Convert all data frame character columns to factors

user_827280 0
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Given a (pre-existing) data frame that has columns of various types, what is the simplest way to convert all its character columns to factors, without affecting any columns of other types?

Here's an example data.frame:

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)
df
#   A B     C D   E
# 1 A 1  TRUE a A a
# 2 B 2  TRUE b B b
# 3 C 3 FALSE c C c
# 4 D 4 FALSE d D d
# 5 E 5  TRUE e E e
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" ...

I know I can do:

df$D <- as.factor(df$D)
df$E <- as.factor(df$E)

Is there a way to automate this process a bit more?

dataframe r
8 Answers

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)
user_119836010

I noticed "[" indexing columns fails to create levels when iterating:

for ( a_feature in convert.to.factors) {
feature.df[a_feature] <- factor(feature.df[a_feature]) }

It creates, e.g. for the "Status" column:

Status : Factor w/ 1 level "c(\"Success\", \"Fail\")" : NA NA NA ...

Which is remedied by using "[[" indexing:

for ( a_feature in convert.to.factors) {
feature.df[[a_feature]] <- factor(feature.df[[a_feature]]) }

Giving instead, as desired:

. Status : Factor w/ 2 levels "Success", "Fail" : 1 1 2 1 ...

user_11081240

Based on @Roland 's answer and @Paul de Barros 's comments, I observed to the following conclusion:

    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)
   
   df<-as.data.frame(unclass(df),stringsAsFactors=TRUE)
   str(df)

Practically and simply seems to work.

> 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
user_114416100
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