker + region,                data = insurance)ins_model <- lm(expenses ~ ., data = insurance) # this is equivalent to above# see the estimated beta coefficientsins_model## Step 4: Evaluating model performance ----# see more detail about the estimated beta coefficientssummary(ins_model)## Step 5: Improving model performance ----# add a higher-order "age" terminsurance$age2 <- insurance$age^2# add an indicator for BMI >= 30insurance$bmi30 <- ifelse(insurance$bmi >= 30, 1, 0)# create final modelins_model2 <- lm(expenses ~ age + age2 + children + bmi + sex +                   bmi30*smoker + region, data = insurance)summary(ins_model2)