Question

给定（预先存在的）具有各种类型列的数据框，将所有字符列转换为因子的最简单方法是什么，而不影响其他类型的任何列？

以下是data.frame示例：

df <- data.frame(A = factor(LETTERS[1:5]),
                 B = 1:5, C = as.logical(c(1, 1, 0, 0, 1)),
                 D = letters[1:5],
                 E = paste(LETTERS[1:5], letters[1:5]),
                 stringsAsFactors = FALSE)
df
#   A B     C D   E
# 1 A 1  TRUE a A a
# 2 B 2  TRUE b B b
# 3 C 3 FALSE c C c
# 4 D 4 FALSE d D d
# 5 E 5  TRUE e E e
str(df)
# 'data.frame':  5 obs. of  5 variables:
#  $ A: Factor w/ 5 levels "A","B","C","D",..: 1 2 3 4 5
#  $ B: int  1 2 3 4 5
#  $ C: logi  TRUE TRUE FALSE FALSE TRUE
#  $ D: chr  "a" "b" "c" "d" ...
#  $ E: chr  "A a" "B b" "C c" "D d" ...

我知道我能做到：

df$D <- as.factor(df$D)
df$E <- as.factor(df$E)

有没有办法让这个过程自动化一点？

Answer 1

罗兰的答案对于这个具体问题很有帮助，但我想我会分享一种更为通用的方法。

DF <- data.frame(x = letters[1:5], y = 1:5, z = LETTERS[1:5], 
                 stringsAsFactors=FALSE)
str(DF)
# 'data.frame':  5 obs. of  3 variables:
#  $ x: chr  "a" "b" "c" "d" ...
#  $ y: int  1 2 3 4 5
#  $ z: chr  "A" "B" "C" "D" ...

## The conversion
DF[sapply(DF, is.character)] <- lapply(DF[sapply(DF, is.character)], 
                                       as.factor)
str(DF)
# 'data.frame':  5 obs. of  3 variables:
#  $ x: Factor w/ 5 levels "a","b","c","d",..: 1 2 3 4 5
#  $ y: int  1 2 3 4 5
#  $ z: Factor w/ 5 levels "A","B","C","D",..: 1 2 3 4 5

对于转换，assign（DF[sapply(DF, is.character)]）的左侧子集是字符列。在右侧，对于该子集，您使用lapply执行您需要执行的任何转换。 R非常聪明，可以用结果替换原始列。

关于这一点的一个方便的事情是，如果你想采取其他方式或进行其他转换，就像在左边改变你想要的东西一样简单，并在右边指定你想要改变它的东西。 / p>

Answer 2

DF <- data.frame(x=letters[1:5], y=1:5, stringsAsFactors=FALSE)

str(DF)
#'data.frame':  5 obs. of  2 variables:
# $ x: chr  "a" "b" "c" "d" ...
# $ y: int  1 2 3 4 5

（恼人的）as.data.frame默认值是将所有字符列转换为因子列。你可以在这里使用它：

DF <- as.data.frame(unclass(DF))
str(DF)
#'data.frame':  5 obs. of  2 variables:
# $ x: Factor w/ 5 levels "a","b","c","d",..: 1 2 3 4 5
# $ y: int  1 2 3 4 5

Answer 3

正如@Raf Z对此question发表评论，dplyr现在有了mutate_if。超级有用，简单易读。

> str(df)
'data.frame':   5 obs. of  5 variables:
 $ A: Factor w/ 5 levels "A","B","C","D",..: 1 2 3 4 5
 $ B: int  1 2 3 4 5
 $ C: logi  TRUE TRUE FALSE FALSE TRUE
 $ D: chr  "a" "b" "c" "d" ...
 $ E: chr  "A a" "B b" "C c" "D d" ...

> df <- df %>% mutate_if(is.character,as.factor)

> str(df)
'data.frame':   5 obs. of  5 variables:
 $ A: Factor w/ 5 levels "A","B","C","D",..: 1 2 3 4 5
 $ B: int  1 2 3 4 5
 $ C: logi  TRUE TRUE FALSE FALSE TRUE
 $ D: Factor w/ 5 levels "a","b","c","d",..: 1 2 3 4 5
 $ E: Factor w/ 5 levels "A a","B b","C c",..: 1 2 3 4 5

Answer 4

最简单的方法是使用下面给出的代码。它会自动化将所有变量转换为R中数据帧中的因子的整个过程。它对我来说非常好。 food_cat这里是我正在使用的数据集。将其更改为您正在处理的那个。

    for(i in 1:ncol(food_cat)){

food_cat[,i] <- as.factor(food_cat[,i])

