R:为什么我在将列转换为因子后没有得到类型或类 "factor"？

Question

我有以下设置。

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" 

然后我尝试将其中一列转换为“因子”。但是正如您在下面看到的，我没有得到任何“因素”类型或类别。我做错了什么吗？

df[, 1] <- as.factor(df[, 1])

apply(df, 2, typeof)
#         aa          bb 
#"character" "character" 

apply(df, 2, class)
#         aa          bb 
#"character" "character" 

Answer 1

抱歉，我觉得我原来的回答写得不好。为什么我把那个“因素矩阵”放在最开始？这是一个更好的尝试。

来自 ?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’.

因此，在按行或列应用 FUN 之前，通过 as.matrix 将数据框转换为矩阵。

来自 ?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.

我的母语不是英语，我看不懂以下内容(这看起来很重要!)。有人可以澄清一下吗？

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.

来自 ?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"’.

简单地说，只要您在数据框中有一个因子列，as.matrix 就会为您提供一个字符矩阵。

---

我相信这个apply with data frame 问题已经被提出了很多次，上面只是添加了另一个重复的答案。真对不起。我没有仔细阅读 OP 的问题。首先让我感到震惊的是 R 无法构建真正的因子矩阵。

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

虽然这很有趣，但它应该与 OP 的问题不太相关。

再次抱歉把这里搞得一团糟。太糟糕了!!

---

So if I do sapply would it help? Because I have many columns that need to be converted to factor.

实际使用lapply。 sapply 会将结果简化为一个数组，在二维情况下是一个矩阵。这是一个例子:

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" 

关于typeof，因子实际上是以整数形式存储的。

sapply(new_dat, typeof)
#    Girth    Height    Volume 
#"integer" "integer" "integer" 

当您输入一个因素时，您可以看到这一点。例如:

dput(new_dat[[1]])
#structure(1:6, .Label = c("8.3", "8.6", "8.8", "10.5", "10.7", 
#"10.8"), class = "factor")

实际值为 1:6。角色等级只是一个属性。