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+         


        

Answer 1

New function "across" was introduced in dplyr version 1.0.0. The new function will supersede scoped variables (_if, _at, _all). Here's the official documentation

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

Answer 2

If you would use data.table package for the operations on data.frame then the problem is not present.

library(data.table)
dt = data.table(col1 = c("a","b","c"), col2 = 1:3)
sapply(dt, class)
#       col1        col2 
#"character"   "integer" 

If you have a factor columns in you dataset already and you want to convert them to character you can do the following.

library(data.table)
dt = data.table(col1 = factor(c("a","b","c")), col2 = 1:3)
sapply(dt, class)
#     col1      col2 
# "factor" "integer" 
upd.cols = sapply(dt, is.factor)
dt[, names(dt)[upd.cols] := lapply(.SD, as.character), .SDcols = upd.cols]
sapply(dt, class)
#       col1        col2 
#"character"   "integer" 

Answer 3

Another way is to convert it using apply

bob2 <- apply(bob,2,as.character)

And a better one (the previous is of class 'matrix')

bob2 <- as.data.frame(as.matrix(bob),stringsAsFactors=F)

Answer 4

You should use convert in hablar which gives readable syntax compatible with tidyverse pipes:

library(dplyr)
library(hablar)

df <- tibble(a = factor(c(1, 2, 3, 4)),
             b = factor(c(5, 6, 7, 8)))

df %>% convert(chr(a:b))

which gives you:

  a     b    
  <chr> <chr>
1 1     5    
2 2     6    
3 3     7    
4 4     8   

Answer 5

If you want a new data frame bobc where every factor vector in bobf is converted to a character vector, try this:

bobc <- rapply(bobf, as.character, classes="factor", how="replace")

If you then want to convert it back, you can create a logical vector of which columns are factors, and use that to selectively apply factor

f <- sapply(bobf, class) == "factor"
bobc[,f] <- lapply(bobc[,f], factor)

Answer 6

Or you can try transform:

newbob <- transform(bob, phenotype = as.character(phenotype))

Just be sure to put every factor you'd like to convert to character.

Or you can do something like this and kill all the pests with one blow:

newbob_char <- as.data.frame(lapply(bob[sapply(bob, is.factor)], as.character), stringsAsFactors = FALSE)
newbob_rest <- bob[!(sapply(bob, is.factor))]
newbob <- cbind(newbob_char, newbob_rest)

It's not good idea to shove the data in code like this, I could do the sapply part separately (actually, it's much easier to do it like that), but you get the point... I haven't checked the code, 'cause I'm not at home, so I hope it works! =)

This approach, however, has a downside... you must reorganize columns afterwards, while with transform you can do whatever you like, but at cost of "pedestrian-style-code-writting"...

So there... =)