}

Answer 5

使用dplyr

library(dplyr)

df <- data.frame(A = factor(LETTERS[1:5]),
                 B = 1:5, C = as.logical(c(1, 1, 0, 0, 1)),
                 D = letters[1:5],
                 E = paste(LETTERS[1:5], letters[1:5]),
                 stringsAsFactors = FALSE)

str(df)

我们得到：

'data.frame':   5 obs. of  5 variables:
 $ A: Factor w/ 5 levels "A","B","C","D",..: 1 2 3 4 5
 $ B: int  1 2 3 4 5
 $ C: logi  TRUE TRUE FALSE FALSE TRUE
 $ D: chr  "a" "b" "c" "d" ...
 $ E: chr  "A a" "B b" "C c" "D d" ...

现在，我们可以将所有chr转换为factors：

df <- df%>%mutate_if(is.character, as.factor)
str(df)

我们得到：

'data.frame':   5 obs. of  5 variables:
 $ A: Factor w/ 5 levels "A","B","C","D",..: 1 2 3 4 5
 $ B: int  1 2 3 4 5
 $ C: logi  TRUE TRUE FALSE FALSE TRUE
 $ D: chr  "a" "b" "c" "d" ...
 $ E: chr  "A a" "B b" "C c" "D d" ...

我们还提供其他解决方案：

带有基本包装：

df[sapply(df, is.character)] <- lapply(df[sapply(df, is.character)], 
                                                           as.factor)

使用dplyr 1.0.0

df <- df%>%mutate(across(where(is.factor), as.character))

使用purrr软件包：

library(purrr)

df <- df%>% modify_if(is.factor, as.character)

Answer 6

我曾经做过一个简单的for循环。作为@ A5C1D2H2I1M1N2O1R2T1答案，lapply是一个很好的解决方案。但是，如果您转换所有列，则之前需要data.frame，否则最终会得到list。执行时间差异很小。

 mm2N=mm2New[,10:18]
 str(mm2N)
'data.frame':   35487 obs. of  9 variables:
 $ bb    : int  4 6 2 3 3 2 5 2 1 2 ...
 $ vabb  : int  -3 -3 -2 -2 -3 -1 0 0 3 3 ...
 $ bb55  : int  7 6 3 4 4 4 9 2 5 4 ...
 $ vabb55: int  -3 -1 0 -1 -2 -2 -3 0 -1 3 ...
 $ zr    : num  0 -2 -1 1 -1 -1 -1 1 1 0 ...
 $ z55r  : num  -2 -2 0 1 -2 -2 -2 1 -1 1 ...
 $ fechar: num  0 -1 1 0 1 1 0 0 1 0 ...
 $ varr  : num  3 3 1 1 1 1 4 1 1 3 ...
 $ minmax: int  3 0 4 6 6 6 0 6 6 1 ...

 # For solution
 t1=Sys.time()
 for(i in 1:ncol(mm2N)) mm2N[,i]=as.factor(mm2N[,i])
 Sys.time()-t1
Time difference of 0.2020121 secs
 str(mm2N)
'data.frame':   35487 obs. of  9 variables:
 $ bb    : Factor w/ 6 levels "1","2","3","4",..: 4 6 2 3 3 2 5 2 1 2 ...
 $ vabb  : Factor w/ 7 levels "-3","-2","-1",..: 1 1 2 2 1 3 4 4 7 7 ...
 $ bb55  : Factor w/ 8 levels "2","3","4","5",..: 6 5 2 3 3 3 8 1 4 3 ...
 $ vabb55: Factor w/ 7 levels "-3","-2","-1",..: 1 3 4 3 2 2 1 4 3 7 ...
 $ zr    : Factor w/ 5 levels "-2","-1","0",..: 3 1 2 4 2 2 2 4 4 3 ...
 $ z55r  : Factor w/ 5 levels "-2","-1","0",..: 1 1 3 4 1 1 1 4 2 4 ...
 $ fechar: Factor w/ 3 levels "-1","0","1": 2 1 3 2 3 3 2 2 3 2 ...
 $ varr  : Factor w/ 5 levels "1","2","3","4",..: 3 3 1 1 1 1 4 1 1 3 ...
 $ minmax: Factor w/ 7 levels "0","1","2","3",..: 4 1 5 7 7 7 1 7 7 2 ...

