R语言基本数据分析

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
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。