Why Does As.Factor Return a Character When Used Inside Apply

Why does as.factor return a character when used inside apply?

apply converts your data.frame to a character matrix. Use lapply:

lapply(a, class)
# $x1
# [1] "numeric"
# $x2
# [1] "factor"
# $x3
# [1] "factor"

In second command apply converts result to character matrix, using lapply:

a2 <- lapply(a, as.factor)
lapply(a2, class)
# $x1
# [1] "factor"
# $x2
# [1] "factor"
# $x3
# [1] "factor"

But for simple lookout you could use str:

str(a)
# 'data.frame':   100 obs. of  3 variables:
#  $ x1: num  -1.79 -1.091 1.307 1.142 -0.972 ...
#  $ x2: Factor w/ 2 levels "a","b": 2 1 1 1 2 1 1 1 1 2 ...
#  $ x3: Factor w/ 2 levels "a","b": 1 1 1 1 1 1 1 1 1 1 ...

Additional explanation according to comments:

Why does the lapply work while apply doesn't?

The first thing that apply does is to convert an argument to a matrix. So apply(a) is equivalent to apply(as.matrix(a)). As you can see str(as.matrix(a)) gives you:

chr [1:100, 1:3] " 0.075124364" "-1.608618269" "-1.487629526" ...
- attr(*, "dimnames")=List of 2
  ..$ : NULL
  ..$ : chr [1:3] "x1" "x2" "x3"

There are no more factors, so class return "character" for all columns.

lapply works on columns so gives you what you want (it does something like class(a$column_name) for each column).

You can see in help to apply why apply and as.factor doesn't work :

In all cases the result is coerced by
as.vector to one of the basic vector
types before the dimensions are set,
so that (for example) factor results
will be coerced to a character array.

Why sapply and as.factor doesn't work you can see in help to sapply:

Value (...) An atomic vector or matrix
or list of the same length as X (...)
If simplification occurs, the output
type is determined from the highest
type of the return values in the
hierarchy NULL < raw < logical <
integer < real < complex < character <
list < expression, after coercion of
pairlists to lists.

You never get matrix of factors or data.frame.

How to convert output to data.frame?

Simple, use as.data.frame as you wrote in comment:

a2 <- as.data.frame(lapply(a, as.factor))
str(a2)
'data.frame':   100 obs. of  3 variables:
 $ x1: Factor w/ 100 levels "-2.49629293159922",..: 60 6 7 63 45 93 56 98 40 61 ...
 $ x2: Factor w/ 2 levels "a","b": 1 1 2 2 2 2 2 1 2 2 ...
 $ x3: Factor w/ 2 levels "a","b": 1 1 1 1 1 1 1 1 1 1 ...

But if you want to replace selected character columns with factor there is a trick:

a3 <- data.frame(x1=letters, x2=LETTERS, x3=LETTERS, stringsAsFactors=FALSE)
str(a3)
'data.frame':   26 obs. of  3 variables:
 $ x1: chr  "a" "b" "c" "d" ...
 $ x2: chr  "A" "B" "C" "D" ...
 $ x3: chr  "A" "B" "C" "D" ...

columns_to_change <- c("x1","x2")
a3[, columns_to_change] <- lapply(a3[, columns_to_change], as.factor)
str(a3)
'data.frame':   26 obs. of  3 variables:
 $ x1: Factor w/ 26 levels "a","b","c","d",..: 1 2 3 4 5 6 7 8 9 10 ...
 $ x2: Factor w/ 26 levels "A","B","C","D",..: 1 2 3 4 5 6 7 8 9 10 ...
 $ x3: chr  "A" "B" "C" "D" ...

You could use it to replace all columns using:

a3 <- data.frame(x1=letters, x2=LETTERS, x3=LETTERS, stringsAsFactors=FALSE)
a3[, ] <- lapply(a3, as.factor)
str(a3)
'data.frame':   26 obs. of  3 variables:
 $ x1: Factor w/ 26 levels "a","b","c","d",..: 1 2 3 4 5 6 7 8 9 10 ...
 $ x2: Factor w/ 26 levels "A","B","C","D",..: 1 2 3 4 5 6 7 8 9 10 ...
 $ x3: Factor w/ 26 levels "A","B","C","D",..: 1 2 3 4 5 6 7 8 9 10 ...

Why is.factor() used in apply() and sapply() returns different values?

From the reference of apply:

Returns a vector or array or list of values obtained by applying a
function to margins of an array or matrix.

Therefore, it converts your input object to a matrix (array) first which must have the same atomic data type. This means that your data get coerced to character, because factor is not an atomic vector type.

