可以使用unclass()显示因子变量的数值：

Type_peau<-as.factor(c("Mixte","Normale","Sèche","Mixte","Normale","Mixte"))
Type_peau
unclass(Type_peau)

要对所有分类变量执行此操作，可以使用sapply()：

must_convert<-sapply(M,is.factor)       # logical vector telling if a variable needs to be displayed as numeric
M2<-sapply(M[,must_convert],unclass)    # data.frame of all categorical variables now displayed as numeric
out<-cbind(M[,!must_convert],M2)        # complete data.frame with all variables put together

编辑：A5C1D2H2I1M1N2O1R2T1's solution一步工作：

out<-data.matrix(M)

它只适用于你的 data.frame 不包含任何字符变量（否则，它们将被放入 NA）。

也许你在data.matrix之后。从函数的描述：

返回通过将数据框中的所有变量转换为数字模式，然后将它们绑定在一起作为矩阵的列而获得的矩阵。因子和有序因子由其内部代码替换。

例子：

mydf <- data.frame(A = letters[1:5],
                   B = LETTERS[1:5],
                   C = month.abb[1:5],
                   D = 1:5)
str(mydf)
# 'data.frame': 5 obs. of  4 variables:
#  $ A: Factor w/ 5 levels "a","b","c","d",..: 1 2 3 4 5
#  $ B: Factor w/ 5 levels "A","B","C","D",..: 1 2 3 4 5
#  $ C: Factor w/ 5 levels "Apr","Feb","Jan",..: 3 2 4 1 5
#  $ D: int  1 2 3 4 5
data.matrix(mydf)
#      A B C D
# [1,] 1 1 3 1
# [2,] 2 2 2 2
# [3,] 3 3 4 3
# [4,] 4 4 1 4
# [5,] 5 5 5 5

将其全部替换为：

mydf[] <- data.matrix(mydf)
mydf
#   A B C D
# 1 1 1 3 1
# 2 2 2 2 2
# 3 3 3 4 3
# 4 4 4 1 4
# 5 5 5 5 5

当然，如果您有更多列类型，则必须首先决定如何处理它们。例如，有一个问题是，如果有一个character列，data.matrix将导致一列NA值，这是正确的。但是，正确的问题应该是“您想如何处理character列？

这里有两个选项。您可以类似地扩展其他列类型的逻辑。

mydf <- data.frame(A = letters[1:5],
                   B = LETTERS[1:5],
                   C = month.abb[1:5],
                   D = 1:5)
mydf$E <- state.abb[1:5]
str(mydf)
# 'data.frame': 5 obs. of  5 variables:
#  $ A: Factor w/ 5 levels "a","b","c","d",..: 1 2 3 4 5
#  $ B: Factor w/ 5 levels "A","B","C","D",..: 1 2 3 4 5
#  $ C: Factor w/ 5 levels "Apr","Feb","Jan",..: 3 2 4 1 5
#  $ D: int  1 2 3 4 5
#  $ E: chr  "AL" "AK" "AZ" "AR" ...
## You want to convert everything to numeric
data.matrix(data.frame(unclass(mydf))) 
#      A B C D E
# [1,] 1 1 3 1 2
# [2,] 2 2 2 2 1
# [3,] 3 3 4 3 4
# [4,] 4 4 1 4 3
# [5,] 5 5 5 5 5
## You only want to convert factors to numeric
mydf[sapply(mydf, is.factor)] <- data.matrix(mydf[sapply(mydf, is.factor)])
mydf
#   A B C D  E
# 1 1 1 3 1 AL
# 2 2 2 2 2 AK
# 3 3 3 4 3 AZ
# 4 4 4 1 4 AR
# 5 5 5 5 5 CA

library(dplyr)
mydf <- data.frame(A = letters[1:5],
                   B = LETTERS[1:5],
                   C = month.abb[1:5],
                   D = 1:5)
glimpse(mydf)
# Observations: 5
# Variables: 4
# $ A <fctr> a, b, c, d, e
# $ B <fctr> A, B, C, D, E
# $ C <fctr> Jan, Feb, Mar, Apr, May
# $ D <int> 1, 2, 3, 4, 5

在dplyr中使用谓词函数

mydf %>% mutate_if(is.factor, as.numeric)
#  A B C D
# 1 1 1 3 1
# 2 2 2 2 2
# 3 3 3 4 3
# 4 4 4 1 4
# 5 5 5 5 5

as.numeric也是这样做的。

df <- iris
df$newgroup <- as.factor(rep(c(letters[1:10]))) # just another factor
str(df) # Species and newgroup are categorial variables
as.numeric(df$Species) # this returns the levels (numeric) of Species.
                       # Now, we want to apply this automatically to all
                       # categorical variables
# using lapply
i <- sapply(df, is.factor)
df[i] <- lapply(df[i], as.numeric)
str(df)
# using dplyr
#(load df again)
library(dplyr)
df2 <- df %>% mutate_if(is.factor, as.numeric)
str(df2)
# using purrr
library(purrr)
df3 <- df %>% map_if(is.factor, as.numeric)
str(df3)

如果您还想创建假人变量，请尝试

library(dummies)
df.4 <- dummy.data.frame(df, sep = ".")