Why Use As.Factor() Instead of Just 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

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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.

Different result in the case of `factor` or `as.factor`?

Problem is as stated in the error message. You are passing arguments which are not present for as.factor. If you read ?as.factor you see the parameter to as.factor is only x. levels, exclude, ordered, nmax are arguments for factor and not as.factor. Hence, it is giving you error that you are passing arguments which you are not using.

If you remove those arguments and run the function then it works without any error message.

lapply(df['cola'], function(x) as.factor(x))
#$cola
# [1] a    b    c    d    e    e    1    <NA> c    d   
#Levels: 1 a b c d e

OR just

lapply(df['cola'], as.factor)

and if you have just one column no need for lapply

as.factor(df$cola)

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

You want

df[] <- lapply(df, factor)

why do I get a levels error in as.factor() R?

You need

yearsPostEDI<-factor(yearsPostEDI, levels = c("HIV_neg", "<1 Year", "1 Year", "2 Years", "3 Years", "4 Years", ">4 Years"))

factor() not as.factor()

R: use of factor

Factors vs character vectors when doing stats:
In terms of doing statistics, there's no difference in how R treats factors and character vectors. In fact, its often easier to leave factor variables as character vectors.

If you do a regression or ANOVA with lm() with a character vector as a categorical
variable you'll get normal model output but with the message:

Warning message:
In model.matrix.default(mt, mf, contrasts) :
  variable 'character_x' converted to a factor

Factors vs character vectors when manipulating dataframes:
When manipulating dataframes, however, character vectors and factors are treated very differently. Some information on the annoyances of R & factors can be found on the Quantum Forest blog, R pitfall #3: friggin’ factors.

Its useful to use stringsAsFactors = FALSE when reading data in from a .csv or .txt using read.table or read.csv. As noted in another reply you have to make sure that everything in your character vector is consistent, or else every typo will be designated as a different factor. You can use the function gsub() to fix typos.

Here is a worked example showing how lm() gives you the same results with
a character vector and a factor.

A random independent variable:

continuous_x <- rnorm(10,10,3)

A random categorical variable as a character vector:

character_x  <- (rep(c("dog","cat"),5))

Convert the character vector to a factor variable.
factor_x <- as.factor(character_x)

Give the two categories random values:

character_x_value <- ifelse(character_x == "dog", 5*rnorm(1,0,1), rnorm(1,0,2))

Create a random relationship between the indepdent variables and a dependent variable

continuous_y <- continuous_x*10*rnorm(1,0) + character_x_value

Compare the output of a linear model with the factor variable and the character
vector. Note the warning that is given with the character vector.

summary(lm(continuous_y ~ continuous_x + factor_x))
summary(lm(continuous_y ~ continuous_x + character_x))

R considers factor name as a level

Use stringsAsFactors=T when you read data and header = T:

db_nouns <- read.table("Final_Database.txt", stringsAsFactors = T, header = T)

colnames(db_nouns) <- c ("category", "space")

new_order <- c( "Ground", "Building", "Tool_precise_grip", "Tool_power_grip", "Food", "Clothes", "Animal", "Object", "Transport", "Action", "Body_Part", "Sense_Phys", "Sound", "Sense_Emotion", "Intelligence", "Space")

db_nouns$category <- factor(db_nouns$category, levels = new_order)

as.factor not working with INT values on R

Checking the documentation for as.factor (by typing ?as.factor), you'll see it says that the first argument x is "a vector of data, usually taking a small number of distinct values". If you supply multiple columns of a data frame, they are treated as one vector. In your example, as.factor creates a unique factor level for each unique value in the entire vectorized, concatenation of columns 4 through 7 of your data frame above.

You should use:

data[4:7] <- lapply(data[4:7], as.factor)

or (requiring tidyverse packages)

data <- data %>% mutate_at(4:7, as.factor)

Both of these solutions will correctly treat each column supplied, here columns 4, 5, 6, and 7, as their own vectors, individually. Each one is converted to a factor separately, and re-assigned appropriately.

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