> as.matrix(X)
     X1  X2  
[1,] "1" "f1"
[2,] "2" "f2"
[3,] "3" "f3"
[4,] "4" "f4"

Why `as.factor` does not work when applied via `apply` function in R?

I think this is because of how apply simplifies the result to return a matrix. From ?apply:

If ‘X’ is not an array but an object of a class with a non-null
    ‘dim’ value (such as a data frame), ‘apply’ attempts to coerce it
    to an array via ‘as.matrix’ if it is two-dimensional (e.g., a data
    frame) or via ‘as.array’.

In fact your original data frame is as you wish. Try str(df) or sapply(df, is.factor) to verify it. Basically character vectors are always coerced to factors, unless stringsAsFactors=FALSE.

lapply(x, as.factor) returning just one level

You want

df[] <- lapply(df, factor)

Why use as.factor() instead of just factor()

as.factor is a wrapper for factor, but it allows quick return if the input vector is already a factor:

function (x) 
{
    if (is.factor(x)) 
        x
    else if (!is.object(x) && is.integer(x)) {
        levels <- sort(unique.default(x))
        f <- match(x, levels)
        levels(f) <- as.character(levels)
        if (!is.null(nx <- names(x))) 
        names(f) <- nx
        class(f) <- "factor"
        f
    }
else factor(x)
}

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Comment from Frank: it's not a mere wrapper, since this "quick return" will leave factor levels as they are while factor() will not:

f = factor("a", levels = c("a", "b"))
#[1] a
#Levels: a b

factor(f)
#[1] a
#Levels: a

as.factor(f)
#[1] a
#Levels: a b

---

Expanded answer two years later, including the following:

• What does the manual say?
• Performance: as.factor > factor when input is a factor
• Performance: as.factor > factor when input is integer
• Unused levels or NA levels
• Caution when using R's group-by functions: watch for unused or NA levels

---

What does the manual say?

The documentation for ?factor mentions the following:

‘factor(x, exclude = NULL)’ applied to a factor without ‘NA’s is a
 no-operation unless there are unused levels: in that case, a
 factor with the reduced level set is returned.

 ‘as.factor’ coerces its argument to a factor.  It is an
 abbreviated (sometimes faster) form of ‘factor’.

Performance: as.factor > factor when input is a factor

The word "no-operation" is a bit ambiguous. Don't take it as "doing nothing"; in fact, it means "doing a lot of things but essentially changing nothing". Here is an example:

set.seed(0)
## a randomized long factor with 1e+6 levels, each repeated 10 times
f <- sample(gl(1e+6, 10))

system.time(f1 <- factor(f))  ## default: exclude = NA
#   user  system elapsed 
#  7.640   0.216   7.887 

system.time(f2 <- factor(f, exclude = NULL))
#   user  system elapsed 
#  7.764   0.028   7.791 

system.time(f3 <- as.factor(f))
#   user  system elapsed 
#      0       0       0 

identical(f, f1)
#[1] TRUE

identical(f, f2)
#[1] TRUE

identical(f, f3)
#[1] TRUE

as.factor does give a quick return, but factor is not a real "no-op". Let's profile factor to see what it has done.

Rprof("factor.out")
f1 <- factor(f)
Rprof(NULL)
summaryRprof("factor.out")[c(1, 4)]
#$by.self
#                      self.time self.pct total.time total.pct
#"factor"                   4.70    58.90       7.98    100.00
#"unique.default"           1.30    16.29       4.42     55.39
#"as.character"             1.18    14.79       1.84     23.06
#"as.character.factor"      0.66     8.27       0.66      8.27
#"order"                    0.08     1.00       0.08      1.00
#"unique"                   0.06     0.75       4.54     56.89
#
#$sampling.time
#[1] 7.98

It first sort the unique values of the input vector f, then converts f to a character vector, finally uses factor to coerces the character vector back to a factor. Here is the source code of factor for confirmation.

function (x = character(), levels, labels = levels, exclude = NA, 
    ordered = is.ordered(x), nmax = NA) 
{
    if (is.null(x)) 
        x <- character()
    nx <- names(x)
    if (missing(levels)) {
        y <- unique(x, nmax = nmax)
        ind <- sort.list(y)
        levels <- unique(as.character(y)[ind])
    }
    force(ordered)
    if (!is.character(x)) 
        x <- as.character(x)
    levels <- levels[is.na(match(levels, exclude))]
    f <- match(x, levels)
    if (!is.null(nx)) 
        names(f) <- nx
    nl <- length(labels)
    nL <- length(levels)
    if (!any(nl == c(1L, nL))) 
        stop(gettextf("invalid 'labels'; length %d should be 1 or %d", 
            nl, nL), domain = NA)
    levels(f) <- if (nl == nL) 
        as.character(labels)
    else paste0(labels, seq_along(levels))
    class(f) <- c(if (ordered) "ordered", "factor")
    f
}

