Convert data.frame columns from factors to characters

Question

I have a data frame. Let's call him bob:

> head(bob)                  phenotype                         exclusion GSM399350 3- 4- 8- 25- 44+ 11b- 11c- 19- NK1.1- Gr1- TER119- GSM399351 3- 4- 8- 25- 44+ 11b- 11c- 19- NK1.1- Gr1- TER119- GSM399352 3- 4- 8- 25- 44+ 11b- 11c- 19- NK1.1- Gr1- TER119- GSM399353 3- 4- 8- 25+ 44+ 11b- 11c- 19- NK1.1- Gr1- TER119- GSM399354 3- 4- 8- 25+ 44+ 11b- 11c- 19- NK1.1- Gr1- TER119- GSM399355 3- 4- 8- 25+ 44+ 11b- 11c- 19- NK1.1- Gr1- TER119- 

I'd like to concatenate the rows of this data frame (this will be another question). But look:

> class(bob$phenotype) [1] "factor" 

Bob's columns are factors. So, for example:

> as.character(head(bob)) [1] "c(3, 3, 3, 6, 6, 6)"       "c(3, 3, 3, 3, 3, 3)"       [3] "c(29, 29, 29, 30, 30, 30)" 

I don't begin to understand this, but I guess these are indices into the levels of the factors of the columns (of the court of king caractacus) of bob? Not what I need.

Strangely I can go through the columns of bob by hand, and do

bob$phenotype <- as.character(bob$phenotype) 

which works fine. And, after some typing, I can get a data.frame whose columns are characters rather than factors. So my question is: how can I do this automatically? How do I convert a data.frame with factor columns into a data.frame with character columns without having to manually go through each column?

Bonus question: why does the manual approach work?

Answer 1

Just following on Matt and Dirk. If you want to recreate your existing data frame without changing the global option, you can recreate it with an apply statement:

bob <- data.frame(lapply(bob, as.character), stringsAsFactors=FALSE) 

This will convert all variables to class "character", if you want to only convert factors, see Marek's solution below.

As @hadley points out, the following is more concise.

bob[] <- lapply(bob, as.character) 

In both cases, lapply outputs a list; however, owing to the magical properties of R, the use of [] in the second case keeps the data.frame class of the bob object, thereby eliminating the need to convert back to a data.frame using as.data.frame with the argument stringsAsFactors = FALSE.

Answer 2

To replace only factors:

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

In package dplyr in version 0.5.0 new function mutate_if was introduced:

library(dplyr) bob %>% mutate_if(is.factor, as.character) -> bob 

...and in version 1.0.0 was replaced by across:

library(dplyr) bob %>% mutate(across(where(is.factor), as.character)) -> bob 

Package purrr from RStudio gives another alternative:

library(purrr) bob %>% modify_if(is.factor, as.character) -> bob