Question

我在R中编写了一个函数，它将包含字母等级的数据框转换为数字等级。然后，我在数据框的每一列上使用sapply（）。有没有更简单的方法来做到这一点，不需要三次单独调用sapply？有没有办法将函数应用于数据框的每个元素而不是每个行或列？

源数据＆＃34;成绩＆＃34;，如下所示：

grades <- read.table("Grades.txt", header = TRUE)
head(grades)
  final_exam quiz_avg homework_avg
1          C        A            A
2         C-       B-            A
3         D+       B+            A
4         B+       B+            A
5          F       B+            A
6          B       A-            A

我的＆＃34; convert_grades＆＃34;功能看起来像这样：

convert_grades <- function(x) {
    if (x == "A+") {
        x <- 4.3
    } else if (x == "A") {
        x <- 4
    } else if (x == "A-") {
        x <- 3.7
    } else if (x == "B+") {
        x <- 3.3
    } else if (x == "B") {
        x <- 3
    } else if (x == "B-") {
        x <- 2.7
    } else if (x == "C+") {
        x <- 2.3
    } else if (x == "C") {
        x <- 2
    } else if (x == "C-") {
        x <- 1.7
    } else if (x == "D+") {
        x <- 1.3
    } else if (x == "D") {
        x <- 1
    } else if (x == "D-") {
        x <- 0.7
    } else if (x == "F") {
        x <- 0
    } else {
        x <- NA
    }
    return(x)
}

我目前的做法如下：

num_grades <- grades
num_grades[, 1] <- sapply(grades[, 1], convert_grades)
num_grades[, 2] <- sapply(grades[, 2], convert_grades)
num_grades[, 3] <- sapply(grades[, 3], convert_grades)

head(num_grades)
   final_exam quiz_avg homework_avg
 1        2.0      4.0            4
 2        1.7      2.7            4
 3        1.3      3.3            4
 4        3.3      3.3            4
 5        0.0      3.3            4
 6        3.0      3.7            4

Answer 1

我会按如下方式重写您的convert_grades函数：

convert_grades <- function(x) {
  A <- factor(x, levels=c("A+", "A", "A-",
                     "B+", "B", "B-",
                     "C+", "C", "C-",
                     "D+", "D", "D-", "F"))
  values <- c(4.3, 4, 3.7, 
              3.3, 3, 2.7,
              2.3, 2, 1.7,
              1.3, 1, 0.7, 0)
  values[A]
}

然后，我会像这样进行转换：

num_grades <- grades
num_grades[] <- lapply(num_grades, convert_grades)
num_grades
  final_exam quiz_avg homework_avg
1        2.0      4.0            4
2        1.7      2.7            4
3        1.3      3.3            4
4        3.3      3.3            4
5        0.0      3.3            4
6        3.0      3.7            4

Answer 2

首先对您的函数进行矢量化：您可以使用ifelse或：

执行此操作

grvec <- c("A+"=4.3,"A"=4,"A-"=3.7,"B+"=3.3,"B"=3,"B-"=2.7,
           "C+"=2.3,"C"=2,"C-"=1.7,"D+"=1.3,"D"=1,"D-"= 0.7,
       "F"=0)

grades <- data.frame(final_exam=c("C","C-","D+"),
                     quiz_avg=c("A","B-","Q"))

##   final_exam quiz_avg
## 1          C        A
## 2         C-       B-
## 3         D+        Q

num_grades <- apply(grades,2,function(x) grvec[as.character(x)])
##      final_exam quiz_avg
## [1,]        2.0      4.0
## [2,]        1.7      2.7
## [3,]        1.3       NA

Answer 3

这是一种非常快速的哈希方法，可以让你获得更多的成绩：

library(qdap)

grvec <- list("A+"=4.3,"A"=4,"A-"=3.7,"B+"=3.3,"B"=3,"B-"=2.7,
    "C+"=2.3,"C"=2,"C-"=1.7,"D+"=1.3,"D"=1,"D-"= 0.7, "F"=0)

dat[] <- lapply(dat, lookup, list2df(grvec)[, c(2:1)])

##   final_exam quiz_avg homework_avg
## 1        2.0      4.0            4
## 2        1.7      2.7            4
## 3        1.3      3.3            4
## 4        3.3      3.3            4
## 5        0.0      3.3            4
## 6        3.0      3.7            4

R：简化将字母等级转换为数字等级的代码

3 个答案: