Coerce multiple columns to factors at once

Question

I have a sample data frame like below:

data <- data.frame(matrix(sample(1:40), 4, 10, dimnames = list(1:4, LETTERS[1:10]))) 

I want to know how can I select multiple columns and convert them together to factors. I usually do it in the way like data$A = as.factor(data$A). But when the data frame is very large and contains lots of columns, this way will be very time consuming. Does anyone know of a better way to do it?

Answer 1

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"  

Answer 2

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 

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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]