Question

Recently I was surprised by the fact that is.logical and is.factor weren’t able to run out of the box with apply – at least they did not the deliver the correct result.

Here’s my little reproducible example:

# generate a dataset that contains a couple of modes
someDf <- data.frame(fac1=gl(2,3,12),
                 int=1:12,
                 char=letters[1:12],
                 logi=rep(c(T,F),6),
                 fac2=gl(3,2,12))

# hooray, this did work, got factors, int,
# characters and logical
str(someDf)

# I expected this to work, but it didn't
# everything is just FALSE
apply(someDf,2,is.logical)

I didn’t give up and found a way to sneak around this.

unlist(lapply(names(someDf),function(x) is.logical(someDf[,x])))

Though this delivers the correct result I wonder why it has to be that complicated and whether there’s a simpler to solution to it. Any ideas?

Hint: I have seen that

apply(someDf,2,class)

delivers all characters. This is unexpected, too. Maybe something with eval does the trick I couldn’t find.

Answer 1

Since a data.frame is a list, you should use lapply or sapply:

sapply(someDf, is.logical)
 fac1   int  char  logi  fac2 
FALSE FALSE FALSE  TRUE FALSE 

The reason that your code doesn’t work is because apply needs a matrix as its argument, and coerces to a matrix if you provide a data frame. Since a matrix can only have elements of a single class, your values get converted to character. Try it:

as.matrix(someDf)
      fac1 int  char logi    fac2
 [1,] "1"  " 1" "a"  " TRUE" "1" 
 [2,] "1"  " 2" "b"  "FALSE" "1" 
 [3,] "1"  " 3" "c"  " TRUE" "2" 
 [4,] "2"  " 4" "d"  "FALSE" "2" 
 [5,] "2"  " 5" "e"  " TRUE" "3" 
 [6,] "2"  " 6" "f"  "FALSE" "3" 
 [7,] "1"  " 7" "g"  " TRUE" "1" 
 [8,] "1"  " 8" "h"  "FALSE" "1" 
 [9,] "1"  " 9" "i"  " TRUE" "2" 
[10,] "2"  "10" "j"  "FALSE" "2" 
[11,] "2"  "11" "k"  " TRUE" "3" 
[12,] "2"  "12" "l"  "FALSE" "3"