I have the following setup.
df <- data.frame(aa = rnorm(1000), bb = rnorm(1000))
apply(df, 2, typeof)
# aa bb
#"double" "double"
apply(df, 2, class)
# aa bb
#"numeric" "numeric"
Then I try to convert one of the columns to "factor". But as you can see below, I am not getting any "factor" type or classes. Am I doing anything wrong ?
df[, 1] <- as.factor(df[, 1])
apply(df, 2, typeof)
# aa bb
#"character" "character"
apply(df, 2, class)
# aa bb
#"character" "character"
Sorry I felt my original answer badly written. Why did I put that "matrix of factors" in the very beginning? Here is a better try.
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’.
So a data frame is converted to a matrix by as.matrix, before FUN is applied row-wise or column-wise.
From ?as.matrix:
‘as.matrix’ is a generic function. The method for data frames
will return a character matrix if there is only atomic columns and
any non-(numeric/logical/complex) column, applying ‘as.vector’ to
factors and ‘format’ to other non-character columns. Otherwise,
the usual coercion hierarchy (logical < integer < double <
complex) will be used, e.g., all-logical data frames will be
coerced to a logical matrix, mixed logical-integer will give a
integer matrix, etc.
The default method for ‘as.matrix’ calls ‘as.vector(x)’, and hence
e.g. coerces factors to character vectors.
I am not a native English speaker and I can't read the following (which looks rather important!). Can someone clarify it?
The method for data frames will return a character matrix if there is only atomic columns and any non-(numeric/logical/complex) column, applying ‘as.vector’ to factors and ‘format’ to other non-character columns.
From ?as.vector:
Note that factors are _not_ vectors; ‘is.vector’ returns ‘FALSE’
and ‘as.vector’ converts a factor to a character vector for ‘mode
= "any"’.
Simply put, as long as you have a factor column in a data frame, as.matrix gives you a character matrix.
I believed this apply with data frame problem has been raised many times and the above just adds another duplicate answer. Really sorry. I failed to read OP's question carefully. What hit me in the first instance is that R can not build a true matrix of factors.
f <- factor(letters[1:4])
matrix(f, 2, 2)
# [,1] [,2]
#[1,] "a" "c"
#[2,] "b" "d"
## a sneaky way to get a matrix of factors by setting `dim` attribute
dim(f) <- c(2, 2)
# [,1] [,2]
#[1,] a c
#[2,] b d
#Levels: a b c d
is.matrix(f)
#[1] TRUE
class(f)
#[1] "factor" ## not a true matrix with "matrix" class
While this is interesting, it should be less-relevant to OP's question.
Sorry again for making a mess here. So bad!!
So if I do
sapplywould it help? Because I have many columns that need to be converted to factor.
Use lapply actually. sapply would simplify the result to an array, which is a matrix in 2D case. Here is an example:
dat <- head(trees)
sapply(dat, as.factor)
# Girth Height Volume
#[1,] "8.3" "70" "10.3"
#[2,] "8.6" "65" "10.3"
#[3,] "8.8" "63" "10.2"
#[4,] "10.5" "72" "16.4"
#[5,] "10.7" "81" "18.8"
#[6,] "10.8" "83" "19.7"
new_dat <- data.frame(lapply(dat, as.factor))
str(new_dat)
#'data.frame': 6 obs. of 3 variables:
# $ Girth : Factor w/ 6 levels "8.3","8.6","8.8",..: 1 2 3 4 5 6
# $ Height: Factor w/ 6 levels "63","65","70",..: 3 2 1 4 5 6
# $ Volume: Factor w/ 5 levels "10.2","10.3",..: 2 2 1 3 4 5
sapply(new_dat, class)
# Girth Height Volume
#"factor" "factor" "factor"
apply(new_dat, 2, class)
# Girth Height Volume
#"character" "character" "character"
Regarding typeof, factors are actually stored as integers.
sapply(new_dat, typeof)
# Girth Height Volume
#"integer" "integer" "integer"
When you dput a factor you can see this. For example:
dput(new_dat[[1]])
#structure(1:6, .Label = c("8.3", "8.6", "8.8", "10.5", "10.7",
#"10.8"), class = "factor")
The real values are 1:6. Character levels are just an attribute.