将分数转换为R

Question

我有一个非常简单的问题。假设，我给100名学生打了这样的分数：

set.seed(1234)
Marks <- rnorm(100, 55, 10)

z <- runif(100)
Gender <- ifelse(z < 0.5, "M", "F")

#Creating Data frame
Df <- data.frame(SNo = 1:100, Marks, Gender)
head(Df)

现在，我需要为学生提供成绩，但是男女的评分标准不同。评分标准是：

我设法解决了这个问题，但是我发现我的方法并不吸引人。我这样尝试过：

#1 Method
Grade = ifelse(Df$Gender == "M", cut(Df$Marks, breaks = c(0, 35, 45, 55, 101), labels = FALSE), 
                        cut(Df$Marks, breaks = c(0, 40, 50, 60, 101), labels = FALSE)) 
Grade <- as.character(factor(Grade, labels = LETTERS[4:1]))

#2. Method
Gradef <- function(x, cp = c(35, 45, 55)) {
  ifelse(x < cp[1], "D", ifelse(x < cp[2], "C", ifelse(x < cp[3], "B", "A")))
}

Grade2 <- ifelse(Df$Gender == "M", Gradef(Df$Marks), Gradef(Df$Marks, c(40, 50, 60)))
sum(Grade == Grade2)  #both method give same grade

Df$Grade <- Grade

有人可以建议我更好的方法来解决同一问题吗？我不想在R中使用任何外部软件包。

谢谢

Answer 1

mylist = list(F = c(35, 45, 55), M = c(40, 50, 60))
grades = c("D", "C", "B", "A")
Df$Grade = grades[1 + sapply(1:NROW(Df), function(i)
    findInterval(Df$Marks[i], mylist[[Df$Gender[i]]]))]
head(Df, 10)
#   SNo    Marks Gender Grade
#1    1 42.92934      F     C
#2    2 57.77429      F     A
#3    3 65.84441      M     A
#4    4 31.54302      F     D
#5    5 59.29125      F     A
#6    6 60.06056      F     A
#7    7 49.25260      M     C
#8    8 49.53368      M     C
#9    9 49.35548      M     C
#10  10 46.09962      F     B

Answer 2

鉴于您对高效的定义是更少的代码行，我想这就是您想要的，使用方法1，我们只需要第二行即可：

Grade = ifelse(Df$Gender == "M", as.vector(cut(Df$Marks, breaks = c(0, 34, 45, 56, 101), labels = c("D", "C", "B", "A"))), 
           as.vector(cut(Df$Marks, breaks = c(0, 39, 50, 61, 101), labels = c("D", "C", "B", "A"))))

> head(Grade)
[1] "C" "B" "A" "D" "B" "A"

因此需要一行代码。

注意：您可以通过替换代码中的每一段来使代码更加灵活，例如

labs <- c("D", "C", "B", "A")

然后将labs变量放入代码中，这样，您现在只需在顶部更改一部分代码，然后将您的函数重用于不同的评分系统等即可。

使用的代码：

set.seed(1234)
Marks <- rnorm(100, 55, 10)
z <- runif(100)
Gender <- ifelse(z < 0.5, "M", "F")
Df <- data.frame(SNo = 1:100, Marks, Gender)

Answer 3

使用带有标签的Cut是我的诀窍，上面的@ hector-haffenden很类似。不过，这是一步一步来的。

set.seed(1234)
#Marks <- rnorm(100, 55, 10)

Marks <- 1:100  #for verification 


z <- runif(100)
Gender <- ifelse(z < 0.5, "M", "F")

#Creating Data frame
Df <- data.frame(SNo = 1:100, Marks, Gender)
head(Df)

cutsF<- cut(Df$Marks,breaks = c(0,35,45,55,100),labels = c('D','C','B','A') , right=F )
cutsM<- cut(Df$Marks,breaks = c(0,40,50,60,100),labels = c('D','C','B','A') , right=F )

Df$Grades= ifelse(Df$Gender=='F' , as.character(cutsF)  ,as.character(cutsM ) )

# For sake of Verification : 
Df$CutsF=cutsF
Df$cutsM= cutsM

head(Df ,20)

编辑：我已经编辑了代码，并将include.lowest替换为right=False。这将关闭左侧的组并满足小于35的条件。但是，这不适用于55/60。您可能需要改用54和59。