#第一种方法
per <- data.frame(name = c("张三","李四","王五","赵六"),
age = c(23,45,34,1000))
per
per$age[per$age == 1000] <- NA #设置缺失值
per$age1[per$age <30] = "young" #生成新变量
per$age1[per$age >= 30 &per$age<50] <- "middle age"
per
#第二种方法
per <- data.frame(name = c("张三","李四","王五","赵六"),
age = c(23,45,34,1000))
per <- within(per,{
age1 <- NA
age1[age <30] <- "young"
age1[age>=30 &age<50] <- "middle age"
})
per
用简单r语言中的for循环进行编码
>a=c(2,6,9,12,33)
>for(i in 1:length(a)) cat("a=",a[i],", a^2=",a[i]*a[i],", a^3=", a[i]*a[i]*a[i], "\n")
a= 2 , a^2= 4 , a^3= 8
a= 6 , a^2= 36 , a^3= 216
a= 9 , a^2= 81 , a^3= 729
a= 12 , a^2= 144 , a^3= 1728
a= 33 , a^2= 1089 , a^3= 35937
>
简单码?
点菜单的编辑,替换,查找处输入 2010,替换处输入 2011,全部替换。 由于担心把不必要的2010替换为2011,可以输入类似 2010.xls 和 2011.xls,如果适用的话。