So function factor is really designed to work with a character vector and it applies as.character to its input to ensure that. We can at least learn two performance-related issues from above:

1. For a data frame DF, lapply(DF, as.factor) is much faster than lapply(DF, factor) for type conversion, if many columns are readily factors.
2. That function factor is slow can explain why some important R functions are slow, say table: R: table function suprisingly slow

Performance: as.factor > factor when input is integer

A factor variable is the next of kin of an integer variable.

unclass(gl(2, 2, labels = letters[1:2]))
#[1] 1 1 2 2
#attr(,"levels")
#[1] "a" "b"

storage.mode(gl(2, 2, labels = letters[1:2]))
#[1] "integer"

This means that converting an integer to a factor is easier than converting a numeric / character to a factor. as.factor just takes care of this.

x <- sample.int(1e+6, 1e+7, TRUE)

system.time(as.factor(x))
#   user  system elapsed 
#  4.592   0.252   4.845 

system.time(factor(x))
#   user  system elapsed 
# 22.236   0.264  22.659 

Unused levels or NA levels

Now let's see a few examples on factor and as.factor's influence on factor levels (if the input is a factor already). Frank has given one with unused factor level, I will provide one with NA level.

f <- factor(c(1, NA), exclude = NULL)
#[1] 1    <NA>
#Levels: 1 <NA>

as.factor(f)
#[1] 1    <NA>
#Levels: 1 <NA>

factor(f, exclude = NULL)
#[1] 1    <NA>
#Levels: 1 <NA>

factor(f)
#[1] 1    <NA>
#Levels: 1

There is a (generic) function droplevels that can be used to drop unused levels of a factor. But NA levels can not be dropped by default.

## "factor" method of `droplevels`
droplevels.factor
#function (x, exclude = if (anyNA(levels(x))) NULL else NA, ...) 
#factor(x, exclude = exclude)

droplevels(f)
#[1] 1    <NA>
#Levels: 1 <NA>

droplevels(f, exclude = NA)
#[1] 1    <NA>
#Levels: 1

Caution when using R's group-by functions: watch for unused or NA levels

R functions doing group-by operations, like split, tapply expect us to provide factor variables as "by" variables. But often we just provide character or numeric variables. So internally, these functions need to convert them into factors and probably most of them would use as.factor in the first place (at least this is so for split.default and tapply). The table function looks like an exception and I spot factor instead of as.factor inside. There might be some special consideration which is unfortunately not obvious to me when I inspect its source code.

Since most group-by R functions use as.factor, if they are given a factor with unused or NA levels, such group will appear in the result.

x <- c(1, 2)
f <- factor(letters[1:2], levels = letters[1:3])

split(x, f)
#$a
#[1] 1
#
#$b
#[1] 2
#
#$c
#numeric(0)

tapply(x, f, FUN = mean)
# a  b  c 
# 1  2 NA 

Interestingly, although table does not rely on as.factor, it preserves those unused levels, too:

table(f)
#a b c 
#1 1 0 

Sometimes this kind of behavior can be undesired. A classic example is barplot(table(f)):

If this is really undesired, we need to manually remove unused or NA levels from our factor variable, using droplevels or factor.

Hint:

1. split has an argument drop which defaults to FALSE hence as.factor is used; by drop = TRUE function factor is used instead.
2. aggregate relies on split, so it also has a drop argument and it defaults to TRUE.
3. tapply does not have drop although it also relies on split. In particular the documentation ?tapply says that as.factor is (always) used.

Why gsub automatically changes a Factor into Character

Yes, gsub performs as.character. If you type gsub in the console you can see the function

function (pattern, replacement, x, ignore.case = FALSE, perl = FALSE, 
fixed = FALSE, useBytes = FALSE) 
{
    if (!is.character(x)) 
        x <- as.character(x)
    .Internal(gsub(as.character(pattern), as.character(replacement), 
         x, ignore.case, perl, fixed, useBytes))
}

And no, it will not convert to integer directly as it always returns a character vector. From ?gsub

sub and gsub return a character vector of the same length and with the same attributes as x (after possible coercion to character).

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