R：为什么我在将列转换为因子后没有得到类型或 class "factor"？
R: Why am I not getting type or class "factor" after converting columns to factor?
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 "factor"。但是正如您在下面看到的，我没有得到任何 "factor" 类型或 类。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"
抱歉，我觉得我原来的回答写得不好。为什么我把 "matrix of factors" 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，在按行或按列应用 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（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，只要你在数据框中有一个因子列，as.matrix就会给你一个字符矩阵。
我相信这个 apply 数据框问题已被多次提出，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，因子实际上存储为整数。
sapply(new_dat, typeof)
# Girth Height Volume
#"integer" "integer" "integer"
当你 dput 一个因素时你可以看到这个。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。
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 "factor"。但是正如您在下面看到的，我没有得到任何 "factor" 类型或 类。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"
抱歉，我觉得我原来的回答写得不好。为什么我把 "matrix of factors" 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，在按行或按列应用 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（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，只要你在数据框中有一个因子列，as.matrix就会给你一个字符矩阵。
我相信这个 apply 数据框问题已被多次提出，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
sapplywould 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，因子实际上存储为整数。
sapply(new_dat, typeof)
# Girth Height Volume
#"integer" "integer" "integer"
当你 dput 一个因素时你可以看到这个。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。