manager<-c(1,2,3,4,5)
data<-c("10/24/08","10/28/08","10/1/08","10/12/08","5/1/09")
country<-c("US","US","UK","UK","UK")
gender<-c("M","F","F","M","F")
age<-c(32,45,25,39,99)
q1<-c(5,3,3,3,2)
q2<-c(4,5,5,3,2)
q3<-c(5,2,5,4,1)
q4<-c(5,5,5,NA,2)
q5<-c(5,5,2,NA,1)
leadership<-data.frame(manager,data,country,gender,age,q1,q2,q3,q4,q5,stringASFactors=FALSE)

leadership within <-(leadership,{
  agecat <-NA
  agecat[age>75]<-"Elder"
  agecat[age>=55&age<=75]<-"Middle Age"
  agecat[age<55]<-"Young" })

fix(leadship)#交互的方式修改变量的名称

ibrary(reshape)
leadership<-rename(leadership,
                   c(manage="mamanageID",data="testdata")
                   )#通过reshape库将名字更改

names(leadship)[2]<"testdata"

> y<-c(1,2,3,NA)
> is.na(y)
[1] FALSE FALSE FALSE TRUE #检测y内是否有空值存在

 x<-(1,2,3,NA)
y<-sum(x,na.rm = TRUE)#na.rm计算之前排除缺失值，并使用剩余值进行计算

newdata<-na.omit(leadership)#使用na.omit 函数

newdata<-leadership[order(leadership$age),]#按照年龄排序

attach(leadership)
newdata<-leadership[order(gender,-age),]
detach(leadership)#性别 年龄降序排列

attach(leadership)
newdata<-leadership[which(gender==‘M‘&age>30),]
detach(leadership)
newdata #挑选性别为男且年龄大于30岁的人

 mysample<-leadership[sample(1:nrow(leadership),3,replace=FALSE),]