Coerce multiple columns to factors at once

Choose some columns to coerce to factors:

cols <- c("A", "C", "D", "H")

Use lapply() to coerce and replace the chosen columns:

data[cols] <- lapply(data[cols], factor)  ## as.factor() could also be used

Check the result:

sapply(data, class)
#        A         B         C         D         E         F         G 
# "factor" "integer"  "factor"  "factor" "integer" "integer" "integer" 
#        H         I         J 
# "factor" "integer" "integer" 

Here is an option using dplyr. The %<>% operator from magrittr update the lhs object with the resulting value.

library(magrittr)
library(dplyr)
cols <- c("A", "C", "D", "H")

data %<>%
       mutate_each_(funs(factor(.)),cols)
str(data)
#'data.frame':  4 obs. of  10 variables:
# $ A: Factor w/ 4 levels "23","24","26",..: 1 2 3 4
# $ B: int  15 13 39 16
# $ C: Factor w/ 4 levels "3","5","18","37": 2 1 3 4
# $ D: Factor w/ 4 levels "2","6","28","38": 3 1 4 2
# $ E: int  14 4 22 20
# $ F: int  7 19 36 27
# $ G: int  35 40 21 10
# $ H: Factor w/ 4 levels "11","29","32",..: 1 4 3 2
# $ I: int  17 1 9 25
# $ J: int  12 30 8 33

---

Or if we are using data.table, either use a for loop with set

setDT(data)
for(j in cols){
  set(data, i=NULL, j=j, value=factor(data[[j]]))
}

Or we can specify the 'cols' in .SDcols and assign (:=) the rhs to 'cols'

setDT(data)[, (cols):= lapply(.SD, factor), .SDcols=cols]

The more recent tidyverse way is to use the mutate_at function:

library(tidyverse)
library(magrittr)
set.seed(88)

data <- data.frame(matrix(sample(1:40), 4, 10, dimnames = list(1:4, LETTERS[1:10])))
cols <- c("A", "C", "D", "H")

data %<>% mutate_at(cols, factor)
str(data)
 $ A: Factor w/ 4 levels "5","17","18",..: 2 1 4 3   
 $ B: int  36 35 2 26
 $ C: Factor w/ 4 levels "22","31","32",..: 1 2 4 3
 $ D: Factor w/ 4 levels "1","9","16","39": 3 4 1 2
 $ E: int  3 14 30 38
 $ F: int  27 15 28 37
 $ G: int  19 11 6 21
 $ H: Factor w/ 4 levels "7","12","20",..: 1 3 4 2
 $ I: int  23 24 13 8
 $ J: int  10 25 4 33