我有一个庞大的数据库，我有很多分类变量。 你可以在这里观看：
> M=data.frame(Type_peau,PEAU_CORPS,SENSIBILITE,IMPERFECTIONS,BRILLANCE ,GRAIN_PEAU,RIDES_VISAGE,ALLERGIES,MAINS,
+ INTERET_ALIM_NATURELLE,INTERET_ORIGINE_GEO,INTERET_VACANCES,INTERET_COMPOSITION,DataQuest1,Priorite2,
+ Priorite1,DataQuest4,Age,Nbre_gift,w,Nbre_achat)
> # pour voir s'il y a des données manquantes
> str(M)
'data.frame': 836 obs. of 21 variables:
$ Type_peau : Factor w/ 5 levels "","Grasse","Mixte",..: 3 4 5 3 4 3 3 3 2 3 ...
$ PEAU_CORPS : Factor w/ 4 levels "","Normale","Sèche",..: 2 3 3 2 2 2 3 2 3 2 ...
$ SENSIBILITE : Factor w/ 4 levels "","Aucune","Fréquente",..: 4 4 4 2 4 3 4 2 4 4 ...
$ IMPERFECTIONS : Factor w/ 4 levels "","Fréquente",..: 3 4 3 4 3 2 3 4 3 3 ...
$ BRILLANCE : Factor w/ 4 levels "","Aucune","Partout",..: 4 2 2 4 4 4 4 4 3 4 ...
$ GRAIN_PEAU : Factor w/ 4 levels "","Dilaté","Fin",..: 4 4 4 2 4 2 4 4 2 4 ...
$ RIDES_VISAGE : Factor w/ 4 levels "","Aucune","Très visibles",..: 2 2 2 4 4 2 4 2 4 2 ...
$ ALLERGIES : Factor w/ 4 levels "","Non","Oui",..: 2 2 2 2 2 2 2 2 2 2 ...
$ MAINS : Factor w/ 4 levels "","Moites","Normales",..: 3 4 4 3 3 3 3 4 4 4 ...
$ INTERET_ALIM_NATURELLE: Factor w/ 4 levels "","Beaucoup",..: 2 4 4 4 2 2 2 4 4 2 ...
$ INTERET_ORIGINE_GEO : Factor w/ 5 levels "","Beaucoup",..: 2 4 2 5 2 2 2 2 2 2 ...
$ INTERET_VACANCES : Factor w/ 6 levels "","À la mer",..: 3 4 2 2 3 2 3 2 3 2 ...
$ INTERET_COMPOSITION : Factor w/ 4 levels "","Beaucoup",..: 2 2 2 4 2 2 2 2 4 2 ...
$ DataQuest1 : Factor w/ 4 levels "-20","20-30",..: 4 3 4 4 4 3 3 2 3 2 ...
$ Priorite2 : Factor w/ 7 levels "éclatante","hydratée",..: 3 1 3 4 3 2 7 1 4 6 ...
$ Priorite1 : Factor w/ 7 levels "éclatante","hydratée",..: 4 6 1 5 1 6 1 2 6 4 ...
$ DataQuest4 : Factor w/ 2 levels "nature","urbain": 2 2 2 2 2 1 2 2 2 2 ...
$ Age : int 32 37 23 44 33 30 43 43 60 31 ...
$ Nbre_gift : int 1 4 1 1 2 1 1 1 1 1 ...
$ w : num 0.25 0.25 0.5 0.25 0.5 0 0 0 0 0.75 ...
$ Nbre_achat : int 3 4 7 3 6 9 22 13 7 16 ...
我需要自动将所有分类变量转换为数字。例如，对于变量 Type_peau ，它是：
head(Type_peau)
[1] Mixte Normale Sèche Mixte Normale Mixte
Levels: Grasse Mixte Normale Sèche
我想要它：
head(Type_peau)
[1] 2 3 4 2 3 2
Levels: 1 2 3 4
如何自动为所有分类变量执行此操作？
答案 0 :(得分：5)
您可以使用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，您可以使用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一步到位：
out<-data.matrix(M)
仅当您的data.frame不包含任何字符变量时才会起作用（否则，它们将被置于NA）。
答案 1 :(得分：4)
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获得的矩阵，然后将它们绑定在一起作为矩阵的列。因素和有序因素被其内部代码所取代。
示例：
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
使用以下方法全部替换：
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，有人担心如果有character列，data.matrix会产生一列NA值，这是正确的。但是，正确的问题应该是“您希望如何处理character列？
以下是两个选项。您可以类似地为其他列类型扩展逻辑。
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
答案 2 :(得分：4)
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
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
答案 3 :(得分：2)
as.numeric也完成了这项工作。
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)
如果您还想创建虚拟变量，请尝试
library(dummies)
df.4 <- dummy.data.frame(df, sep = ".")
答案 4 :(得分：0)
最好和最快的方法是使用以下代码：
DataFrameYouWant <- data.frame(yourData)
DataFrameYouWant[] <- lapply(DataFrameYouWant, as.integer)
上面的代码会自动将数据中的所有因子变量转换为数字，将数据转换为数据框。您可以指定要转换为数字的列/变量。
答案 5 :(得分：0)
仅添加到已发布的答案中，此link提供了一些示例，这些示例将分类数据转换为数值，但是如果您对默认转换不满意，还可以将这些数字映射到指定值。
答案 6 :(得分：0)
这也可以使用因子函数一步完成。
{
"name": "laravel/laravel",
"type": "project",
"keywords": [
"framework",
"laravel"
],
"config": {
"optimize-autoloader": true,
"preferred-install": "dist",
"sort-packages": true
},
"extra": {
"laravel": {
"dont-discover": []
}
},
"autoload": {
"psr-0": {
"":["database/seeds"]
}
}
}
注意：它将替换该列。