How do I convert certain columns of a data frame to become factors? [duplicate]

Answer 1

Here's an example:

#Create a data frame
> d<- data.frame(a=1:3, b=2:4)
> d
  a b
1 1 2
2 2 3
3 3 4

#currently, there are no levels in the `a` column, since it's numeric as you point out.
> levels(d$a)
NULL

#Convert that column to a factor
> d$a <- factor(d$a)
> d
  a b
1 1 2
2 2 3
3 3 4

#Now it has levels.
> levels(d$a)
[1] "1" "2" "3"

You can also handle this when reading in your data. See the colClasses and stringsAsFactors parameters in e.g. readCSV().

Note that, computationally, factoring such columns won't help you much, and may actually slow down your program (albeit negligibly). Using a factor will require that all values are mapped to IDs behind the scenes, so any print of your data.frame requires a lookup on those levels -- an extra step which takes time.

Factors are great when storing strings which you don't want to store repeatedly, but would rather reference by their ID. Consider storing a more friendly name in such columns to fully benefit from factors.

Answer 2

Given the following sample

myData <- data.frame(A=rep(1:2, 3), B=rep(1:3, 2), Pulse=20:25)  

then

myData$A <-as.factor(myData$A)
myData$B <-as.factor(myData$B)

or you could select your columns altogether and wrap it up nicely:

# select columns
cols <- c("A", "B")
myData[,cols] <- data.frame(apply(myData[cols], 2, as.factor))

levels(myData$A) <- c("long", "short")
levels(myData$B) <- c("1kg", "2kg", "3kg")

To obtain

> myData
      A   B Pulse
1  long 1kg    20
2 short 2kg    21
3  long 3kg    22
4 short 1kg    23
5  long 2kg    24
6 short 3kg    25