R，使用字符对象标注标签列

Question

我有一个data.frame，其中的列需要标记。

df <- structure(list(q1 = c("5", "6", "5", "5", "7", "5", "5", "5", 
"5", "6", "5", "6", "6", "6", "7", "6", "5", "6", "5", "6", "6", 
"5", "7", "5", "6", "6", "5", "6", "6", "5", "5", "5", "5", "5", 
"5", "5", "4", "5", "5", "4", "4", "5", "4", "4", "5", "4", "5", 
"5", "4", "5"), q2 = c("2", "2", "1", "1", "2", "1", "1", "2", 
"1", "1", "1", "2", "1", "1", "2", "1", "2", "1", "2", "2", "2", 
"1", "2", "2", "2", "2", "2", "1", "1", "2", "2", "2", "2", "2", 
"2", "1", "2", "1", "1", "1", "1", "2", "1", "1", "1", "1", "2", 
"2", "1", "2"), q3 = c("3", "3", "3", "3", "3", "3", "3", "3", 
"3", "3", "3", "3", "3", "3", "3", "3", "3", "3", "3", "3", "3", 
"3", "3", "3", "3", "3", "3", "3", "3", "3", "3", "3", "3", "3", 
"3", "2", "2", "2", "2", "2", "2", "2", "2", "2", "2", "2", "2", 
"2", "2", "2")), row.names = c(NA, -50L), class = c("tbl_df", 
"tbl", "data.frame"), na.action = structure(c(`71` = 71L, `78` = 78L, 
`96` = 96L, `250` = 250L, `393` = 393L, `488` = 488L, `644` = 644L, 
`847` = 847L, `862` = 862L, `1083` = 1083L, `1120` = 1120L, `1149` = 1149L, 
`1322` = 1322L, `1357` = 1357L), class = "omit"))

每个列需要一个不同的标签。这些标签位于单独的对象中，如下所示。

Q1_Label <- c("12 y",  "13 y",  "14 y", "15 y",  "16 y",  "17 y",  "18 y")

Q2_Label <-  c("Female", "Male" )

Q3_Label <- c("9th", "10th", "11th",  "12th", "Ung"

如何在较少的代码行中使用字符对象来标记数据框列？

下面是尝试执行此操作的代码，但我无法获得sapply结构的名称。

感谢您的帮助。

a_df <- sapply(X = df, FUN = function(x) factor(x, 
                   levels = 1:length(table(x)), 
                   labels = get(paste(toupper(names(x)), "_Label", sep = "")) # This line is where I get the problem
))

Answer 1

sapply(
  X = colnames(df), 
  FUN = function(x) factor(
    df[[x]], 
    levels = 1:length(get(paste0(toupper(x), "_Label"))),
    labels = get(paste0(toupper(x), "_Label"))
  )
)

输出

      q1     q2       q3    
 [1,] "16 y" "Male"   "11th"
 [2,] "17 y" "Male"   "11th"
 [3,] "16 y" "Female" "11th"
 [4,] "16 y" "Female" "11th"
 [5,] "18 y" "Male"   "11th"
 [6,] "16 y" "Female" "11th"
 [7,] "16 y" "Female" "11th"
 [8,] "16 y" "Male"   "11th"
 [9,] "16 y" "Female" "11th"
[10,] "17 y" "Female" "11th"
[11,] "16 y" "Female" "11th"
[12,] "17 y" "Male"   "11th"
[13,] "17 y" "Female" "11th"
[14,] "17 y" "Female" "11th"
[15,] "18 y" "Male"   "11th"
[16,] "17 y" "Female" "11th"
[17,] "16 y" "Male"   "11th"
[18,] "17 y" "Female" "11th"
[19,] "16 y" "Male"   "11th"
[20,] "17 y" "Male"   "11th"
[21,] "17 y" "Male"   "11th"
[22,] "16 y" "Female" "11th"
[23,] "18 y" "Male"   "11th"
[24,] "16 y" "Male"   "11th"
[25,] "17 y" "Male"   "11th"
[26,] "17 y" "Male"   "11th"
[27,] "16 y" "Male"   "11th"
[28,] "17 y" "Female" "11th"
[29,] "17 y" "Female" "11th"
[30,] "16 y" "Male"   "11th"
[31,] "16 y" "Male"   "11th"
[32,] "16 y" "Male"   "11th"
[33,] "16 y" "Male"   "11th"
[34,] "16 y" "Male"   "11th"
[35,] "16 y" "Male"   "11th"
[36,] "16 y" "Female" "10th"
[37,] "15 y" "Male"   "10th"
[38,] "16 y" "Female" "10th"
[39,] "16 y" "Female" "10th"
[40,] "15 y" "Female" "10th"
[41,] "15 y" "Female" "10th"
[42,] "16 y" "Male"   "10th"
[43,] "15 y" "Female" "10th"
[44,] "15 y" "Female" "10th"
[45,] "16 y" "Female" "10th"
[46,] "15 y" "Female" "10th"
[47,] "16 y" "Male"   "10th"
[48,] "16 y" "Male"   "10th"
[49,] "15 y" "Female" "10th"
[50,] "16 y" "Male"   "10th"

Answer 2

我们可以使用Map：

Map(function(x, y) factor(x, labels = y[1:length(unique(x))]), 
                    df,list(Q1_Label, Q2_Label, Q3_Label))