Convert data.frame columns from factors to characters
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?
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 .
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
Package purrr from RStudio gives another alternative:
library(purrr)
library(dplyr)
bob %>% map_if(is.factor, as.character) %>% as_data_frame -> bob
(keep in mind it's fresh package)
The global option
stringsAsFactors: The default setting for arguments of data.frame and read.table.
may be something you want to set to FALSE in your startup files (eg ~/.Rprofile). Please see help(options) .
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