R语言实战（四）——基本数据管理

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

> mydatatest <- data.frame(x1,x2)
> mydatatest$sumx <- mydatatest$x1 + mydatatest$x2
> mydatatest$meanx <- (mydatatest$x1+mydatatest$x2)/2

> leadership$age[leadership$age == 99] <- NA

> fix(leadership)

> library(reshape)
> leadership <- rename(leadership,c(manager="managerID",date="testDate"))

names(leadership)[3] <- "myCountry"

xx <- c(1,2,NA,3)
zz <- sum(xx,na.rm=TRUE)

> newdata <- na.omit(leadership)
> newdata
  managerID testDate myCountry gender age q1 q2 q3 q4 q5 agecat
1         1 10/24/08        US      M  32  5  4  5  5  5  Young
2         2 10/28/08        US      F  45  3  5  2  5  5  Young
3         3  10/1/08        UK      F  25  3  5  5  5  2  Young

> newdata <- leadership[order(leadership$age),]