 #lapply solution
 mm2N=mm2New[,10:18]
 t1=Sys.time()
 mm2N <- lapply(mm2N, as.factor)
 Sys.time()-t1
Time difference of 0.209012 secs
 str(mm2N)
List of 9
 $ bb    : Factor w/ 6 levels "1","2","3","4",..: 4 6 2 3 3 2 5 2 1 2 ...
 $ vabb  : Factor w/ 7 levels "-3","-2","-1",..: 1 1 2 2 1 3 4 4 7 7 ...
 $ bb55  : Factor w/ 8 levels "2","3","4","5",..: 6 5 2 3 3 3 8 1 4 3 ...
 $ vabb55: Factor w/ 7 levels "-3","-2","-1",..: 1 3 4 3 2 2 1 4 3 7 ...
 $ zr    : Factor w/ 5 levels "-2","-1","0",..: 3 1 2 4 2 2 2 4 4 3 ...
 $ z55r  : Factor w/ 5 levels "-2","-1","0",..: 1 1 3 4 1 1 1 4 2 4 ...
 $ fechar: Factor w/ 3 levels "-1","0","1": 2 1 3 2 3 3 2 2 3 2 ...
 $ varr  : Factor w/ 5 levels "1","2","3","4",..: 3 3 1 1 1 1 4 1 1 3 ...
 $ minmax: Factor w/ 7 levels "0","1","2","3",..: 4 1 5 7 7 7 1 7 7 2 ...

 #data.frame lapply solution
 mm2N=mm2New[,10:18]
 t1=Sys.time()
 mm2N <- data.frame(lapply(mm2N, as.factor))
 Sys.time()-t1
Time difference of 0.2010119 secs
 str(mm2N)
'data.frame':   35487 obs. of  9 variables:
 $ bb    : Factor w/ 6 levels "1","2","3","4",..: 4 6 2 3 3 2 5 2 1 2 ...
 $ vabb  : Factor w/ 7 levels "-3","-2","-1",..: 1 1 2 2 1 3 4 4 7 7 ...
 $ bb55  : Factor w/ 8 levels "2","3","4","5",..: 6 5 2 3 3 3 8 1 4 3 ...
 $ vabb55: Factor w/ 7 levels "-3","-2","-1",..: 1 3 4 3 2 2 1 4 3 7 ...
 $ zr    : Factor w/ 5 levels "-2","-1","0",..: 3 1 2 4 2 2 2 4 4 3 ...
 $ z55r  : Factor w/ 5 levels "-2","-1","0",..: 1 1 3 4 1 1 1 4 2 4 ...
 $ fechar: Factor w/ 3 levels "-1","0","1": 2 1 3 2 3 3 2 2 3 2 ...
 $ varr  : Factor w/ 5 levels "1","2","3","4",..: 3 3 1 1 1 1 4 1 1 3 ...
 $ minmax: Factor w/ 7 levels "0","1","2","3",..: 4 1 5 7 7 7 1 7 7 2 ...

Answer 7

我注意到“ []索引列在迭代时无法创建级别：

for（convert.to.factors中的a_feature）{
feature.df [a_feature] <-factor（feature.df [a_feature]）}

它创建例如对于“状态”列：

状态：因子为1级“ c（\“成功\”，\“失败\”）“：不适用不适用...

通过使用“ [[”索引”来补救：

for（convert.to.factors中的a_feature）{
feature.df [[a_feature]] <-factor（feature.df [[a_feature]]）}

根据需要提供：

。状态：具有2个级别的“成功”，“失败”因子：1 1 2 1 ...

Answer 8

根据@Roland 的回答和@Paul de Barros 的评论，我得出以下结论：

    df <- data.frame(A = factor(LETTERS[1:5]),
                 B = 1:5, C = as.logical(c(1, 1, 0, 0, 1)),
                 D = letters[1:5],
                 E = paste(LETTERS[1:5], letters[1:5]),
                 stringsAsFactors = FALSE)
   
   df<-as.data.frame(unclass(df),stringsAsFactors=TRUE)
   str(df)

实际上而且看起来很简单。

> str(df)
'data.frame':   5 obs. of  5 variables:
 $ A: Factor w/ 5 levels "A","B","C","D",..: 1 2 3 4 5
 $ B: int  1 2 3 4 5
 $ C: logi  TRUE TRUE FALSE FALSE TRUE
 $ D: Factor w/ 5 levels "a","b","c","d",..: 1 2 3 4 5
 $ E: Factor w/ 5 levels "A a","B b","C c",..: 1 2 3 4 5

将所有数据框字符列转换为因子

8 个答案: