將所有數據框字符列轉換為因子

Question

給定具有各種類型列的（預先存在的）數據框，將其所有字符列轉換為因子而不影響任何其他類型的列的最簡單方法是什么？

這是一個示例data.frame ：

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" ...

我知道我可以做到：

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

有沒有辦法讓這個過程自動化一點？

Answer 1

Roland 的回答非常適合這個特定問題，但我想我會分享一種更通用的方法。

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

對於轉換，賦值的左側（ DF[sapply(DF, is.character)] ）對字符列進行子集化。 在右側，對於該子集，您可以使用lapply執行您需要執行的任何轉換。 R 足夠聰明，可以用結果替換原始列。

這樣做的方便之處在於，如果您想走另一條路或進行其他轉換，只需在左側更改您要查找的內容並在右側指定要更改的內容即可。

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

as.data.frame的（煩人的）默認值是將所有字符列轉換為因子列。 你可以在這里使用它：

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 對這個問題的評論，dplyr 現在有 mutate_if。 超級有用，簡單易讀。

> 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

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)

我們得到：

'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" ...

現在，我們可以將所有chr轉換為factors ：

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

我們得到：

'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" ...

讓我們也提供其他解決方案：

帶基礎包：

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包：

library(purrr)

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

Answer 5

最簡單的方法是使用下面給出的代碼。 它會自動完成將所有變量轉換為 R 中數據幀中的因子的整個過程。它對我來說非常好。 food_cat 這里是我正在使用的數據集。 將其更改為您正在處理的那個。

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

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

}

Answer 6

我曾經做過一個簡單的for循環。 正如@A5C1D2H2I1M1N2O1R2T1 的回答， lapply是一個不錯的解決方案。 但是如果你轉換了所有的列，你之前需要一個data.frame ，否則你最終會得到一個list 。 執行時間差異很小。

 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“[”索引列在迭代時無法創建級別：

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

它創建，例如“狀態”列：

狀態：因子 w/ 1 級 "c(\\"Success\\", \\"Fail\\")" : NA NA NA ...

這是通過使用“[[”索引來補救的：

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

根據需要給予：

. 狀態：具有 2 個級別“成功”、“失敗”的因素：1 1 2 1 ...

Answer 8

根據@Roland 的回答和@Paul de Barros 的評論，我得出以下結論：

    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)

實際上而且簡單地似乎有效。

> 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