R语言做数据分析的基本方法

本文基于R语言进行基本数据统计分析,包括基本作图,线性拟合,逻辑回归,bootstrap采样和Anova方差分析的实现及应用。

不多说,直接上代码,代码中有注释。

1. 基本作图(盒图,qq图)


#basic plot

boxplot(x)

qqplot(x,y)

2. 线性拟合

#linear regression

n = 10

x1 = rnorm(n)#variable 1

x2 = rnorm(n)#variable 2

y = rnorm(n)*3

mod = lm(y~x1+x2)

model.matrix(mod) #erect the matrix of mod

plot(mod) #plot residual and fitted of the solution, Q-Q plot and cook distance

summary(mod) #get the statistic information of the model

hatvalues(mod) #very important, for abnormal sample detection

3. 逻辑回归

#logistic regression

x <- c(0, 1, 2, 3, 4, 5)

y <- c(0, 9, 21, 47, 60, 63) # the number of successes

n <- 70 #the number of trails

z <- n – y #the number of failures

b <- cbind(y, z) # column bind

fitx <- glm(b~x,family = binomial) # a particular type of generalized linear model

print(fitx)

plot(x,y,xlim=c(0,5),ylim=c(0,65)) #plot the points (x,y)

beta0 <- fitx$coef[1]

beta1 <- fitx$coef[2]

fn <- function(x) n*exp(beta0+beta1*x)/(1+exp(beta0+beta1*x))

par(new=T)

curve(fn,0,5,ylim=c(0,60)) # plot the logistic regression curve

R语言做数据分析的基本方法

3. Bootstrap采样

# bootstrap

# Application: 随机采样,获取最大eigenvalue占所有eigenvalue和之比,并画图显示distribution

dat = matrix(rnorm(100*5),100,5)

no.samples = 200 #sample 200 times

# theta = matrix(rep(0,no.samples*5),no.samples,5)

theta =rep(0,no.samples*5);

for (i in 1:no.samples)

{

j = sample(1:100,100,replace = TRUE)#get 100 samples each time

datrnd = dat[j,]; #select one row each time

lambda = princomp(datrnd)$sdev^2; #get eigenvalues

# theta[i,] = lambda;

theta[i] = lambda[1]/sum(lambda); #plot the ratio of the biggest eigenvalue

}

# hist(theta[1,]) #plot the histogram of the first(biggest) eigenvalue

hist(theta); #plot the percentage distribution of the biggest eigenvalue

sd(theta)#standard deviation of theta

#上面注释掉的语句,可以全部去掉注释并将其下一条语句注释掉,完成画最大eigenvalue分布的功能

4. ANOVA方差分析

#Application:判断一个自变量是否有影响 (假设我们喂3种维他命给3头猪,想看喂维他命有没有用)

#

y = rnorm(9); #weight gain by pig(Yij, i is the treatment, j is the pig_id), 一般由用户自行输入

#y = matrix(c(1,10,1,2,10,2,1,9,1),9,1)

Treatment <- factor(c(1,2,3,1,2,3,1,2,3)) #each {1,2,3} is a group

mod = lm(y~Treatment) #linear regression

print(anova(mod))

#解释:Df(degree of freedom)

#Sum Sq: deviance (within groups, and residuals) 总偏差和

# Mean Sq: variance (within groups, and residuals) 平均方差和

# compare the contribution given by Treatment and Residual

#F value: Mean Sq(Treatment)/Mean Sq(Residuals)

#Pr(>F): p-value. 根据p-value决定是否接受Hypothesis H0:多个样本总体均数相等(检验水准为0.05)

qqnorm(mod$residual) #plot the residual approximated by mod

#如果qqnorm of residual像一条直线,说明residual符合正态分布,也就是说Treatment带来的contribution很小,也就是说Treatment无法带来收益(多喂维他命少喂维他命没区别)

如下面两图分别是

(左)用 y = matrix(c(1,10,1,2,10,2,1,9,1),9,1)和

(右)y = rnorm(9);

的结果。可见如果给定猪吃维他命2后体重特别突出的数据结果后,qq图种residual不在是一条直线,换句话说residual不再符合正态分布,i.e., 维他命对猪的体重有影响。

R语言做数据分析的基本方法
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