install.packages("ggplot2")
install.packages("ggpubr")
install.packages("vegan")
计算Shannon-香农指数和Simpson-辛普森指数的命令在vegan包中,计算各组显著性的命令在ggpubr包中;画图使用ggplot命令,在行使每个命令之前一定要加载相应的包,如下:
library(ggplot2)
library(ggpubr)
library(vegan)
拿到一个otu表格,要先计算香农指数和辛普森指数,操作如下:
otu=read.table('D:/r-working/feature-table.taxonomy.txt',row.names = 1,skip=1,header=T,comment.char ='',sep='\t')
#读取out表格
#'D:/feature table.taxonomy.txt'为文件路径,注意斜线方向
#row.names = 1指定第一列为行名
#skip=1跳过第一行不读
#header=T指定第一个有效行为列名
#sep='\t'表示指定制表符为分隔符
#comment.char=''表示设置注释符号为空字符‘’,这样#后面的内容就不会被省略
otu=otu[,-ncol(otu)]
#去除表格的最后一列,无用信息
otu=t(otu)
#表格转置,必须将样品名作为行名
shannon=diversity(otu,"shannon")
#计算香农指数,先加载vegan包
shannon
#查看香农指数
simpson=diversity(otu,"simpson")
#计算辛普森指数,先加载vegan包
simpson
#查看辛普森指数
alpha=data.frame(shannon,simpson,check.names=T)
#合并两个指数
write.table(alpha,"D:/r-working/alpha-summary.xls",sep='\t',quote=F)
#存储数据,注意路径使用反斜杠
将各样本进行分组,并进行画图,操作如下:
map<-read.table('D:/r-working/mapping_file.txt',row.names = 1,header=T,comment.char ='',sep='\t',check.names=F)
#读取分组表格
group<-map["Group1"]
#提取需要的分组,'Group1'是表中的分组列名,包括A,B,C三组
alpha<-alpha[match(rownames(group),rownames(alpha)),]
#重排alpha的行的顺序,使其与group的样本id(行名)一致
data<-data.frame(group,alpha,check.rows=T)
#合并两个表格.'<-'与'='同属赋值的含义.
p=ggplot(data=data,aes(x=Group1,y=shannon))+geom_boxplot(fill=rainbow(7)[2])
#data = data指定数据表格
#x=Group1指定作为x轴的数据列名
#y=shannon指定作为y轴的数据列名
#geom_boxplot()表示画箱线图
#fill=rainbow(7)[2]指定填充色
此处用到ggplot2包画箱线图,将画图函数赋值给p后,可以用‘+’不断进行图层叠加,给图片p增加新的特性
p
#查看p
mycompare=list(c('A','B'),c('A','C'),c('B','C'))
#指定多重比较的分组对
mycompare
p<-p+stat_compare_means(comparisons=mycompare,label = "p.signif",method = 'wilcox')
#添加显著性标记的第一种方法,在此之前先加载ggpubr包
p<-p+ylim(2,5.5)
#调整图像的外观
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。