Convert all data frame character columns to factors

As @Raf Z commented on this question, dplyr now has mutate_if. Super useful, simple and readable.

> str(df)
'data.frame':   5 obs. of  5 variables:
 $ A: Factor w/ 5 levels "A","B","C","D",..: 1 2 3 4 5
 $ B: int  1 2 3 4 5
 $ C: logi  TRUE TRUE FALSE FALSE TRUE
 $ D: chr  "a" "b" "c" "d" ...
 $ E: chr  "A a" "B b" "C c" "D d" ...

> df <- df %>% mutate_if(is.character,as.factor)

> str(df)
'data.frame':   5 obs. of  5 variables:
 $ A: Factor w/ 5 levels "A","B","C","D",..: 1 2 3 4 5
 $ B: int  1 2 3 4 5
 $ C: logi  TRUE TRUE FALSE FALSE TRUE
 $ D: Factor w/ 5 levels "a","b","c","d",..: 1 2 3 4 5
 $ E: Factor w/ 5 levels "A a","B b","C c",..: 1 2 3 4 5

Roland's answer is great for this specific problem, but I thought I would share a more generalized approach.

DF <- data.frame(x = letters[1:5], y = 1:5, z = LETTERS[1:5], 
                 stringsAsFactors=FALSE)
str(DF)
# 'data.frame':  5 obs. of  3 variables:
#  $ x: chr  "a" "b" "c" "d" ...
#  $ y: int  1 2 3 4 5
#  $ z: chr  "A" "B" "C" "D" ...

## The conversion
DF[sapply(DF, is.character)] <- lapply(DF[sapply(DF, is.character)], 
                                       as.factor)
str(DF)
# 'data.frame':  5 obs. of  3 variables:
#  $ x: Factor w/ 5 levels "a","b","c","d",..: 1 2 3 4 5
#  $ y: int  1 2 3 4 5
#  $ z: Factor w/ 5 levels "A","B","C","D",..: 1 2 3 4 5

For the conversion, the left hand side of the assign (DF[sapply(DF, is.character)]) subsets the columns that are character. In the right hand side, for that subset, you use lapply to perform whatever conversion you need to do. R is smart enough to replace the original columns with the results.

The handy thing about this is if you wanted to go the other way or do other conversions, it's as simple as changing what you're looking for on the left and specifying what you want to change it to on the right.

DF <- data.frame(x=letters[1:5], y=1:5, stringsAsFactors=FALSE)

str(DF)
#'data.frame':  5 obs. of  2 variables:
# $ x: chr  "a" "b" "c" "d" ...
# $ y: int  1 2 3 4 5

You can use as.data.frame to turn all character columns into factor columns:

DF <- as.data.frame(unclass(DF),stringsAsFactors=TRUE)
str(DF)
#'data.frame':  5 obs. of  2 variables:
# $ x: Factor w/ 5 levels "a","b","c","d",..: 1 2 3 4 5
# $ y: int  1 2 3 4 5

Working with dplyr

library(dplyr)

df <- data.frame(A = factor(LETTERS[1:5]),
                 B = 1:5, C = as.logical(c(1, 1, 0, 0, 1)),
                 D = letters[1:5],
                 E = paste(LETTERS[1:5], letters[1:5]),
                 stringsAsFactors = FALSE)

str(df)

we get:

'data.frame':   5 obs. of  5 variables:
 $ A: Factor w/ 5 levels "A","B","C","D",..: 1 2 3 4 5
 $ B: int  1 2 3 4 5
 $ C: logi  TRUE TRUE FALSE FALSE TRUE
 $ D: chr  "a" "b" "c" "d" ...
 $ E: chr  "A a" "B b" "C c" "D d" ...

Now, we can convert all chr to factors:

df <- df%>%mutate_if(is.character, as.factor)
str(df)

And we get:

'data.frame':   5 obs. of  5 variables:
 $ A: Factor w/ 5 levels "A","B","C","D",..: 1 2 3 4 5
 $ B: int  1 2 3 4 5
 $ C: logi  TRUE TRUE FALSE FALSE TRUE
 $ D: chr  "a" "b" "c" "d" ...
 $ E: chr  "A a" "B b" "C c" "D d" ...

Let's provide also other solutions:

With base package:

df[sapply(df, is.character)] <- lapply(df[sapply(df, is.character)], 
                                                           as.factor)

With dplyr 1.0.0

df <- df%>%mutate(across(where(is.factor), as.character))

With purrr package:

library(purrr)

df <- df%>% modify_if(is.factor, as.character)