How to concatenate factors, without them being converted to integer level?

From the R Mailing list:

unlist(list(facs[1 : 3], facs[4 : 5]))

To 'cbind' factors, do

data.frame(facs[1 : 3], facs[4 : 5])

An alternate workaround is to convert the factor to be a character vector, then convert back when you are finshed concatenating.

cfacs <- as.character(facs)
x <- c(cfacs[1:3], cfacs[4:5]) 

# Now choose between
factor(x)
# and
factor(x, levels = levels(facs))

Use fct_c from the forcats package (part of the tidyverse).

> library(forcats)
> facs <- as.factor(c("i", "want", "to", "be", "a", "factor", "not", "an", "integer"))
> fct_c(facs[1:3], facs[4:5])
[1] i    want to   be   a
Levels: a an be factor i integer not to want

fct_c isn't fooled by concatenations of factors with discrepant numerical codings:

> x <- as.factor(c('c', 'z'))
> x
[1] c z
Levels: c z
> y <- as.factor(c('a', 'b', 'z'))
> y
[1] a b z
Levels: a b z
> c(x, y)
[1] 1 2 1 2 3
> fct_c(x, y)
[1] c z a b z
Levels: c z a b
> as.numeric(fct_c(x, y))
[1] 1 2 3 4 2

Wow, I never realized it did that. Here is a work-around:

x <- c(facs[1 : 3], facs[4 : 5]) 
x <- factor(x, levels=1:nlevels(facs), labels=levels(facs))
x

With the output:

[1] i    want to   be   a   
Levels: a an be factor i integer not to want

It will only work if the two vectors have the same levels as here.

This is a really bad R gotcha. Along those lines, here's one that just swallowed several hours of my time.

x <- factor(c("Yes","Yes","No", "No", "Yes", "No"))
y <- c("Yes", x)

> y
[1] "Yes" "2"   "2"   "1"   "1"   "2"   "1"  
> is.factor(y)
[1] FALSE

It appears to me the better fix is Richie's, which coerces to character.

> y <- c("Yes", as.character(x))
> y
[1] "Yes" "Yes" "Yes" "No"  "No"  "Yes" "No" 
> y <- as.factor(y)
> y
[1] Yes Yes Yes No  No  Yes No 
Levels: No Yes

As long as you get the levels set properly, as Richie mentions.