> dat <- data.frame(y=rnorm(10), x1=1:10, x2=2:11, x3=3:12)
> str(dat)
'data.frame':   10 obs. of  4 variables:
$ y : num  1.318 0.731 0.274 -0.488 -0.431 ...
$ x1: int  1 2 3 4 5 6 7 8 9 10
$ x2: int  2 3 4 5 6 7 8 9 10 11
$ x3: int  3 4 5 6 7 8 9 10 11 12
>
> myfun <- function(dat, xvar) {
+     for (colid in xvar) {
+         dat[, colid] <- as.factor(dat[, colid])
+ }
+     return(dat)
+ }
>
> dat <- myfun(dat, c("x1", "x2", "x3"))
> str(dat)
'data.frame':   10 obs. of  4 variables:
$ y : num  1.318 0.731 0.274 -0.488 -0.431 ...
$ x1: Factor w/ 10 levels "1","2","3","4",..: 1 2 3 4 5 6 7 8 9 10
$ x2: Factor w/ 10 levels "2","3","4","5",..: 1 2 3 4 5 6 7 8 9 10
$ x3: Factor w/ 10 levels "3","4","5","6",..: 1 2 3 4 5 6 7 8 9 10

myfunc <- function(data, xvar) {
        xv <- strsplit(xvar, " ")[[1]]
        xvfactor <- as.data.frame(lapply(xv, function(x) factor(eval(as.name(x), envir=data))))
        names(xvfactor) <- paste(xv, ".factor", sep="")
        cbind(data, xvfactor)
}


data <- data.frame(y=rnorm(100), x1=sample(1:3, 100, replace=T),
                         x2=sample(2:5, 100, replace=T), x3=sample(1:4, 100, replace=T))

newdata <- myfunc(data, "x1 x2 x3")

head(newdata)

newdata$x1.factor

如果要覆盖原来 x1 x2 x3的话

myfunc <- function(data, xvar) {
        xv <- strsplit(xvar, " ")[[1]]
        xvfactor <- as.data.frame(lapply(xv, function(x) factor(eval(as.name(x), envir=data))))
        data[, xv] <- xvfactor
        data
}


data <- data.frame(y=rnorm(100), x1=sample(1:3, 100, replace=T),
                         x2=sample(2:5, 100, replace=T), x3=sample(1:4, 100, replace=T))

newdata <- myfunc(data, "x1 x2 x3")

head(newdata)

newdata$x1

xv <- strsplit(xvar, " ")[[1]]
[[1]] 能否解释一下呢？谢谢！

strsplit是把"x1 x2 x3"变成 "x1" "x2" "x3", 这个取决你想如何输入xvar
如果直接是个vector，比如 c("x1", "x2", "x3") 那这个strsplit就没必要了
如果是象你帖子里面输入"x1 x2 x3"那就需要先提取哪些变量