r - 如何使用 data.table 高效地创建新变量并分配列名？

Question

我需要根据如下数据计算新列:

structure(list(english_score = c(3L, 4L, 3L, 3L, 4L, 3L, 4L, 
2L, 4L, 2L, 3L, 3L, 2L, 2L, 3L, 4L, 3L, 3L, 4L, 3L, 4L, 3L, 2L
), math_score = c(4L, 4L, 3L, 4L, 4L, 4L, 3L, 2L, 3L, 3L, 4L, 
2L, 4L, 2L, 4L, 2L, 3L, 3L, 2L, 2L, 2L, 4L, 2L), science_score = c(3L, 
4L, 4L, 4L, 3L, 4L, 4L, 3L, 3L, 2L, 3L, 4L, 4L, 4L, 4L, 4L, 4L, 
2L, 3L, 2L, 3L, 3L, 4L)), row.names = c(NA, -23L), class = c("data.table", 
"data.frame"), .internal.selfref = <pointer: 0x000002478ee34d50>)

我想制作这样的东西:

structure(list(english_score = c(3L, 4L, 3L, 3L, 4L, 3L, 4L, 
2L, 4L, 2L, 3L, 3L, 2L, 2L, 3L, 4L, 3L, 3L, 4L, 3L, 4L, 3L, 2L
), math_score = c(4L, 4L, 3L, 4L, 4L, 4L, 3L, 2L, 3L, 3L, 4L, 
2L, 4L, 2L, 4L, 2L, 3L, 3L, 2L, 2L, 2L, 4L, 2L), science_score = c(3L, 
4L, 4L, 4L, 3L, 4L, 4L, 3L, 3L, 2L, 3L, 4L, 4L, 4L, 4L, 4L, 4L, 
2L, 3L, 2L, 3L, 3L, 4L), english_level = c("Level C", "Level D", 
"Level C", "Level C", "Level D", "Level C", "Level D", "Level B", 
"Level D", "Level B", "Level C", "Level C", "Level B", "Level B", 
"Level C", "Level D", "Level C", "Level C", "Level D", "Level C", 
"Level D", "Level C", "Level B"), math_level = c("Level D", "Level D", 
"Level C", "Level D", "Level D", "Level D", "Level C", "Level B", 
"Level C", "Level C", "Level D", "Level B", "Level D", "Level B", 
"Level D", "Level B", "Level C", "Level C", "Level B", "Level B", 
"Level B", "Level D", "Level B"), science_level = c("Level C", 
"Level D", "Level D", "Level D", "Level C", "Level D", "Level D", 
"Level C", "Level C", "Level B", "Level C", "Level D", "Level D", 
"Level D", "Level D", "Level D", "Level D", "Level B", "Level C", 
"Level B", "Level C", "Level C", "Level D")), row.names = c(NA, 
-23L), class = c("data.table", "data.frame"), .internal.selfref = <pointer: 
0x000002478ee34d50>)

到目前为止，我的方法是使用函数来计算新变量的水平...

myfunction<-function(x){case_when(x<2~"Level A",
                              x>1 & x<3~"Level B",
                              x>2 & x<4~"Level C",
                              x>3~"Level D")}

....然后，创建新变量并一一指定它们的名称。

DT[, english_level:=lapply(.SD, myfunction), .SDcols='english_score']

DT[, math_level:=lapply(.SD, myfunction), .SDcols='math_score']

DT[, science_level:=lapply(.SD, myfunction), .SDcols='science_score']

如何简化此过程，最好使用 data.table？

Answer 1

我会这样做(我将你的数据称为DT，因为utils::data()是一个基本的R函数):

score_cols  <- grep("_score$", names(DT), value = TRUE)
level_cols  <- sub("_score", "_level", score_cols)

DT[, 
    (level_cols) := lapply(.SD, myfunction),
    .SDcols = score_cols
]

此外，您的 myfunction() 使用 dplyr::case_when()。这可以工作，但某些 dplyr 函数与 data.table 发生冲突( Between() 、first() 和 last() 与我当前拥有的版本)。您可以将其替换为 data.table::fcase()。

myfunction <- function(x) {
    fcase(
        x == 1, "Level A",
        x == 2, "Level B",
        x == 3, "Level C",
        x == 4, "Level D"
    )
}

这应该也比 dplyr 版本更快。

此外，使用此特定函数，您实际上可以通过将字母表中的第 n 个字母指定为等级来替换键入逻辑时的大小写:

assign_letter_grade  <- function(n) {
    paste("Level", LETTERS[n])
}